dr. zyad ahmed tawfik email : [email protected] ... · know how to conduct a series of...
TRANSCRIPT
1
physics course
Dr Zyad Ahmed Tawfik
Email zmohammedinayaedusa
Website zyadinayawordpresscom
2
Course Description This course serves as an introduction to the basic principles of physics and also this
course is designed for students in Health Science to enable them to appreciate the
basic concepts of Physics which are relevant to their further studies
Student Learning Outcomes
Upon completion of this course the students are expected to
1 understand the essential elements of physics needed by premedical students
2 Recognize the basic principles of physics in the branches of mechanics movement forces fluid mechanics electric and magnetic phenomena and
radiation 3 Describe the nature phenomena by using the language of physics
4 develop the ability to solve problems and think critically by applying the acquired knowledge of physics to the various problems
5 know how to conduct a series of practical experiments for the study of
physical phenomena related to some previous knowledge
3
Week 2 Daily Objectives SLO
Assignments amp Activities
Week 2
Newtonrsquos laws of Motion
Definition of force Equilibrium state Newtonrsquos laws of motion
Fundamental forces Weight
and friction
11
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 3 Daily Objectives SLO
Assignments amp Activities
Week 3
Review and resolve quizzes on vector
and Newtons laws
11
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 4 Daily Objectives SLO
Assignments amp Activities
Week 4
Work Energy and Power
Work
Kinetic energy
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 5 Daily Objectives SLO
Assignments amp Activities
Weekly Outline of Curriculum
Week 1 Daily Objectives SLO
Assignments amp Activities
Week 1
Vectors
-Addition Geometrical method amp
Analytical method
-Product of vector Scalar and Cross
product
11
- Lecturing - Team work - Exercises - Self-learning - Group Discussions
4
Week 5
Work Energy and Power
Conservative forces and potential energy
Observations of work and energy
Power
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 6 Daily Objectives SLO Assignments amp Activities
Week 6
Work Energy and Power
Review and resolve quizzes on work
and energy
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 7 Daily Objectives SLO Assignments amp Activities
Week 7
Exam1_Examination 11
12
Comprehensive Examination
Good luck
5
Week 8 Daily Objectives SLO Assignments amp Activities
Week 8
Mechanics of non-Viscous Fluid
The Equation of continuity Stream line flow
Bernoullirsquos equation and its static consequences
13
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 9 Daily Objectives SLO Assignments amp Activities
Week 9
Mechanics of non-Viscous Fluid
Role of gravity in blood circulation
Pressure measurement using manometer
13 - Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 10 Daily Objectives SLO Assignments amp Activities
Week 10
Direct Electric Current (DC)
Basic concept of DC Electric current
Ohmrsquos law Electric safety
14
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 11 Daily Objectives SLO Assignments amp Activities
Week 11
Nerve Conduction
Structure of nerve
Electric characteristics of
axon
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 12 Daily Objectives SLO Assignments amp Activities
Week 12
Exam2_Examination
13
14
15
Comprehensive Examination
Good luck
Week 13 Daily Objectives SLO Assignments amp Activities
6
Week 13
Nerve Conduction
Ionic concentration and the resting potential
Response to weak stimuli The action potential
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 14 Daily Objectives SLO
Assignments amp Activities
Week 14
Wave properties of light
The Index of Refraction Reflection of Light Refraction of Light
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 15 Daily Objectives SLO
Assignments amp Activities
Week 15
Wave properties of light
Ionizing Radiation
The Interaction of Radiation with Matter
Chronic Radiation Exposure
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 16 Daily Objectives SLO
Assignments amp Activities
Week 16
Final Examination-Practical Lab
Week 17 Daily Objectives SLO
Assignments amp Activities
Week 17
Final Examination-
Comprehensive
11
12
13
14
15
16
22
23
24
31
32
Comprehensive Final Examination
Must be passed with a score of 60 or
better to pass course
7
Vocabulary
ENGLISH ARABIC
Acceleration تسارع ndashعجلة
Activity نشاط أشعاعى
Air pressure ضغط الهواء
Ampere أمبير
Analytical تحليلى
Analytical Method الطريق التحليلية
Angle زاوية
Angle of Deviation زاوية االنحراف
Angle of Incidence زواية السقوط
Angle of reflection زاوية االنعكاس
Angle of refraction زاوية األنكسار
Archimedess Principle مبدأ أرخميدس
Atmospheric pressure الضغط الجوي
Atom الذرة
Axoplasm جبلة المحوار
Axons محاور عصبية
Balanced Force القوة المتزنة
Bernoullis Principle law مبدأ )قانون( برنولي
Binding Energy طاقة الربط
Blood pressure ضغط الدم
blood vessel وعاء دموى
Buoyancy الطفو
Buoyant force قوة الطفو
8
Capacitance سعة المكثف
Capacitor المكثف
Coefficient of friction معامل األحتكاك
Charge شحنة
Circuit دائرة
cohesion قوة التماسك بين جزئيات السائل
components of a vector مكونات متجه
compression ضغط كباس أنضغاط
conduction توصيل
conductivity معامل الموصلية الحرارية
Conductor الموصل
Conservation of Energy حفظ الطاقة
conservation of momentum حفظ كمية الحركة
consumed مستهلك
Coulomb كولوم
cube مكعب
Current تيار
deceleration تباطؤ
Density الكثافة
Dendrites التشعبات العصبية
diffraction حيود الضوء
dimensions االبعاد
Dispersion تشتت تفرق
direction إتجاه
directly proportional يتناسب طرديا
displacement اإلزاحة
9
distance المسافة
Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك
عبره(
drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل
dynamics الديناميكا
efficiency كفاءة
effort arm ذراع القوة
Electric Charge شحنة كهربائية
Electric Circuit دائرة كهربائية
Electric Current التيار الكهربى
Electric Energy طاقة كهربائية
Electric field المجال الكهربي
Electrical Conductivity توصيل كهربائى
Electric Potential جهد كهربائى
Electrical Resistance مقاومة كهربائية
electricity الكهرباء
electrode قطب كهربائى
Electromagnetic Field مجال كهرومغناطيسي
electromagnetic induction الحث الكهرومغناطيسي
Electromotive Force القوة الدافعة الكهربية
Electron اإللكترون
Electron Diffraction حيود اإللكترون
Energy الطاقة
Energy Level مستويات الطاقة
Energy Transformations تحوالت الطاقة
Equation of continuity معادلة اإلستمرارية
Equilibrium إتزان
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
2
Course Description This course serves as an introduction to the basic principles of physics and also this
course is designed for students in Health Science to enable them to appreciate the
basic concepts of Physics which are relevant to their further studies
Student Learning Outcomes
Upon completion of this course the students are expected to
1 understand the essential elements of physics needed by premedical students
2 Recognize the basic principles of physics in the branches of mechanics movement forces fluid mechanics electric and magnetic phenomena and
radiation 3 Describe the nature phenomena by using the language of physics
4 develop the ability to solve problems and think critically by applying the acquired knowledge of physics to the various problems
5 know how to conduct a series of practical experiments for the study of
physical phenomena related to some previous knowledge
3
Week 2 Daily Objectives SLO
Assignments amp Activities
Week 2
Newtonrsquos laws of Motion
Definition of force Equilibrium state Newtonrsquos laws of motion
Fundamental forces Weight
and friction
11
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 3 Daily Objectives SLO
Assignments amp Activities
Week 3
Review and resolve quizzes on vector
and Newtons laws
11
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 4 Daily Objectives SLO
Assignments amp Activities
Week 4
Work Energy and Power
Work
Kinetic energy
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 5 Daily Objectives SLO
Assignments amp Activities
Weekly Outline of Curriculum
Week 1 Daily Objectives SLO
Assignments amp Activities
Week 1
Vectors
-Addition Geometrical method amp
Analytical method
-Product of vector Scalar and Cross
product
11
- Lecturing - Team work - Exercises - Self-learning - Group Discussions
4
Week 5
Work Energy and Power
Conservative forces and potential energy
Observations of work and energy
Power
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 6 Daily Objectives SLO Assignments amp Activities
Week 6
Work Energy and Power
Review and resolve quizzes on work
and energy
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 7 Daily Objectives SLO Assignments amp Activities
Week 7
Exam1_Examination 11
12
Comprehensive Examination
Good luck
5
Week 8 Daily Objectives SLO Assignments amp Activities
Week 8
Mechanics of non-Viscous Fluid
The Equation of continuity Stream line flow
Bernoullirsquos equation and its static consequences
13
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 9 Daily Objectives SLO Assignments amp Activities
Week 9
Mechanics of non-Viscous Fluid
Role of gravity in blood circulation
Pressure measurement using manometer
13 - Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 10 Daily Objectives SLO Assignments amp Activities
Week 10
Direct Electric Current (DC)
Basic concept of DC Electric current
Ohmrsquos law Electric safety
14
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 11 Daily Objectives SLO Assignments amp Activities
Week 11
Nerve Conduction
Structure of nerve
Electric characteristics of
axon
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 12 Daily Objectives SLO Assignments amp Activities
Week 12
Exam2_Examination
13
14
15
Comprehensive Examination
Good luck
Week 13 Daily Objectives SLO Assignments amp Activities
6
Week 13
Nerve Conduction
Ionic concentration and the resting potential
Response to weak stimuli The action potential
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 14 Daily Objectives SLO
Assignments amp Activities
Week 14
Wave properties of light
The Index of Refraction Reflection of Light Refraction of Light
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 15 Daily Objectives SLO
Assignments amp Activities
Week 15
Wave properties of light
Ionizing Radiation
The Interaction of Radiation with Matter
Chronic Radiation Exposure
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 16 Daily Objectives SLO
Assignments amp Activities
Week 16
Final Examination-Practical Lab
Week 17 Daily Objectives SLO
Assignments amp Activities
Week 17
Final Examination-
Comprehensive
11
12
13
14
15
16
22
23
24
31
32
Comprehensive Final Examination
Must be passed with a score of 60 or
better to pass course
7
Vocabulary
ENGLISH ARABIC
Acceleration تسارع ndashعجلة
Activity نشاط أشعاعى
Air pressure ضغط الهواء
Ampere أمبير
Analytical تحليلى
Analytical Method الطريق التحليلية
Angle زاوية
Angle of Deviation زاوية االنحراف
Angle of Incidence زواية السقوط
Angle of reflection زاوية االنعكاس
Angle of refraction زاوية األنكسار
Archimedess Principle مبدأ أرخميدس
Atmospheric pressure الضغط الجوي
Atom الذرة
Axoplasm جبلة المحوار
Axons محاور عصبية
Balanced Force القوة المتزنة
Bernoullis Principle law مبدأ )قانون( برنولي
Binding Energy طاقة الربط
Blood pressure ضغط الدم
blood vessel وعاء دموى
Buoyancy الطفو
Buoyant force قوة الطفو
8
Capacitance سعة المكثف
Capacitor المكثف
Coefficient of friction معامل األحتكاك
Charge شحنة
Circuit دائرة
cohesion قوة التماسك بين جزئيات السائل
components of a vector مكونات متجه
compression ضغط كباس أنضغاط
conduction توصيل
conductivity معامل الموصلية الحرارية
Conductor الموصل
Conservation of Energy حفظ الطاقة
conservation of momentum حفظ كمية الحركة
consumed مستهلك
Coulomb كولوم
cube مكعب
Current تيار
deceleration تباطؤ
Density الكثافة
Dendrites التشعبات العصبية
diffraction حيود الضوء
dimensions االبعاد
Dispersion تشتت تفرق
direction إتجاه
directly proportional يتناسب طرديا
displacement اإلزاحة
9
distance المسافة
Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك
عبره(
drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل
dynamics الديناميكا
efficiency كفاءة
effort arm ذراع القوة
Electric Charge شحنة كهربائية
Electric Circuit دائرة كهربائية
Electric Current التيار الكهربى
Electric Energy طاقة كهربائية
Electric field المجال الكهربي
Electrical Conductivity توصيل كهربائى
Electric Potential جهد كهربائى
Electrical Resistance مقاومة كهربائية
electricity الكهرباء
electrode قطب كهربائى
Electromagnetic Field مجال كهرومغناطيسي
electromagnetic induction الحث الكهرومغناطيسي
Electromotive Force القوة الدافعة الكهربية
Electron اإللكترون
Electron Diffraction حيود اإللكترون
Energy الطاقة
Energy Level مستويات الطاقة
Energy Transformations تحوالت الطاقة
Equation of continuity معادلة اإلستمرارية
Equilibrium إتزان
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
3
Week 2 Daily Objectives SLO
Assignments amp Activities
Week 2
Newtonrsquos laws of Motion
Definition of force Equilibrium state Newtonrsquos laws of motion
Fundamental forces Weight
and friction
11
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 3 Daily Objectives SLO
Assignments amp Activities
Week 3
Review and resolve quizzes on vector
and Newtons laws
11
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 4 Daily Objectives SLO
Assignments amp Activities
Week 4
Work Energy and Power
Work
Kinetic energy
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 5 Daily Objectives SLO
Assignments amp Activities
Weekly Outline of Curriculum
Week 1 Daily Objectives SLO
Assignments amp Activities
Week 1
Vectors
-Addition Geometrical method amp
Analytical method
-Product of vector Scalar and Cross
product
11
- Lecturing - Team work - Exercises - Self-learning - Group Discussions
4
Week 5
Work Energy and Power
Conservative forces and potential energy
Observations of work and energy
Power
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 6 Daily Objectives SLO Assignments amp Activities
Week 6
Work Energy and Power
Review and resolve quizzes on work
and energy
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 7 Daily Objectives SLO Assignments amp Activities
Week 7
Exam1_Examination 11
12
Comprehensive Examination
Good luck
5
Week 8 Daily Objectives SLO Assignments amp Activities
Week 8
Mechanics of non-Viscous Fluid
The Equation of continuity Stream line flow
Bernoullirsquos equation and its static consequences
13
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 9 Daily Objectives SLO Assignments amp Activities
Week 9
Mechanics of non-Viscous Fluid
Role of gravity in blood circulation
Pressure measurement using manometer
13 - Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 10 Daily Objectives SLO Assignments amp Activities
Week 10
Direct Electric Current (DC)
Basic concept of DC Electric current
Ohmrsquos law Electric safety
14
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 11 Daily Objectives SLO Assignments amp Activities
Week 11
Nerve Conduction
Structure of nerve
Electric characteristics of
axon
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 12 Daily Objectives SLO Assignments amp Activities
Week 12
Exam2_Examination
13
14
15
Comprehensive Examination
Good luck
Week 13 Daily Objectives SLO Assignments amp Activities
6
Week 13
Nerve Conduction
Ionic concentration and the resting potential
Response to weak stimuli The action potential
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 14 Daily Objectives SLO
Assignments amp Activities
Week 14
Wave properties of light
The Index of Refraction Reflection of Light Refraction of Light
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 15 Daily Objectives SLO
Assignments amp Activities
Week 15
Wave properties of light
Ionizing Radiation
The Interaction of Radiation with Matter
Chronic Radiation Exposure
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 16 Daily Objectives SLO
Assignments amp Activities
Week 16
Final Examination-Practical Lab
Week 17 Daily Objectives SLO
Assignments amp Activities
Week 17
Final Examination-
Comprehensive
11
12
13
14
15
16
22
23
24
31
32
Comprehensive Final Examination
Must be passed with a score of 60 or
better to pass course
7
Vocabulary
ENGLISH ARABIC
Acceleration تسارع ndashعجلة
Activity نشاط أشعاعى
Air pressure ضغط الهواء
Ampere أمبير
Analytical تحليلى
Analytical Method الطريق التحليلية
Angle زاوية
Angle of Deviation زاوية االنحراف
Angle of Incidence زواية السقوط
Angle of reflection زاوية االنعكاس
Angle of refraction زاوية األنكسار
Archimedess Principle مبدأ أرخميدس
Atmospheric pressure الضغط الجوي
Atom الذرة
Axoplasm جبلة المحوار
Axons محاور عصبية
Balanced Force القوة المتزنة
Bernoullis Principle law مبدأ )قانون( برنولي
Binding Energy طاقة الربط
Blood pressure ضغط الدم
blood vessel وعاء دموى
Buoyancy الطفو
Buoyant force قوة الطفو
8
Capacitance سعة المكثف
Capacitor المكثف
Coefficient of friction معامل األحتكاك
Charge شحنة
Circuit دائرة
cohesion قوة التماسك بين جزئيات السائل
components of a vector مكونات متجه
compression ضغط كباس أنضغاط
conduction توصيل
conductivity معامل الموصلية الحرارية
Conductor الموصل
Conservation of Energy حفظ الطاقة
conservation of momentum حفظ كمية الحركة
consumed مستهلك
Coulomb كولوم
cube مكعب
Current تيار
deceleration تباطؤ
Density الكثافة
Dendrites التشعبات العصبية
diffraction حيود الضوء
dimensions االبعاد
Dispersion تشتت تفرق
direction إتجاه
directly proportional يتناسب طرديا
displacement اإلزاحة
9
distance المسافة
Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك
عبره(
drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل
dynamics الديناميكا
efficiency كفاءة
effort arm ذراع القوة
Electric Charge شحنة كهربائية
Electric Circuit دائرة كهربائية
Electric Current التيار الكهربى
Electric Energy طاقة كهربائية
Electric field المجال الكهربي
Electrical Conductivity توصيل كهربائى
Electric Potential جهد كهربائى
Electrical Resistance مقاومة كهربائية
electricity الكهرباء
electrode قطب كهربائى
Electromagnetic Field مجال كهرومغناطيسي
electromagnetic induction الحث الكهرومغناطيسي
Electromotive Force القوة الدافعة الكهربية
Electron اإللكترون
Electron Diffraction حيود اإللكترون
Energy الطاقة
Energy Level مستويات الطاقة
Energy Transformations تحوالت الطاقة
Equation of continuity معادلة اإلستمرارية
Equilibrium إتزان
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
4
Week 5
Work Energy and Power
Conservative forces and potential energy
Observations of work and energy
Power
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 6 Daily Objectives SLO Assignments amp Activities
Week 6
Work Energy and Power
Review and resolve quizzes on work
and energy
12
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 7 Daily Objectives SLO Assignments amp Activities
Week 7
Exam1_Examination 11
12
Comprehensive Examination
Good luck
5
Week 8 Daily Objectives SLO Assignments amp Activities
Week 8
Mechanics of non-Viscous Fluid
The Equation of continuity Stream line flow
Bernoullirsquos equation and its static consequences
13
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 9 Daily Objectives SLO Assignments amp Activities
Week 9
Mechanics of non-Viscous Fluid
Role of gravity in blood circulation
Pressure measurement using manometer
13 - Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 10 Daily Objectives SLO Assignments amp Activities
Week 10
Direct Electric Current (DC)
Basic concept of DC Electric current
Ohmrsquos law Electric safety
14
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 11 Daily Objectives SLO Assignments amp Activities
Week 11
Nerve Conduction
Structure of nerve
Electric characteristics of
axon
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 12 Daily Objectives SLO Assignments amp Activities
Week 12
Exam2_Examination
13
14
15
Comprehensive Examination
Good luck
Week 13 Daily Objectives SLO Assignments amp Activities
6
Week 13
Nerve Conduction
Ionic concentration and the resting potential
Response to weak stimuli The action potential
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 14 Daily Objectives SLO
Assignments amp Activities
Week 14
Wave properties of light
The Index of Refraction Reflection of Light Refraction of Light
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 15 Daily Objectives SLO
Assignments amp Activities
Week 15
Wave properties of light
Ionizing Radiation
The Interaction of Radiation with Matter
Chronic Radiation Exposure
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 16 Daily Objectives SLO
Assignments amp Activities
Week 16
Final Examination-Practical Lab
Week 17 Daily Objectives SLO
Assignments amp Activities
Week 17
Final Examination-
Comprehensive
11
12
13
14
15
16
22
23
24
31
32
Comprehensive Final Examination
Must be passed with a score of 60 or
better to pass course
7
Vocabulary
ENGLISH ARABIC
Acceleration تسارع ndashعجلة
Activity نشاط أشعاعى
Air pressure ضغط الهواء
Ampere أمبير
Analytical تحليلى
Analytical Method الطريق التحليلية
Angle زاوية
Angle of Deviation زاوية االنحراف
Angle of Incidence زواية السقوط
Angle of reflection زاوية االنعكاس
Angle of refraction زاوية األنكسار
Archimedess Principle مبدأ أرخميدس
Atmospheric pressure الضغط الجوي
Atom الذرة
Axoplasm جبلة المحوار
Axons محاور عصبية
Balanced Force القوة المتزنة
Bernoullis Principle law مبدأ )قانون( برنولي
Binding Energy طاقة الربط
Blood pressure ضغط الدم
blood vessel وعاء دموى
Buoyancy الطفو
Buoyant force قوة الطفو
8
Capacitance سعة المكثف
Capacitor المكثف
Coefficient of friction معامل األحتكاك
Charge شحنة
Circuit دائرة
cohesion قوة التماسك بين جزئيات السائل
components of a vector مكونات متجه
compression ضغط كباس أنضغاط
conduction توصيل
conductivity معامل الموصلية الحرارية
Conductor الموصل
Conservation of Energy حفظ الطاقة
conservation of momentum حفظ كمية الحركة
consumed مستهلك
Coulomb كولوم
cube مكعب
Current تيار
deceleration تباطؤ
Density الكثافة
Dendrites التشعبات العصبية
diffraction حيود الضوء
dimensions االبعاد
Dispersion تشتت تفرق
direction إتجاه
directly proportional يتناسب طرديا
displacement اإلزاحة
9
distance المسافة
Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك
عبره(
drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل
dynamics الديناميكا
efficiency كفاءة
effort arm ذراع القوة
Electric Charge شحنة كهربائية
Electric Circuit دائرة كهربائية
Electric Current التيار الكهربى
Electric Energy طاقة كهربائية
Electric field المجال الكهربي
Electrical Conductivity توصيل كهربائى
Electric Potential جهد كهربائى
Electrical Resistance مقاومة كهربائية
electricity الكهرباء
electrode قطب كهربائى
Electromagnetic Field مجال كهرومغناطيسي
electromagnetic induction الحث الكهرومغناطيسي
Electromotive Force القوة الدافعة الكهربية
Electron اإللكترون
Electron Diffraction حيود اإللكترون
Energy الطاقة
Energy Level مستويات الطاقة
Energy Transformations تحوالت الطاقة
Equation of continuity معادلة اإلستمرارية
Equilibrium إتزان
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
5
Week 8 Daily Objectives SLO Assignments amp Activities
Week 8
Mechanics of non-Viscous Fluid
The Equation of continuity Stream line flow
Bernoullirsquos equation and its static consequences
13
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 9 Daily Objectives SLO Assignments amp Activities
Week 9
Mechanics of non-Viscous Fluid
Role of gravity in blood circulation
Pressure measurement using manometer
13 - Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 10 Daily Objectives SLO Assignments amp Activities
Week 10
Direct Electric Current (DC)
Basic concept of DC Electric current
Ohmrsquos law Electric safety
14
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 11 Daily Objectives SLO Assignments amp Activities
Week 11
Nerve Conduction
Structure of nerve
Electric characteristics of
axon
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 12 Daily Objectives SLO Assignments amp Activities
Week 12
Exam2_Examination
13
14
15
Comprehensive Examination
Good luck
Week 13 Daily Objectives SLO Assignments amp Activities
6
Week 13
Nerve Conduction
Ionic concentration and the resting potential
Response to weak stimuli The action potential
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 14 Daily Objectives SLO
Assignments amp Activities
Week 14
Wave properties of light
The Index of Refraction Reflection of Light Refraction of Light
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 15 Daily Objectives SLO
Assignments amp Activities
Week 15
Wave properties of light
Ionizing Radiation
The Interaction of Radiation with Matter
Chronic Radiation Exposure
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 16 Daily Objectives SLO
Assignments amp Activities
Week 16
Final Examination-Practical Lab
Week 17 Daily Objectives SLO
Assignments amp Activities
Week 17
Final Examination-
Comprehensive
11
12
13
14
15
16
22
23
24
31
32
Comprehensive Final Examination
Must be passed with a score of 60 or
better to pass course
7
Vocabulary
ENGLISH ARABIC
Acceleration تسارع ndashعجلة
Activity نشاط أشعاعى
Air pressure ضغط الهواء
Ampere أمبير
Analytical تحليلى
Analytical Method الطريق التحليلية
Angle زاوية
Angle of Deviation زاوية االنحراف
Angle of Incidence زواية السقوط
Angle of reflection زاوية االنعكاس
Angle of refraction زاوية األنكسار
Archimedess Principle مبدأ أرخميدس
Atmospheric pressure الضغط الجوي
Atom الذرة
Axoplasm جبلة المحوار
Axons محاور عصبية
Balanced Force القوة المتزنة
Bernoullis Principle law مبدأ )قانون( برنولي
Binding Energy طاقة الربط
Blood pressure ضغط الدم
blood vessel وعاء دموى
Buoyancy الطفو
Buoyant force قوة الطفو
8
Capacitance سعة المكثف
Capacitor المكثف
Coefficient of friction معامل األحتكاك
Charge شحنة
Circuit دائرة
cohesion قوة التماسك بين جزئيات السائل
components of a vector مكونات متجه
compression ضغط كباس أنضغاط
conduction توصيل
conductivity معامل الموصلية الحرارية
Conductor الموصل
Conservation of Energy حفظ الطاقة
conservation of momentum حفظ كمية الحركة
consumed مستهلك
Coulomb كولوم
cube مكعب
Current تيار
deceleration تباطؤ
Density الكثافة
Dendrites التشعبات العصبية
diffraction حيود الضوء
dimensions االبعاد
Dispersion تشتت تفرق
direction إتجاه
directly proportional يتناسب طرديا
displacement اإلزاحة
9
distance المسافة
Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك
عبره(
drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل
dynamics الديناميكا
efficiency كفاءة
effort arm ذراع القوة
Electric Charge شحنة كهربائية
Electric Circuit دائرة كهربائية
Electric Current التيار الكهربى
Electric Energy طاقة كهربائية
Electric field المجال الكهربي
Electrical Conductivity توصيل كهربائى
Electric Potential جهد كهربائى
Electrical Resistance مقاومة كهربائية
electricity الكهرباء
electrode قطب كهربائى
Electromagnetic Field مجال كهرومغناطيسي
electromagnetic induction الحث الكهرومغناطيسي
Electromotive Force القوة الدافعة الكهربية
Electron اإللكترون
Electron Diffraction حيود اإللكترون
Energy الطاقة
Energy Level مستويات الطاقة
Energy Transformations تحوالت الطاقة
Equation of continuity معادلة اإلستمرارية
Equilibrium إتزان
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
6
Week 13
Nerve Conduction
Ionic concentration and the resting potential
Response to weak stimuli The action potential
15
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 14 Daily Objectives SLO
Assignments amp Activities
Week 14
Wave properties of light
The Index of Refraction Reflection of Light Refraction of Light
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 15 Daily Objectives SLO
Assignments amp Activities
Week 15
Wave properties of light
Ionizing Radiation
The Interaction of Radiation with Matter
Chronic Radiation Exposure
16
- Lecturing
- Team work
- Exercises
- Self-learning
- Group Discussions
Week 16 Daily Objectives SLO
Assignments amp Activities
Week 16
Final Examination-Practical Lab
Week 17 Daily Objectives SLO
Assignments amp Activities
Week 17
Final Examination-
Comprehensive
11
12
13
14
15
16
22
23
24
31
32
Comprehensive Final Examination
Must be passed with a score of 60 or
better to pass course
7
Vocabulary
ENGLISH ARABIC
Acceleration تسارع ndashعجلة
Activity نشاط أشعاعى
Air pressure ضغط الهواء
Ampere أمبير
Analytical تحليلى
Analytical Method الطريق التحليلية
Angle زاوية
Angle of Deviation زاوية االنحراف
Angle of Incidence زواية السقوط
Angle of reflection زاوية االنعكاس
Angle of refraction زاوية األنكسار
Archimedess Principle مبدأ أرخميدس
Atmospheric pressure الضغط الجوي
Atom الذرة
Axoplasm جبلة المحوار
Axons محاور عصبية
Balanced Force القوة المتزنة
Bernoullis Principle law مبدأ )قانون( برنولي
Binding Energy طاقة الربط
Blood pressure ضغط الدم
blood vessel وعاء دموى
Buoyancy الطفو
Buoyant force قوة الطفو
8
Capacitance سعة المكثف
Capacitor المكثف
Coefficient of friction معامل األحتكاك
Charge شحنة
Circuit دائرة
cohesion قوة التماسك بين جزئيات السائل
components of a vector مكونات متجه
compression ضغط كباس أنضغاط
conduction توصيل
conductivity معامل الموصلية الحرارية
Conductor الموصل
Conservation of Energy حفظ الطاقة
conservation of momentum حفظ كمية الحركة
consumed مستهلك
Coulomb كولوم
cube مكعب
Current تيار
deceleration تباطؤ
Density الكثافة
Dendrites التشعبات العصبية
diffraction حيود الضوء
dimensions االبعاد
Dispersion تشتت تفرق
direction إتجاه
directly proportional يتناسب طرديا
displacement اإلزاحة
9
distance المسافة
Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك
عبره(
drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل
dynamics الديناميكا
efficiency كفاءة
effort arm ذراع القوة
Electric Charge شحنة كهربائية
Electric Circuit دائرة كهربائية
Electric Current التيار الكهربى
Electric Energy طاقة كهربائية
Electric field المجال الكهربي
Electrical Conductivity توصيل كهربائى
Electric Potential جهد كهربائى
Electrical Resistance مقاومة كهربائية
electricity الكهرباء
electrode قطب كهربائى
Electromagnetic Field مجال كهرومغناطيسي
electromagnetic induction الحث الكهرومغناطيسي
Electromotive Force القوة الدافعة الكهربية
Electron اإللكترون
Electron Diffraction حيود اإللكترون
Energy الطاقة
Energy Level مستويات الطاقة
Energy Transformations تحوالت الطاقة
Equation of continuity معادلة اإلستمرارية
Equilibrium إتزان
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
7
Vocabulary
ENGLISH ARABIC
Acceleration تسارع ndashعجلة
Activity نشاط أشعاعى
Air pressure ضغط الهواء
Ampere أمبير
Analytical تحليلى
Analytical Method الطريق التحليلية
Angle زاوية
Angle of Deviation زاوية االنحراف
Angle of Incidence زواية السقوط
Angle of reflection زاوية االنعكاس
Angle of refraction زاوية األنكسار
Archimedess Principle مبدأ أرخميدس
Atmospheric pressure الضغط الجوي
Atom الذرة
Axoplasm جبلة المحوار
Axons محاور عصبية
Balanced Force القوة المتزنة
Bernoullis Principle law مبدأ )قانون( برنولي
Binding Energy طاقة الربط
Blood pressure ضغط الدم
blood vessel وعاء دموى
Buoyancy الطفو
Buoyant force قوة الطفو
8
Capacitance سعة المكثف
Capacitor المكثف
Coefficient of friction معامل األحتكاك
Charge شحنة
Circuit دائرة
cohesion قوة التماسك بين جزئيات السائل
components of a vector مكونات متجه
compression ضغط كباس أنضغاط
conduction توصيل
conductivity معامل الموصلية الحرارية
Conductor الموصل
Conservation of Energy حفظ الطاقة
conservation of momentum حفظ كمية الحركة
consumed مستهلك
Coulomb كولوم
cube مكعب
Current تيار
deceleration تباطؤ
Density الكثافة
Dendrites التشعبات العصبية
diffraction حيود الضوء
dimensions االبعاد
Dispersion تشتت تفرق
direction إتجاه
directly proportional يتناسب طرديا
displacement اإلزاحة
9
distance المسافة
Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك
عبره(
drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل
dynamics الديناميكا
efficiency كفاءة
effort arm ذراع القوة
Electric Charge شحنة كهربائية
Electric Circuit دائرة كهربائية
Electric Current التيار الكهربى
Electric Energy طاقة كهربائية
Electric field المجال الكهربي
Electrical Conductivity توصيل كهربائى
Electric Potential جهد كهربائى
Electrical Resistance مقاومة كهربائية
electricity الكهرباء
electrode قطب كهربائى
Electromagnetic Field مجال كهرومغناطيسي
electromagnetic induction الحث الكهرومغناطيسي
Electromotive Force القوة الدافعة الكهربية
Electron اإللكترون
Electron Diffraction حيود اإللكترون
Energy الطاقة
Energy Level مستويات الطاقة
Energy Transformations تحوالت الطاقة
Equation of continuity معادلة اإلستمرارية
Equilibrium إتزان
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
8
Capacitance سعة المكثف
Capacitor المكثف
Coefficient of friction معامل األحتكاك
Charge شحنة
Circuit دائرة
cohesion قوة التماسك بين جزئيات السائل
components of a vector مكونات متجه
compression ضغط كباس أنضغاط
conduction توصيل
conductivity معامل الموصلية الحرارية
Conductor الموصل
Conservation of Energy حفظ الطاقة
conservation of momentum حفظ كمية الحركة
consumed مستهلك
Coulomb كولوم
cube مكعب
Current تيار
deceleration تباطؤ
Density الكثافة
Dendrites التشعبات العصبية
diffraction حيود الضوء
dimensions االبعاد
Dispersion تشتت تفرق
direction إتجاه
directly proportional يتناسب طرديا
displacement اإلزاحة
9
distance المسافة
Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك
عبره(
drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل
dynamics الديناميكا
efficiency كفاءة
effort arm ذراع القوة
Electric Charge شحنة كهربائية
Electric Circuit دائرة كهربائية
Electric Current التيار الكهربى
Electric Energy طاقة كهربائية
Electric field المجال الكهربي
Electrical Conductivity توصيل كهربائى
Electric Potential جهد كهربائى
Electrical Resistance مقاومة كهربائية
electricity الكهرباء
electrode قطب كهربائى
Electromagnetic Field مجال كهرومغناطيسي
electromagnetic induction الحث الكهرومغناطيسي
Electromotive Force القوة الدافعة الكهربية
Electron اإللكترون
Electron Diffraction حيود اإللكترون
Energy الطاقة
Energy Level مستويات الطاقة
Energy Transformations تحوالت الطاقة
Equation of continuity معادلة اإلستمرارية
Equilibrium إتزان
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
9
distance المسافة
Drag السحب )المقاومة )اللزوجية( التي يبديها المائع لجسم متحرك
عبره(
drift )انسياق )حركة حامالت التيار الكهربائي في شبه الموصل
dynamics الديناميكا
efficiency كفاءة
effort arm ذراع القوة
Electric Charge شحنة كهربائية
Electric Circuit دائرة كهربائية
Electric Current التيار الكهربى
Electric Energy طاقة كهربائية
Electric field المجال الكهربي
Electrical Conductivity توصيل كهربائى
Electric Potential جهد كهربائى
Electrical Resistance مقاومة كهربائية
electricity الكهرباء
electrode قطب كهربائى
Electromagnetic Field مجال كهرومغناطيسي
electromagnetic induction الحث الكهرومغناطيسي
Electromotive Force القوة الدافعة الكهربية
Electron اإللكترون
Electron Diffraction حيود اإللكترون
Energy الطاقة
Energy Level مستويات الطاقة
Energy Transformations تحوالت الطاقة
Equation of continuity معادلة اإلستمرارية
Equilibrium إتزان
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
10
ev إلكترون فولت
Farad فولتالفاراد )وحدة السعة الكهربة( كولوم لكل
Field مجال
Fluid المائع السائل
Fluid Dynamics ديناميكا الموائع
flow rate معدل السريان
Force قوة
force exerted القوة المبذولة
Frequency التردد
Friction اإلحتكاك
Friction forces قوة االحتكاك
fusion دمج -إنصهار
Geometric Method الرسم الهندسى او البيانىطريق
graph الرسم البيانى
Gravitational Force قوة الجاذبية
Gravitational potential energy الطاقة الكامنة لمجال الجذب الكوني
Gravity الجاذبية
Heat حرارة
Heat Energy الطاقة الحرارية
Heat Transfer انتقال الحرارة
Heavy Water الثقيلالماء
History of Physics تاريخ الفيزياء
Hookes Law قانون هوك
horizontal أفقى
Impedance المقاومه الكهربائيه
Ideal Gas Law قانون الغاز المثالي
inclined مائل
inertia القصور الذاتى
Infinity النهائية
intensity الشدة
interference التداخل
International System of Units نظام الوحدات الدولي
inversely proportional يتناسب عكسى
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
11
Ion أيون
ionic concentration التركيز األيوني
Ionizing التأين
Ionizing Radiation أشعة مؤينة
Joule )الجول )الوحدة الدولية لقياس الطاقة
Kelvin المطلقةكلفن درجة الحرارة
Kinetics علم الحركة
Kinetic Energy الطاقة الحركية
kinetic friction االحتكاك الحركي
Laminar flow تدفق المنار
laser الليزر
law of conservation of mechanical energy
قانون حفظ الطاقة الميكانيكية
laws of motion قوانين الحركة
leakage resistance مقاومة التسرب
Light year السنة الضوئية
Light الضوء
Liquid السائل
longitudinal wave الموجه الطوليه
Luminosity سطوع
Magnetic Field مجال مغناطيسي
Magnetic Flux تدفق مغناطيسي
Magnetic Moment عزم مغناطيسي
magnitude معيار amp قيمة
manometer ضغط الدم مقياس
mass الكتلة
matter مادة
mechanics ميكانيكا
mechanical energy الطاقة الميكانيكية
medium وسط
metal معدن
molecules جزئيات
Motion حركة
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
12
Movement حركة
net force محصلة القوة
neutron النيوترون
Nerve عصب
Nerve cells (neuron) خاليا عصبية
Nerve conduction التوصيل العصبى
newton نيوتن
Newtons first law قانون نيوتن األول
Newtons first law of motion قانون نيوتن األول للحركة
Newtons first law of motion (Inertia)
قانون نيوتن األول للحرآة قانون القصور الذاتي
newtons law of gravitation قانون نيوتن للجاذبية
newtons second law of motion قانون نيوتن الثاني للحركة
newtons third law of motion قانون نيوتن الثالث للحركة
normal force قوة عمودية
nuclear radiation اإلشعاع النووي
nucleus نواة
Ohm أوم
Ohms Law قانون
Optics البصريات
particle جسيم
pendulum بندول
photoelectric effect التأثير الكهروضوئي
photon فوتون
physics الفيزياء
pipeline خط انابيب
position موضع
Potential Difference فرق الجهد
Potential Energy طاقة الوضع
power القدرة
pressure الضغط
pressure in liquids في السوائل الضغط
Proton بروتون
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
13
Pulses ومضات
pull سحب
pulling force قوة السحب
quantity كمية
radiation اإلشعاع
radius نصف القطر
reaction رد الفعل
reflection أنعكاس
refraction األنكسار
refractive index معامل االنكسار الضوئي
relativity النسبية
resistance المقاومة
resistance force قوة مقاومة
resistivity المقاومة النوعية
resistor المقاوم
Rest Energy طاقة السكون
Resultant ناتج
Resultant Force القوة الناتجة
Scalar قياسى
Scalar quantities كميات قياسية
smooth أملس -ناعم
Smooth horizontal surface سطح أفقى املس
sound الصوت
space الفضاء
speed السرعة المطلقة
sphygmomanometer مقياس ضغط الدم
static electricity الكهريبة الساكنة
surface السطح
surface tension التوتر السطحى
tension توتر -شد
terminal velocity سرعة الوصول
Temperature درجة الحرارة
Thermal Physics الفيزياء الحرارية
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
14
thermometer مقياس الحرارة
Turbulence اضطراب
turbulent flow تدفق مضطرب
Transformers المحوالت
Transient Energy الطاقة الزائلة
Transverse Waves الموجات المستعرضة
Ultrasound فوق صوتى
ultraviolet ray االشعة فوق البنفسجية
unbalanced forces قوى غير متزنة
uniform motion حركة منتظمة
Vacuum الفراغ
valence electron إلكترونات التكافؤ
vector متجه
vectors متجهات
Vectors Geometry هندسة المتجهات
Velocity السرعة المتجهة
Viscosity اللزوجة
visual angle زاوية اإلبصار
volt فولت وحدة قياس فرق الجهد الكهربي
voltage الجهد الكهربي
voltmeter الفولتميتر جهاز قياس فرق الجهد الكهربي
Volume الحجم
water pipe انابيب مياه
water pressure ضغط الماء
watt W وحدة قياس القدرة الكهربية -واط
Wave موجة
Wave Function الدالة الموجية
wave length الطول الموجي
Wave Interference التداخل الموجى
Wave Motion الحركة الموجية
Wave Phenomena الظواهر الموجية
wave speed سرعة الموجة
Wave Superposition التراكب الموجى
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
15
weakcohesive
Weight الوزن
work الشغل
Work done الشغل المبذول
X-Ray )اشعة إكس ) السينية
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
16
Important unit conversions in physics course
Cm x 10-2 m
m m x 10-3m
μ m x 10-6m
Litter x 10-3m3
Gram x 10-3 kg
Symbol
L ndash length (m)
A ndash cross sectional area (m2)
r- radius (m)
R ndash resistance (Ω)
ρ (rho)ndash resistivity (Ω m) specific electrical
resistance
ρ=mv The density of a fluid(kgm3)
P ndash Pressure (Nm2 called the Pascal (Pa)
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
17
Ch 1 (11 Vector)
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
18
Part 1 Define scalar and vector quantity
Part 2 Adding vector
There are three methods to adding Vector
1- Graphical or called (Geometrical Method)
2- Pythagorean Theorem
3- Analytical Method or called Components Method
1- Graphical or called (Geometrical Method)
Add vectors A and B graphically by drawing them together in a head to
tail arrangement
Draw vector A first and then draw vector B such that its tail is on
the head of vector A
Then draw the sum or resultant vector by drawing a vector from the
tail of A to the head of B
Measure the magnitude and direction of the resultant vector
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
19
Example 1
A man walks at40 meters East and 30 meters north Find the magnitude
of resultant displacement and its vector angle Use Graphical Method
Answer
Given
A = 40 meters East B = 30 meters North
Resultant (R) = Angle θ =
So from this
Resultant (R) =50 amp Angle θ = 37
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
20
2- Pythagorean Theorem
The Pythagorean Theorem is a useful method for determining the result
of adding two (and only two) vectors and must be the angle between
this two vector equal =90
Example2
A man walks at 40 meters East and 30 meters North Find the magnitude
of resultant displacement and its vector angle Use Pythagorean
Theorem
Answer
_____________________________________________________
22 BABAR
)BA (1Tan
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
21
Notes(1) To calculate the magnitude A+B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A+B
Solution
1- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A+B
Notes(2) To calculate the magnitude A-B with angle degree 90 o or 90 o
We use the equation
Example
Given A = 5 and θA = 120o and B = 7θB = 60o find the magnitude A-B
Solution
2- we find the total angle θ =θA-θB SO θ =120-60 = 60
2-We use the equation
So A-B
3-Analytical Method or called Components Method
First to calculate the components and magnitude of vector for example
the components of vector A are
Ax = A Cos θ and Ay = A sin θ
COSABBABA 222
COSABBABA 222
44106075275 22 COSxx
COSABBABA 222
COSABBABA 222
2466075275 22 COSxx
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
22
Example 1
Find the components of the vector A If A = 2 and the angle θ = 30o
Solution
Since Ax = A Cos θ and cos 30 = 0866 so Ax = 2 cos 30 = 2 x 0866 = 173
Also Ay = A sin θ and sin 30 = 0500 so Ay = A sin 30 = 2 x 05 = 1
Example 2
Given A = 3 and θ = 90o find Ax and Ay
Solution
Since Ax = A Cos θ and cos 90 = 0 so Ax = 3 cos 90 = 3 x 0= 0
Also Ay = A sin θ and sin 90 = 1 Ay = A sin 90 = 3 x 1 = 3
Second To calculate the magnitude of vector for example magnitude vector A and
direction angle
We use the equation and
Example
If the components of a vector are defined by Ax =346 and Ay =2 find the
magnitude and direction angle of the vector A
Solution
1-We use the equation to find the magnitude vector
So the magnitude vector A=399
2- To find the direction angle we use the equation
30o So the direction angle θ=30o
22
yx AAA ) AA( xy1Tan
22
yx AAA
993)2()463( 22 A
) AA( xy1Tan
) 3462(1Tan
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
23
Third To calculate the resultant vector by component method
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
24
Example If A= 25 and θA = 50 B=4 and θB = 150 C=6 and θC = 265
1- Calculate the Resultant magnitude by using component method
2- Calculate the Resultant angle direction
Answer
solution (1) We use the last equations So
By using equation so use the equation
solution (2) we use the equation
so
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
25
Part 3 Unit Vector Notation and product of vector
Unit Vector Notation
A unit vector is a vector that has a magnitude of one unit and can have any
direction
1-Traditionally i^ (read ldquoi hatrdquo) is the unit vector in the x direction
2- j^ (read ldquoj hatrdquo) is the unit vector in the y direction |i^|=1 and | j^|=1 this
in two dimensions
3-and motion in three dimensions with ˆk (ldquok hatrdquo) as the unit vector in the z
direction
Notes
If AampB are two vectors where
A = axi + ayj + azkamp B = bxi + byj + bzk Then the
1- To findA+B and A B
A+B= (ax +bx)i + (ay +by)j + (az +bz)k
A B= (axbx)i + (ayby)j + (azbz)k
Example
Two vector A = 3i +2j +3K and B = 5i + 4j +3k find A+B and A B
Solution
1- According the equation A+B= (ax +bx)i + (ay +by)j + (az +bz)k
So A+B= (3 +5)i + (2 +4)j + (3 +3)k =8i + 6j + 6k
2-According the equation A B= (ax bx)i + (ay by)j + (az bz)k
So A B= (35)i + (24)j + (33)k = -2i ndash2j + 0k= -2i-2j
_____________________________________________________
2-To find the magnitude of A+B and A B
Example 222 )()()( zzyyxx bababaBA
222 )()()( zzyyxx bababaBA
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
26
Two vector A = 3i +2j +3K and B = 5i + 4j +3kfind the magnitude for A+B and A B
Solution
1- To find the magnitude for A+B
According the equation
So =1166
2- To find the magnitude for AB
According the equation
So =282
2-the magnitude of vector in Unit Vector Notation
If A is vectoring where A = axi + ayj + azk Then the
To find magnitude of vector Awe use the equation
Example
vector A = 3i +2j +3Kfind magnitude of vector A
Solution
According the last equation
So
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222 )()()( zzyyxx bababaBA
222 )33()42()53( BA
222
zyx aaaA
694323 222 A
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
27
Product of Vectors
There are two kinds of vector product
1 The first one is called scalar product or dot product because the result of
the product is a scalar quantity
2 The second is called vector product or cross product because the result is a
vector perpendicular to the plane of the two vectors
Example on the dot(scalar)and cross product
1- If the magnitude of A is A=4θA = 35o and the magnitude of B is B=5 and θB = 70o
Find a) A B c) A x B
Solution
Θ=θBθA = 70o35o= 35o
So A B= A B COSθ = 4 x5 x COS 35o=1638
A x B= A B Sinθ = 4 x5 x Sin 35o=1147
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
28
Notes on the scalar product
If A amp B are two vectors where
A = Axi + Ayj + Azk ampB = Bxi + Byj + Bzk
Then their Scalar Product is defined as
AB = AxBx + AyBy + AzBz Where
amp
Example
Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the scalar product A B
Solution
According the last equation
So AB =(2x5)+(3x2)+(4x6)=10+6+24= 40
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
29
Summary low in the chapter
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
30
Quizzes
1- If the magnitude of A is A=4 θA = 35o and the magnitude of B is B=5 and θB = 70o
find a) A +b b) A - b c) A x B d) A B
2- Two vector A = 2i +3j +4K and B = 5i + 2j +6k find the magnitude of a) AB b) A+B c) A-B
3- A man walks at 20 meters East and 15 meters north Find the magnitude of
resultant displacement and its vector angle Use Graphical Method and
Pythagorean Theorem
4- If the magnitude of A is A=2 magnitude of B is B=3 and θ =30o
Find a) A +b b) A - b c) A x B d) A B
5- Two vector A= 5i -7j+10k and B= 2i +3j-2k find AB
6- Vector A has a magnitude of 5 units and direction angle ΘA = 30
find Ax and Ay
7- the components of a vector are defined by Ax =346 and Ay =2 find the magnitude and direction angle of the vector A
8- If A= 10 and θA = 30 B=7 and θB = 70 C=8 and θC = 240
Calculate the Resultant magnitude by using component method
Calculate the Resultant angle direction
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
31
Choose the correct answer
Which of the following is a physical quantity that has a magnitude but no direction
A Vector B Resultant C Scalar D None
Which of the following is an example of a vector quantity
A Temperature B Velocity C Volume D Mass
Which of the following is a physical quantity that has a magnitude and direction
A Vector B Resultant C Scalar D None
Given |A|=6 and ӨA =60 Find the Ax and Ay A Ax= 23 Ay=19 B Ax= 2 Ay=3 C Ax= 3 Ay=52 D Ax= 51 Ay=17
The magnitude of the resultant of the vectors shown in Figure is A 2 N B 12 N C 35 N D minus2 N
Given |A|= 5 ӨA =120o and |B|=7 ӨB =60o Find the magnitude |B +A|
A |B +A|= 5
B |B +A|=72
C |B +A|=1044
D |B +A|=86
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
32
A car travels 90 meters due north in 15 seconds Then the car turns around and
travels 40 meters due south What is the magnitude and direction of the cars
resultant displacement
A 40 meters South
B 50 meters South C 50 meters North
D 40 meters North
A car moved 60 km East and 90 km West What is the distance it traveled A 30 km West B 60 km East C 90 km D 150 km
What is magnitude
A The direction that describes a quantity
B A numerical value C A unit of force
150N weight hanging DOWN from a rope Vector or scalar
A Scalar
B Vector
What type of quantity is produced by the dot product of two vectors
A scalar
B vector
Tow vectors A= 3i +5j-2k and B= 4i -3j Find the scalar product AB
A - 6
B - 8
C -2
D -3
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
33
Ch1 (12Newtons laws)
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
34
Facts about FORCE
Force unit is the NEWTON (N)
Its definition a push or a pull
What change the state of object is called ldquoforcerdquo
Means that we can control the magnitude of the
applied force and also its direction so force is a vector
quantity just like velocity and acceleration
Adding Forces
Forces are vectors (They have both magnitude and direction)
and so add as follows
1-Adding Forces In one dimension
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
35
2-Adding Forces In two dimensions
a) The angle between them is 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
2
2
2
1 FFF
NF 252015 22
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
36
B) The angle between them is or 90deg
Example
In this figure shown find the resultant (Net) force
Solution
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
37
Newtonrsquos First Law
An object at rest tends to stay at rest and an object in motion tends to
stay in motion with the same speed and in the same direction unless
an external force is acting on it
Or in other words
Everybody continuous in its state of rest or in uniform motion Unless
an external force is acting on it
Notes Newtonrsquos First Law is also called the Law of Inertia
So
Inertia is a term used to measure the ability of an object to
resist a change in its state of motion
An object with a lot of inertia takes a lot of force to start or
stop an object with a small amount of inertia requires a small
amount of force to start or stop
------------------------------------------------------------------------------------------------
Weight
Weight (W) or Force due to Gravity is how our MASS (m) is effected by gravity (g)
mgW
Inability of an object to change its position by itself is called Inertia
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
38
Normal Forces Fn
Normal force this force acts in the direction perpendicular to the contact
surface and opposite the weight
Friction Forces Ff
Is opposing force caused by the interaction between two surfaces
Calculate the Friction Force and Normal Force
a) With out angle
Notes
If the surface is smooth the friction force Ff= 0
N
mg
F
N
w
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
39
Example 1
A man is pulling 20Kg suitcase with constant speed on a horizontal rough
floor show figure The pulling force F1 action is unknown Find The pulling
force F1 and normal force FN
Solution
From figure
F1= F2 = 20 N So the pulling force F1 action is 20 N
FN =m g where m= 20 and g=10 So FN = 20 x10=200N
Example 2
In this figure shown the object is at rest Find normal force FN
Solution
From figure
FN + F2 = F1 FN = F1F2 =2510=15 N
So the normal force FN =15 N
b) With angle
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
40
Example
An object of mass m=5Kg is pulled by a force F on a horizontal floor If the magnitude
of the pulling force F= 16N and its direct 30degabove the horizontal Find
a) Friction force b) The normal force FN
Solution
Given
m=5 Fp =16N θ=30deg and g=10
The pulling force F analysis in x and y direction show figure
a) Friction force
Ff = Fx = F cos θ=16 x cos30deg = 138 N
So Ff Friction force =138 N
b) The normal force FN
FN=mg Fy=mg FSin θ=5x1016xsin 30deg = 42 N
So FN normal force =42 N
---------------------------------------------------------------------
Newtonrsquos Second Law
ldquoForce equals mass times accelerationrdquo
F = ma
What does F = ma mean
Force is directly proportional to mass and acceleration
Notes
Newtonrsquos second law states that the net force on an object is
proportional to the mass and the acceleration that the object
undergoes
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
41
(a)Acceleration a measurement of how quickly an object is
changing speed a= Fm
Example
Calculate the force required to accelerate a 15Kg block along the floor at 30
ms2
Solution
Given m=5 and a= 30 ms2
According F = ma so F = 5x3=15 N
Net Force
The net force is the vector sum of all the forces acting on a body
321net FFFFF
aF m Example 1
The forces F1=10 N and F2=5N are the action on the block of mass 3 kg with 30deg
Find
1 The net force
2 The acceleration of the block Solution
1 we find the resultant (Net) force
According the equation
So
COSFFFFF 21
2
2
2
1 2
NCOSxxF 514301052105 22
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
42
2 The acceleration of the block (a)
a= Fm where F=145 N amp m=3kg so a= 1453 =483 ms2
Example2 A 10-kg box is being pulled across the table to the right by a rope with an applied force of
50N Calculate the acceleration of the box if a 12 N frictional force acts upon it
Solution
Given m=10 Fa=50 and Ff=12
first we find the resultant (Net) force
So the acceleration of the box
------------------------------------------------------------------------------------------------------------
Newtonrsquos Third Law
ldquoFor every action there is an equal and opposite reactionrdquo
Coefficients of friction
Coefficient of friction is the ratio between friction force and normal force
Symbol is the Greek letter mu (μ)
μ= Ff FN
The coefficient of friction has no units
-----------------------------------------------------------------------------------------------------
Notes
Friction Force = Coefficient of friction Normal Force
Ffriction = Fnormal
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
43
Example1
A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N
Find the coefficient of friction between the bag and the floor
Solution
Given Fp=40 N m=20and g=10
From figure
Ff = Fp = 40 N So Ff action is 40 N
FN = m g where m= 20 and g=10 So FN = 20 x10=200N
So the coefficient of friction ( μ)
μ= Ff FN μ= 40 200 =02
-----------------------------------------------------------------------------------------------
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
44
Example3
A young man of 40kg mass is skidding down an inclined rough surface The surface is inclined at
45deg above horizontal as shown in the figure the coefficient of friction between his skies and the
surface is 01
Find
a) The force of friction
b) the acceleration of the boy
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
45
Solution
Given m=40 kg μ =01 g=10 θ=270-45=225
First from m=40 so w=mg =40x10=400
Weight (W) or Force due to Gravity analysis in x and y direction
So Wx = W Cos θ = 400 Cos 225 = 2828 N
And Wy = W sinθ = 400 sin 225 = 2828 N
from figure FN= Wy= 2828 N
a) So The force of friction According the equation
Ffriction = Fnormal where μ =01 and Fn= 2828 N
Ff = 01 x 2828=2828N
b) the acceleration (a) a= Fm where F=WxFf = 28282828 = 2545 N
And m=40 kg
a=254540 =636 ms2
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
46
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
47
Quizzes 1 Calculate the force required to accelerate a 15Kg block along the floor at 30 ms2 m
2 The forces F1=10 N and F2=5N are the action on the block of mass 3 kg Find the resultant force
and acceleration of the block
3 An object of mass m=3Kg is subject to a force F=9N Find
a) Wight of the object b) the acceleration of the object
4 The forces F1=2 N and F2=4N are the action on the object with 60deg Find the magnitude of the
resultant force
5 An object of mass m=5Kg is pulled by a force F on a smooth horizontal floor If the magnitude of
the force F= 16N and its direct 30degabove the horizontal Find
a) The normal force N b) The acceleration of the object
6 A man is pulling a bag of 20 Kg mass on a horizontal floor The pulling force is 40 N inclined at 30deg
above the horizontal and the coefficient of friction between the bag and the floor is 01
What is the force of friction
What is the acceleration of the suite case 7 A man of 60 Kg sits on a chair while his feet is resting on the ground The ground exerts a force of
350 N on the feet Find the force exerted by the chair on him
8 A man mass is pulling a suitcase of 15Kg on a horizontal rough floor If the coefficient of friction is
02What is the pulling force
9 A man of 80 kg mass is sitting on a chair and his feet is resting against the ground His feet is
experiencing 300 N force applied by the ground Find the force applied on him by the chair
10 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface The force of
friction is Fk = 60 N What is the coefficient of friction microk
11 A lady is pulling a 30 kg mass suit case on a rough horizontal floor The pulling force F=90 N and
the coefficient of friction microk =01
What is the magnitude of the force of friction
What is the acceleration of the suit case
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
48
Choose the correct answer 1 What type of forces do not change the motion of an object
a balanced forces
b unbalanced forces c static forces d accelerating forces
2 If the net force acting on an object is zero then the object will remain at rest or move in a straight line with a constant speed is
a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion d Newtons fourth law of motion
3 What unit do we use to measure force
a Newton b Meter c Pascal d Joule
4 When an unbalanced force acts on an object the force
a changes the motion of the object b is cancelled by another force c does not change the motion of the object d is equal to the weight of the object
5 What is the acceleration of gravity
a 18 ms2 b 5 ms2 c 10 ms2 d 89 ms2
6 An objects resistance to change in motion
b Motion c Inertia d Friction e Mass
7 is the measure of the force of gravity on an object
a mass b weight c density d equation
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
49
8 Forces always act in equal but opposite pairs is a Newtons first law of motion b Newtons second law of motion c Newtons third law of motion
d Newtons fourth law of motion
9 The force of attraction between any two objects that have mass a Energy b Force c Gravity
d Speed
10 When you use a boat paddle to push water backwards the water exerts an opposite force pushing the boat forward This is an example of
a Newtons First Law of Motion b Pascals Law
c Newtons Third Law of Motion d Archimedes Principle
11 Which is the correct equation for Newtons second law (relationship between mass acceleration and force)
a F=ma
b m=Fa c aF=m d m=aF
12 A force that resists motion created by objects rubbing together is a gravity
b friction c speed d force
13 A box of 30 Kg mass is pulled with constant speed on a horizontal rough surface
The force of friction is Fk = 60 N What is the coefficient of friction microk
a) 05 b) 01
c) 03 d) 02
14 In the figure shown find the resultant (Net) force
a) 106 b) 2078
c) 145 d) 304
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
50
15For every action therersquos an equal and opposite reaction
a Newtons First Law
b Newtons Second Law c Newtons Third Law d Force
16The sum of all the forces acting on an object or system a net force b force
c normal force d drag force
17 an opposing force caused by the interaction between two surfaces
a inertia b mass c friction d force
18 state of rest or balance due to the equal action of opposing forces a equilibrium b force c inertia d mass
19The force perpendicular to the surface that pushes up on the object of concern
a normal force
b force c drag force
d net force
20An object of mass 10 kg is accelerated upward at 2 ms2 What force is required a 20 N b 2 N c 5 N
d 0 N
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
51
Ch 2 work and energy
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
52
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
53
Notes on Work
Work = The Scalar Dot Product between Force F
and Displacement d
W = F d
The unit of work is a joule (J) and J = N middot m
Calculate work done on an object
1-Without angle
a) with apply force
The equation used to calculate the work (W) in this case it
W= F d
Example
How much work is done pulling with a 15 N force applied at
distance of 12 m
Solution
Given F=15 N amp d=12m
According the equation W= F d
So W=15x12=180 J
ntdisplacemeForceWork
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
54
b) Also with friction force
The equation used to calculate the work (W) in this case it
W= -Ff d -----------1
But Ffriction = Fnormal so you can write this equation (1)
W= -(Fnormal)d ---------2
But Fnormal= m g so you can write this equation(2)
W= -(mg)d ---------3
-------------------------------------------------------------------
Example
A horizontal force F pulls a 10 kg carton across the floor at
constant speed If the coefficient of sliding friction between the
carton and the floor is 030 how much work is done by F in
moving the carton by 5m
Solution
Given m=10 kg d=5m g=10 and μ=30 W=
The carton moves with constant speed Thus the carton is in
horizontal equilibrium
Fp = Ff = μk N = μk mg
Thus F = 03 x 10 x 10= 30 N
Therefore work done W = F d=30 x 5= 150 J
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
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Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
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95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
55
2-With angle
In this case the work done given by
Example
How much work is done pulling with a 15 N force applied at 20o over
a distance of 12 m
Solution
Given F=15 N θ=20oamp d=12m
According the equation W= F dCos θ
So W=15x12xCos 20o=1691 J
----------------------------------------------------------------------
Example
An Eskimo returning pulls a sled as shown The total mass of the sled is 500 kg
and he exerts a force of 120 times 102 N on the sled by pulling on the rope
a) How much work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) Suppose microk = 0200 How much work done on the sled by friction
c) Calculate the net work if θ = 30deg and he pulls the sled 50 m
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
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87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
56
Solution
Given F=120 times 102 N θ=30deg microk = 0200amp d=5m g=10
a) Calculate work does he do on the sled if θ = 30deg and he pulls the sled 50 m
b) calculate the work done on the sled by friction
c) Calculate the net work
J
mN
dFW
520
)30)(cos05)(10201(
cos
2
J
N
dFmgxN
dFxFW
kk
fffric
440
)5)(30sin10211050)(2000(
)sin(
)180cos(
2
J
WWWWW gNfricFnet
090
00440520
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
57
Kinetic Energy
Kinetic Energy is the energy of a particle due to its motion
KE = frac12 mv2
Where
K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
Example
A 1500 kg car moves down the freeway at 30 ms
Find the Kinetic Energy
Solution
Given m=1500kg v=30ms
According the equation KE = frac12 mv2
So KE = frac12(1500 kg)(30 ms)2= 675000 kgm2s2 = 675 kJ
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
58
Work and Kinetic Energy
When work is done on a system and the only change in the
system is in its speed the work done by the net force equals
the change in kinetic energy of the system
So W = Kf - K0 ------------1
And also W =frac12 mvf2 frac12 m v0
2 ------------2
But W= -Ff d
So -Ff d=frac12 mvf2 frac12 m v0
2 ------------3
From equation (3) you can calculate the friction force
Example
A child of 40kg mass is running with speed 3ms on a
rough horizontal floor skids a distance 4 m till stopped
a) Find the force of friction
b) Find the coefficient of friction
Solution
Given m=40 kg v0=3ms vf=0 d= 4m and g=10
a) Calculate the force of friction
We apply the equation -Ff d=frac12 mvf2 frac12 m v0
2
But vf=0 so frac12 mvf2 =0
-Ff d=0 frac12 m v02 -Ff d=- frac12 m v0
2
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
59
Ff= (frac12 m v02 ) d =(frac12 4032 ) 4= 45 N
So the force of friction = 45 N
b) Calculate the coefficient of friction
According the equation in ch2 μ= Ff FN
Where Ff= 45 N and FN =mg=4010=400
So μ= Ff FN μ= 45 400 μ=01
---------------------------------------------------------------------------- Example
A 60-kg block initially at rest is pulled to the right along a horizontal
frictionless surface by a constant horizontal force of 12 N Find the speed of
the block after it has moved 30 m
Solution
GivenFp= 12 N m=6 kg v0=0 vf=
d= 3m and g=10
W =Fp d =12x 3 = 36J
Δk = w
frac12 mvf2 frac12 m v0
2 = w
But vo=0 so frac12 mv02 =0
frac12 mvf2 = W
frac12 x 6 x vf2 = 36 vf
sm 46312
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
60
Potential Energy
Potential Energy means the work done by gravity on the object
The formula for potential energy (U) due to gravity is U = mgh
PE = mass x height x gravity
The unit of Potential Energy is a joule (J)
----------------------------------------------------------------------------------------
Example
A child of 40 kg mass is sitting at the roof a tower 60m high referenced to the
ground What is the potential energy of child
Solution
Given m=40 kg h= 60m and g=10
According the equation U = mgh
So U = 40 x 10x 60=24000 J
---------------------------------------------------------------------------------------------------
Conservation of Energy
bull Conservation of Mechanical Energy
MEi = MEf
initial mechanical energy = final mechanical energy
SO Ko + Uo = Kf + Uf----------1
SO Uo Uf = KfKo
So KfKo= -(Uf Uo )
K= U-----------------------------------2
So frac12 mvf2 frac12 m v0
2 = mg(hfho)---------------------------3
The equation (123) is very important
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
61
Example
At a construction site a 150 kg brick is dropped from rest and hit the ground
at a speed of 260 ms Assuming air resistance can be ignored calculate the
gravitational potential energy of the brick before it was dropped
Solution
Given m=150 kg v0=0 vf=26 Uf=0 Uo=
According Ko + Uo = Kf + Uf
But vo=0 so Ko =frac12 mv02 =0 and Uf=0
So Uo = Kf Uo=mgho = frac12 mvf2
Uo= frac12 x (15x 26)2= 507 J
-----------------------------------------------------------------------------------------------
Example
A child of 20 kg mass is ON A swing The swing reaches maximum height 3 m
above her lowest position Find her speed at the lowest position
Solution
Given m=20 kg v0=0 vf= hf=0 ho=3 and g= 10
According the equation frac12 mvf2 frac12 m v0
2 = mg(hfho)
But vo=0 so frac12 mv02 =0
frac12 mvf2 = mg(03) frac12 mvf
2 = 3mg frac12 vf2 = 3g
g=10 frac12 vf2 = 30 vf
2 = 60
vf sm 7760
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
62
Power Power is the time rate of energy transfer
P =119882
119905=
119865 119889
119905
Units of Power
Where the unit of work(W) is joule and unit of time(t) is second So The
unit of power is a Watt
where 1 watt = 1 joule second
--------------------------------------------------------------------------------------
Example
A 100 N force is applied to an object in order to lift it a distance of 20 m
within 60 s Find the power
Solution
Given F=100 N d=20 m t=60 s
According the equation P =119882
119905=
119865119889
119905=
100 119857 20
60=
3333 waat
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
63
Example
A woman of 50 Kg mass climb s a mountain 4000 m high
a) Find the work she did against gravitational forces
b) A Kilogram of fat supplies energy of 37x107 J If she converts fat to
energy with efficiency rate of 25 How much fat she consumed in the climb
Solution
Given m=50 kg h=4000 m and g= 10 ms2
a) Calculate the work she did against gravitational forces
W= F d where in this case F= m g and d=h
So W= m g h W= 50 x 10 x 4000=2000000=2 x 106 J
b) Calculate the fat consumed in the climb
According the equation 119862 = 119882119905
where W=2 x 106 J
And the rate (t) = (37x107) x(25)=(37x107) x(025)=9250000=925 x 106
So 119862 = 119882119905
=2 x 106
925 x 106= 0216 kg
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
64
Example
A man of 70 kg mass is lifting up 10 kg mass upwards a vertical distance
of 05m 1000 times
a) Find the work he did against gravitational forces
b) A 1 Kilogram of fat is converted to energy at the rate of 38x107 J If the
man converts energy at 20 efficiency rate How much fat will he
consume in the exercise
Solution
Given mass for man m1=70 kg and he is lifting mass m2 =10
So mass total m= m1+m2=70+10=80 d=05 m and g= 10 ms2
a) Calculate the work he did against gravitational forces
W= F d where in this case F= m g
So W= m g d W= 80 x 10 x 05 =400J He does this 1000 times so the work he did against gravitational forces
W=400x1000=400000 J=4 x 105 J
b) Calculate the fat consumed in the exercise
According the equation 119862 = 119882119905
where W=4 x 105 J
And the rate (t) = (38x107) x(20)=(38x107) x(020)=7600000=76 x 105
So 119862 = 119882119905
=4 x 105
76 x 105 = 00526 kg
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
65
Quizzes 1 Find the potential energy of 20 Kg mass child sitting on a roof 10m above the ground
2 A truck is pulling a box of 20 Kg mass on a horizontal surface a distance of 10 m with a constant
speed The force of friction between the box and the surface is 20 N
Find the work it did against the force of friction
3 A ball of 3 Kg mass was dropped from rest the top of tower 50 m high
Find the speed of the ball 20 m above the base of the tower
4 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the energy loss
5 A boy of 50 Kg mass climbrsquos a wall 500 m high
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If he converts fat to energy with efficiency
rate of 25 How much fat he consumed in the climb
6 A car of 800 Kg mass is travelling at 20 ms speed coasts to a stop in 400 m on a rough horizontal
road Find the force of friction
7 A car of 800 Kg mass is travelling at 20 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car Find the reactive force acting on the car body during the
crash
8 A man raises a 10 Kg mass vertically upwards a distance of 05 m He practices that 1000 times
a) Find the work he did against gravitational forces
b) A Kilogram of fat supplies energy of 37x10^7 J If the man converts fat to energy with
efficiency rate of 25 How much fat he consumed in the exercise
9 A child of 30kg mass is running with speed 5ms on a rough horizontal floor skids a distance 3 m
till stopped Find the force of friction
10 A child 0f 25 kg mass climbs a tower 50m height above the ground Find his potential energy at
the top of the tower
11 A car of 100 Kg mass is travelling at 15 ms speed hits a concrete wall and comes to rest after
smashing 15 meter of the front of the car
a) Find the kinetic energy of the car
b) Find the reactive force acting on the car body during the crash
12 A child of mass 30 kg climbs a tower 50 m high above the ground surface ( given that the
acceleration due to gravity g= 10ms2) Find his potential energy at top of the tower
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
66
Choose the correct answer
Potential energy and kinetic energy are types of A Electrical energy B Magnetic energy C Thermal energy D Mechanical energy
Work done = Force x _______ A distance
B acceleration
C velocity
D speed
1 joule = 1 _______ A N m2 B Kgs2 C N m D N2 m2
The unit of power is _______ 1 watt per second 2 joule 3 kilojoule 4 joule per second
A watt per second B joule C kilojoule D joule per second
A man of mass 50 kg jumps to a height of 1 m His potential energy at the highest point is (g = 10 ms2)
A 50 J
B 60 J
C 500 J
D 600 J
A B C D
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
67
A 1 kg mass has a kinetic energy of 1 joule when its speed is
A 045 ms
B ms
C 14 ms
D 44 ms
Name the physical quantity which is equal to the product of force and
distance
A Work
B energy
C power
D acceleration
An object of mass 1 kg has potential energy of 1 joule relative to the
ground when it is at a height of _______
A 010 m
B 1 m
C 98 m
D 32 m
What is kinetic energy
A When an object is in motion
B When an object is not in motion
C all of the above
D none of the above
It takes 20 N of force to move a box a distance of 10 m How much work is
done on the box A 200 J B 200J
C 2 J D 200 N
Two factors that determine work are
A amount of the force and effort used B amount of the force and type of force
C mass and distance D amount of force and distance moved
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
68
What is energy
A It is measured in watts B It is power
C It is the ability to do work D It is fluid motion
What is work A The product of force and displacement
B Causes a change in potential energy of an object C Does not depend on the path traveled but only starting and ending position
D All of these are true
The law of conservation of energy states
A Energy cannot be created
B Energy cannot be destroyed C Energy can only be transferred
D All of these
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
69
Ch3 Direct currents
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
70
Electric current The electric current in a wire is the rate at which the charge moves in the wire
Definition of the current
The SI Current unit is the ampere (A)
t
QI
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
71
Ohmrsquos Law
For many conductors current depends on
Voltage - more voltage more current
Current is proportional to voltage
Resistance - more resistance less current
Current is inversely proportional to resistance
Example 3
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
72
Example 4
What is the resistance of the heating element in a car lock deicer that
contains a 15-V battery supplying a current of 05 A to the circuit
Resistance (R)
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
73
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
74
According to Ohms law Resistance is equal to to voltage divided by
A potential difference B conduction
C time D current
What is a circuit
A A pathway that electricity flows in It has a load wire and a taco B A pathway that protons flow in It has a load wire and a power source
C A pathway that electricity flows in It has a load wire and a power source
D A pathway that electricity flows in It has a load and wire
What is an Electric Current
A A An Electric Field B B An Ampere
C C The flow of electric charge
What is Ohms Law
A I=VR
B R=VI
C Power= Voltage times Current D AampB
A closed path that electric current follows A Voltage
B Current C Resistance
D Circuit
This is related to the force that causes electric charges to flow
A Voltage B Current
C Resistance D Circuit
What charge does an electron have
A negative (-) B positive (+)
C neutral or no charge (0)
Resistance is affected by a materialrsquos
A temperature B thickness
C length D all of these
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
75
The number of electrons flowing is called
A voltage B power C current D resistance
When the circuit is______ current does not flow
A resistors B heat C closed D open
Electrons leave the ______ of a battery and enter the ______ of the battery
A Positive terminal positive terminal
B Negative terminal negative terminal C Negative terminal positive terminal
D Positive Terminal Negative Terminal
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
76
Ch4 Nerve Conduction
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
77
Nerve Conduction
What is nerve conduction study A nerve conduction study (NCS) also called a nerve conduction velocity (NCV) test--is
a measurement of the speed of conduction of an electrical impulse through a nerve
NCS can determine nerve damage and destruction
A nerve conduction study (NCS) is a medical diagnostic test commonly used to
evaluate the function especially the ability of electrical conduction of the motor and
sensory nerves of the human body
The structure of the nerve cells (neuron)
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
78
Nerve electric properties
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
79
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
80
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
81
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
82
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
83
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
84
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
85
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
86
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
87
Ch 5 THE MECHANICS OF NON-VISCOUS
FLUIDS
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
88
----------------------------------------------------------------------------
What is the Fluids
A fluid is a collection of molecules that are randomly arranged
and held together by weakcohesive forces and by forces exerted
by the Walls of a container
Both liquids and gases fluids
--------------------------------------------------------------
Density and Pressure 1- Density bull The density of a fluid is defined as mass per unit volume
ρ=mv (uniform density)
bullDensity is a scalar the SI unit is kgm3
2-Pressure
P=FA (Pressure of uniform force on flat area)
bull F is the magnitude of the normal force on area Abull The SI unit of pressure is Nm2 called the Pascal (Pa)bull The tire pressure of cars are in kilopascalsbull 1 atm = 1013x 105 Pa = 76 cm Hg = 760mm Hg
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
89
---------------------------------------------------------------
if there is an incompressible fluid completely fills a channel such as a pipe or an artery
Then if more fluid enters one end of the channel So an equal amount must leave the other
end This principle is called
The Equation of Continuity
The Equation of Continuity (STREAMLINE FLOW)
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
90
The flow rate (Q)
119876 is The flow rate which is the volume ΔV of the fluid flowing past a
point in a channel per unit time Δt
The SI unit of the flow rate 119876 is the 119950 3 119956
Example
If the volume of water flows flowing past a point in pipeline in 3
minutes is 5 litters what is the flow rat
Answer
Given
ΔV= 5 litter =5x10-3 119950 3 and Δt=3 minutes=3x60 s= 180 s
So according the last equation
Q = 119881
119905=
5x10minus3
180= 27x10minus5 1198983119852
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
91
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
92
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
93
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
94
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
95
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
96
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
97
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
98
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
99
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
100
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
101
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
102
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
103
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
104
Exercise 1 Questions and Answers
What are fluids A Substances that flow B Liquids and gases
C Aampb D Non of the above
Bernoullis principle states that for streamline motion of an incompressible
non-viscous fluid
A pressure at any part + kinetic energy per unit volume = constant
B kinetic energy per unit volume + potential energy per unit volume = constant
C pressure at any part + potential energy per unit volume = constant
D pressure at any part + kinetic energy per unit volume + potential energy per
unit volume = constant
If layers of fluid has frictional force between them then it is known as
A viscous
B non-viscous
C incompressible
D both a and b
If every particle of fluid has irregular flow then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
if every particle of fluid follow same path then flow is said to be
A laminar flow
B turbulent flow
C fluid flow
D both a and b
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out
105
Which of the following is a fluid
A helium B ice
C iron D gold
Which of the following is NOT a fluid A carbon dioxide
B hydrogen C seawater
D wood
Which of the following properties is NOT a characteristic of an ideal fluid A laminar flow B turbulent flow C nonviscous D incompressable
2 What is the fluid
3 What is the flow rate
4 Write the equation of continuity
5 Write the Bernoullis equation
6 The brain of a man is 05 m above his heart level The blood density ρ =10595
Kgm3What is the blood pressure difference between the brain and the heart
7 Blood flows in to one of an artery of 02Cm radius with a velocity o 3 ms and leaves the
other end of radius 01 Cm find the velocity of blood out