announcements
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Announcements. CAPA assignment #7 due Friday, October 7 at 10 PM CAPA assignment #8 is due Friday, October 14 at 10 PM Read Ch. 6 (Work & Energy): 6.1-6.6 Next week in Section, Assignment #5 : Work & Energy Print and bring to your Lab Section. - PowerPoint PPT PresentationTRANSCRIPT
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Announcements
• CAPA assignment #7 due Friday, October 7 at 10 PM
• CAPA assignment #8 is due Friday, October 14 at 10 PM
• Read Ch. 6 (Work & Energy): 6.1-6.6
• Next week in Section, Assignment #5: Work & Energy Print and bring to your Lab Section.
• Midterm Exam #2 on Tuesday, October 11
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Exam Room Assignments
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D. Ch. 5: Circular Motion; Gravity5-1. Kinematics of Uniform Circular Motion5-2. Dynamics of Uniform Circular Motion5-6. Newton’s Law of Universal Gravitation5-7. Gravity Near Earth’s Surface5-8. Satellites and “Weightlessness”5-9. Kepler’s Laws and Newton’s Synthesis.
A. Recall “Kinematical Equations” for 1D motion.
B. Ch. 3: Kinematics in 2D3-4. Adding Vectors by Components3-5 & 3-6. Projectile Motion & Solving Problems
C. Ch. 4: Newton’s Laws of Motion4-2. First Law4-3. Mass4-4. Second Law4-5. Third Law4-6. Weight & the Normal Force4-7. Solving Problems & Free-Body Diagrams4-8. Friction & Inclines
Exam #2 Topics
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Remember: A large part of what you’ll be expected to do is to apply Newton’s Laws. You can’t just memorize and apply formulas.
Steps for Linear Motion:
1. Draw a free-body diagram identifying and labeling all forces.2. Choose a coordinate system – typically with the forces pointing in the coordinate
directions (x,y) as much as possible.3. Write down Newton’s 2nd law in each coordinate direction (typically), summing
the forces. The equation perpendicular to the direction of motion often allows you to find the Normal Force, which is needed to determine the force of friction.
Σ Fx = max Σ Fy = may (for rectilinear motion)
If a force is not in a coordinate direction, you must find its components in the coordinate directions.
4. If there is more than one mass, then Newton’s 2nd law may be needed for each mass.
Exam #2 Topics
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Remember: A large part of what you’ll be expected to do is to apply Newton’s Laws. You can’t just memorize and apply formulas.
Steps for Uniform Circular Motion:
1. Draw a free-body diagram identifying and labeling all forces.2. Choose a coordinate system – one of the directions will point in the
radial direction. Others directions: tangent direction (T) or vertical (y).3. Write down Newton’s 2nd law in each coordinate direction (typically),
summing the forces. Σ FR = maR = mv2/r Σ FT = maT (for uniform circular motion)
4. If there is more than one mass, then Newton’s 2nd law may be needed for each mass.
Exam #2 Topics
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Incomplete Gallery of Problems Involving Newton’s Laws:
w & w/0 frictionw & w/0 friction
M m+x
F
fc
m
Box 1 Box 2
m
M
m
M1
M2
w & w/0 friction
F
T
w & w/0 gravity
Practice setting up the free-body diagram and Fnet = ma
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Incomplete Gallery of Problems Involving Newton’s Law of Universal Gravitation:
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What is Energy?
All kinds of things come to mind…Things in motion, collisions, electricity, stored energy
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Energy comes in many forms.Energy can be converted from one form to
another.• Kinetic energy (KE) - energy of motion• Thermal energy - energy of atomic vibrations• Potential energy (PE) - energy stored in configuration
- gravitational - electrostatic- elastic- chemical- nuclear
• Radiant (solar) energy – energy of light• Mass energy – Einstein’s E=mc2
What is Energy?
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Definition: Energy is the ability to do “Work”
Of course now we will need a physics definition of “Work”
Whenever work is being done, energy is being changed from one form to another or being transferred from one body to another.
The amount of work done on a system (or by a system) is the change in energy of the system.
Energy is such a useful concept because it is “Conserved”.
What is Energy?
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To understand energy, its inter-conversion to different forms,
and its conservation, we must first define some terms:
WorkKinetic Energy (KE)
“Work-Energy Principle”Potential Energy (PE)
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Consider an object moved by a force FWhile the force is applied, If the object moves a distance d in the direction of the force vector, then the work done is:
Force
|Dx| =d
ntDisplaceme Force W Work Work must have units of Newtons x meters = (kg m/s2) x m = kg m2/s2
We also call this Joules – the SI unit for work and also energy!
Work
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WorkForce
|Dx| =d
What if the force vector is at some angle with
respect to the displacement vector?
dF d )Fcos( d F W Work x
Component of force along the displacement
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Clicker Question Room Frequency BA
Atlas holds up the entire earth for 20 seconds.
What can you say about the work done by Atlas?
A)An enormous amount of work (many, many Joules)B) 20 JoulesC) None
There is no displacement of the earth.
Thus, despite the very large force (Mg) needed to hold up the earth, no work is done.
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Note: Work is a scalar (not a vector), but it does have a sign.
When the component of force in the direction of displacement points along the displacement, Work is positive.
When the force is perpendicular to the displacement, Work is zero.When the component of force in the direction of displacement points
opposite to displacement, Work is negative.
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Clicker Question Room Frequency BA
You are at the gym and observe this fellow lifting weights.
He lifts 10 kg at constant velocity upwards a distance of 0.1 meters.
What are the forces on the barbell in a free body diagram?
Mg
Flift
What is true about the work done by the fellow on the barbell (via Flift)?
A)Work > 0 B) Work < 0 C) Work = 0
What is true about the work done by the force of gravity on the barbell?
A)Work > 0 B) Work < 0 C) Work = 0
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Whenever you talk about “Work Done”, you have to be careful to specify
which force does the work!
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Consider an object displaced Δx = +1 m at a constant speed on a rough table by a constant external force F = 10 N.
N
Fg=mg
Work done by the external force =
Wext = +Fext Δx = (10 N)(1m) = +10 J
Work done by the friction force =
Wfric = -Ffric Δx = -(10 N)(1m) = -10 J
Work done by the normal force =
WN = (0) Δx = 0 J
Work done by gravityWg = (0) Dx = 0 J
Work done by the net force = Wnet = (0) Δx = 0 J
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A rock of mass m is twirled on a string in a horizontal plane.
The work done by the tension in the string on the rock is
The work done by the tension force is zero, because the force of the
tension in the string is perpendicular to the direction of the displacement:
W = F cos 90°= 0
TA) Positive B) NegativeC) Zero
Clicker Question Room Frequency BA
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The kinetic energy (KE) of an object of mass m moving with speed v is:
• KE is the energy of motion.• KE is always positive (KE >= 0).• KE and Work are related by the “Work-Energy Principle”• Units of KE
Kinetic Energy
2
21KE mv
22 ]/][[21KE smkgmv
mskgm
2 ]Joules[ Nm
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A bowling ball of mass m = 7.7 kg (17 lbs) is rolled down the lane with speed v = 7 m/s.
2
21KE mv 2)/7)(7.7(
21 smkg J 190