Mechanics I. 1

BIOPHYSICS – FALL SEMESTER

DENTISTRY PROGRAM

syllabus

CLASSICAL MECHANICS I.

Dr. Beáta Bugyi

associate professor

Andrea Vig

assistant professor

University of Pécs, Medical School

Department of Biophysics

2016

Mechanics I. 2

CLASSICAL MECHANICS I.

KINEMATICS: VELOCITY, ACCELERATION

DYNAMICS: NEWTON’S LAW OF MOTION, FORCE

LINEAR MOMENTUM

WORK, ENERGY

INTRODUCTION

Classical mechanics concerns with the physical laws, which describe the motion of bodies under the

action of a system of forces.

Classical mechanics includes:

Kinematics: study of motions of object without regard of its causes

Dynamics: study of motion of objects and its relation to forces

Statics: study of equilibrium of objects and its relation to forces

KINEMATICS

Kinematics describes the motion of objects without considering its causes.

LINEAR QUANTITIES DESCRIBING MOTION

For simplicity

we analyze the motion of objects without any change in direction, along a straight line (1D, linear

motion)

we assume that the objects are rigid bodies: an ideal solid body in which deformation is neglected,

i.e. the distance between any two points of the body remains constant in time regardless of

external forces acting on it.

Reference (observational) frame

physical concept: set of physical reference points that uniquely fix (locate and orient) the frame

and standardize measurements (the motion is described relative to the reference frame)

mathematical concept: an abstract coordinate system

1D line: 𝑥, 2D plane: 𝑥, 𝑦, 3D: 𝑥, 𝑦, 𝑧

Example

Describe the motion of the driver considering the car or the road as the reference frame.

Mechanics I. 3

(Elapsed) Time [𝑡] = 𝑠

change (∆) or the interval over which motion occurs

difference between the ending time and beginning time

∆𝑡 = 𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑛𝑖𝑡𝑎𝑙 = 𝑡 − 𝑡0

Position

where the object is in space at a particular time

relative position to a convenient reference frame (e.g. Earth, lecture room, blackboard, etc.)

1D line: 𝑥, 2D plane: 𝑥, 𝑦, 3D: 𝑥, 𝑦, 𝑧

Displacement [𝑥] = 𝑚

change (∆) in position of the object due to its movement relative to the reference frame

has both direction and magnitude (vector quantity1)

negative and positive displacement

∆𝑥 = 𝑥𝑓𝑖𝑛𝑎𝑙 − 𝑥𝑖𝑛𝑖𝑡𝑎𝑙 = 𝑥 − 𝑥0

Distance

magnitude of the displacement

Distance traveled

total path length traveled between two positions

distance traveled is not necessarily equals to the magnitude of the displacement

Velocity [𝑣] =𝑚

𝑠= 𝑚𝑠−1

the rate at which displacement changes, i.e. displacement divided by the time of travel

has both direction and magnitude (vector quantity)

can be caused by either a change in the magnitude or the direction of the displacement

negative and positive velocity

average velocity: displacement divided by time over which displacement occurs

o 𝑣𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =∆𝑥

∆𝑡=

𝑥−𝑥0

∆𝑡

instantaneous velocity: velocity at a specific instant, or the average velocity over an

infinitesimal time interval

ADVANCED

𝑣𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑛𝑜𝑢𝑠 = lim∆𝑡→0

∆𝑥

∆𝑡=

𝑑𝑥

𝑑𝑡

1 APPENDIX 1

Mechanics I. 4

Acceleration [𝑎] =𝑚

𝑠2= 𝑚𝑠−2

the rate at which velocity changes, i.e. change in velocity divided by the time

has both direction and magnitude (vector quantity)

can be caused by either a change in the magnitude or the direction of the velocity

average acceleration: change in velocity divided by time over which change occurs

o 𝑎𝑎𝑣𝑒𝑟𝑎𝑔𝑒 =∆𝑣

∆𝑡=

𝑣−𝑣0

∆𝑡

instantaneous acceleration: acceleration at a specific instant

Deceleration always refers to acceleration in the direction opposite to the direction of the

velocity. Deceleration always reduces speed. Negative acceleration, however, is acceleration in

the negative direction in the chosen coordinate system. Negative acceleration may or may not be

deceleration, and deceleration may or may not be considered negative acceleration.,

ADVANCED

𝑎𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑛𝑜𝑢𝑠 = lim∆𝑡→0

∆𝑣

∆𝑡=

𝑑𝑣

𝑑𝑡

Examples

Source: College Physics Page 82. Example 12.

The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on

the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long,

and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance? Give

the velocity in km/h.

𝑥 = 𝑥0 + 𝑣0𝑡 → 𝑡 =𝑥

𝑣𝑎= 0.0611 = 61 ∗ 10−3𝑠 = 61 𝑚𝑠

𝑣 = 18𝑚

𝑠= 18

10−3 𝑘𝑚

13600ℎ

= 64.8𝑘𝑚

ℎ

Source: College Physics Page 83. Example 26.

Blood is accelerated from rest to 30 cm/s in a distance of 1.8 cm at a constant

acceleration by the left ventricle of the heart. How long does the acceleration

take? Give the acceleration of the blood. Is it reasonable when compared with

the time for a heartbeat (consider 70 hbs in 1 min)? What is the blood’s

acceleration, expressed in multiples of g?

𝑣𝑎 =𝑣 + 𝑣0

2= 15

𝑐𝑚

𝑠

𝑥 = 𝑥0 + 𝑣0𝑡 → 𝑡 =𝑥

𝑣𝑎= 0.12 𝑠 = 120 𝑚𝑠

1 heartbeat = 60/70 = 0.85 s = 850 ms, reasonable

Mechanics I. 5

𝑣𝑎 = 𝑣0 +𝑎𝑎𝑡

2→ 𝑎 =

2𝑣𝑎𝑡

= 500𝑐𝑚

𝑠2= 5

𝑚

𝑠2

SUMMARY – MOTION EQUATIONS IN 1D

Elapsed time: ∆𝑡 = 𝑡𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑡 (𝑡𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 0) (Eq. 1)

Displacement: ∆𝑥 = 𝑥𝑓𝑖𝑛𝑎𝑙 − 𝑥𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑥 − 𝑥0 (Eq. 2)

Change in velocity: ∆𝑣 = 𝑣𝑓𝑖𝑛𝑎𝑙 − 𝑣𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑣 − 𝑣0 (Eq. 3)

Average velocity: 𝑣𝑎 =∆𝑥

∆𝑡=

𝑥−𝑥0

𝑡 (Eq. 4)

Average acceleration: 𝑎𝑎 =∆𝑣

∆𝑡=

𝑣−𝑣0

𝑡= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (Eq. 5)

Eq. 4 → final position:

𝒙(𝒕) = 𝒙𝟎 + 𝒗𝒂𝒕 (Eq. 6)

Eq. 5 → final velocity:

𝒗(𝒕) = 𝒗𝟎 + 𝒂𝒂𝒕 (Eq. 7)

If 𝑎𝑎 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 → 𝒗𝒂 =𝒗+𝒗𝟎

𝟐 (Eq. 8)

Eq. 7 & 6 →

𝒗 + 𝒗𝟎

𝟐=

𝟐𝒗𝟎 + 𝒂𝒂𝒕

𝟐

𝒗𝒂 = 𝒗𝟎 +𝒂𝒂𝒕

𝟐

𝒙(𝒕) = 𝒙𝟎 + 𝒗𝟎𝒕 +𝒂𝒂𝒕

𝟐

𝟐

Mechanics I. 6

DYNAMICS

Dynamics describes the motion of objects and its relation to forces.

NEWTON’S LAWS, FORCES

1st law (law of inertia)

A body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted

on by a net external force.

There must be a cause (net external force, force acts from outside the system of interest) for there to be

any change in velocity (either a change in magnitude or direction). If no net external force (Fnet = 0) acts

on a body its velocity cannot change, the object cannot accelerate.

What do we mean by a change in motion? A change in motion is equivalent to a change in velocity. A

change in velocity means that there is an acceleration. Newton’s 1st law says that a net external force

causes a change in motion; thus, a net external force causes acceleration.

net external force → change in motion ~ change in velocity (magnitude, direction) → acceleration

Inertia

The property of a body to remain at rest or to remain in motion with constant velocity is called inertia.

The resistance to any change in the state of motion. The inertia of an object is measured by its inertial

mass. As we know from experience, some objects have more inertia than others. It is obviously more

difficult to change the motion of a large truck than that of a basketball.

It is easy to show experimentally that acceleration is linearly proportional to the net external force…

𝑎~𝐹𝑛𝑒𝑡 and inversely proportional to the mass 𝑎~1

𝑚

Mechanics I. 7

2nd law

The acceleration of a system is directly proportional to and in the same direction as the net external

force acting on the system, and inversely proportional to its mass.

𝒂 =𝑭𝒏𝒆𝒕

𝒎 𝑭𝒏𝒆𝒕 = 𝒎𝒂

Unit of force: 𝑭𝒏𝒆𝒕 = 𝒎𝒂 → [𝐹] = 𝑘𝑔𝑚

𝑠2= 𝑁(𝑁𝑒𝑤𝑡𝑜𝑛)

ADVANCED

The net force on an object is equal to the rate of change (that is, the derivative) of its linear momentum p

in an inertial reference frame:

Original form: 𝐹𝑛𝑒𝑡 =𝑑(𝑚𝑣)

𝑑𝑡= 𝑚

𝑑𝑣

𝑑𝑡+

𝑑𝑚

𝑑𝑡𝑣

Limitations: theory of relativity 𝑚 =𝑚0

√1−𝑣2

𝑐2

3rd law (symmetry in forces, one of the fundamental symmetry principles of the universe, law of equal

and opposite reactions):

Whenever one body (A) exerts a force on a second body (B), the first body experiences a force that is

equal in magnitude and opposite in direction to the force that the first body exerts.

𝑭𝑨𝑩 = −𝑭𝑩𝑨

Example

How many times do you realize Newton’s laws on the videos?

http://video.mit.edu/watch/concept-vignettes-newtons-laws-28734/

https://www.youtube.com/watch?v=C_r5UJrxcck

Mechanics I. 8

LINEAR MOMENTUM

Linear momentum:

𝑝 = 𝑚�⃗�

[𝑝] = 𝑘𝑔𝑚

𝑠

Example

Source: College Physics Page 286. Example 8.1

Calculate the linear momentum of a 110 kg football player running at 8 m/s. Compare the player’s momentum

with the momentum of a hard-thrown 0.4 kg ball that has a speed of 25 m/s.

Linear momentum and Newton’s 2nd law:

�⃗� = 𝑚�⃗� = 𝑚∆𝑣⃗⃗ ⃗⃗ ⃗

∆𝑡=

∆𝑝⃗⃗⃗⃗⃗⃗

∆𝑡

Net external force equals to the change in the linear momentum of a system over the time during which

the change occurs.

CONSERVATION OF LINEAR MOMENTUM

Net force results in a change in the linear momentum of the moving object. If the net force is zero, there

is no change in the linear momentum, it is constant.

�⃗� =∆𝑝⃗⃗⃗⃗⃗⃗

∆𝑡= 0 → ∆𝑝⃗⃗⃗⃗⃗⃗ = 0 → 𝑝 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Example

Source: College Physics Page 310. Example 8.

A car moving at 10 m/s crashes into a tree and stops in 0.26 s. Calculate the force the seat belt exerts on a

passenger (m = 70 kg) in the car to bring him to halt.

𝐹 =∆𝑝

∆𝑡=

𝑚∆𝑣

∆𝑡= 2690𝑁

Source: College Physics Page 310. Example 12.

One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting

Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects,

such as flakes of paint. Calculate the force exerted by a 0.1 mg chip of paint that strikes a spacecraft window

at a relative speed of 4*103 m/s, given the collision lasts 6*10-8 s.

𝐹 =∆𝑝

∆𝑡=

𝑚∆𝑣

∆𝑡=

0.1 ∗ 10−3𝑘𝑔 ∗ 4 ∗ 103𝑚/𝑠

6 ∗ 10−8𝑠= 6.67 ∗ 106 𝑁

Mechanics I. 9

WORK, ENERGY

WORK

The work done on a system by a constant force (𝐹):

𝑊 = �⃗� ∗ 𝑑

magnitude: 𝑊 = 𝐹𝑑𝑐𝑜𝑠𝛼, where 𝐹𝑐𝑜𝑠𝛼 is the component of the force in the direction of the

motion and 𝑑 is the distance through which the force acts

[𝑊] = 𝑁𝑚 = 𝑘𝑔𝑚2

𝑠2= 𝐽 (Joule)

POWER

the rate at which the work is done

𝑃 =𝑊

∆𝑡

[𝑃] =𝐽

𝑠= 𝑊 (Watt)

1 horsepower = 735.5 W (needed to lift 75 kg - the average body mass of a person – by 1 meter

within 1 second)

James Watt determined that a horse could turn a mill wheel 144 times in an hour (or 2.4 times a

minute). The wheel was 3.65 meters in radius; therefore, the horse travelled 2.4·2π·3.65 m =

55.0128 m in one minute. Watt judged that the horse could pull with a force of 800.67 N.

𝑃~800 𝑁 ∗ 55𝑚

60 𝑠~733 𝑊

𝑃 =𝐹∆𝑑

∆𝑡= 𝐹𝑣

Examples

Source: College Physics Page 242. Example 7.1

Investigate how the relative direction of the force and displacement affects the work. (parallel: 𝑐𝑜𝑠0𝑜 = 1,

perpendicular: 𝑐𝑜𝑠90𝑜 = 0).

How much work is done on the lawn mower by the person if he exerts a constant force of 75 N at an angle 35º

below the horizontal and pushes the mower 25 m on level ground? Convert the amount of work from joules

to kilocalories and compare it with this person’s average daily intake of 10000kJ (about 2400 kcal) of food

energy. One calorie (1 cal) of heat is the amount required to warm 1 g of water by 1ºC, and is equivalent to

4.184 J, while one food calorie (1 kcal) is equivalent to 4184 J.

𝑊 = 𝐹𝑑𝑐𝑜𝑠𝛼 = 75 𝑁 ∗ 25 𝑚 ∗ 𝑐𝑜𝑠35𝑜 = 1536 𝐽

Mechanics I. 10

𝑊(𝑐𝑎𝑙) =1536 𝐽

4.184 𝐽= 367 𝑐𝑎𝑙 = 0.367 𝑘𝑐𝑎𝑙

The ratio of the work to the daily consumption:

0.367 𝑘𝑐𝑎𝑙

2400 𝑘𝑐𝑎𝑙= 1.53 ∗ 10−4 = 0.0153 %

This ratio is a tiny fraction of what the person consumes, but it is typical. Very little of the energy released

in the consumption of food is used to do work.

less than 10 % of our food energy intake is used to do work

more than 90 % is converted to thermal energy or stored as chemical energy in fat

Source: College Physics Page 283. Example 56.

The swimmer exerts an average horizontal backward force

of 80.0 N with his arm during each 1.80 m long stroke.

What is his work in each stroke? Calculate the power output

of his arms if he does 120 strokes per minute.

𝑊 = 𝐹 ∗ 𝑑 = 80 𝑁 ∗ 1.8 𝑚 = 144 𝐽

1 stroke is done in 0.5 seconds

𝑃 =𝑊

𝑡=

144 𝑊

0.5 𝑠= 288 𝑊

Work as the area under the F(d) curve.

Mechanics I. 11

Work as the area under the p(V) curve.

Assume constant force, which is parallel to the direction of movement.

𝑊 = 𝐹𝑑𝑐𝑜𝑠𝛼 = 𝐹𝑑 = 𝑝𝐴∆𝑑 = 𝑝∆𝑉

This holds generally, the work can be derived graphically from the area under the p(V) curve.

Example – work of the heart

Source: College Physics Page ?. Example ?.

Give the work done by the heart in one cardiac cycle, considering the p-V diagram of the left ventricle during

a single cardiac cycle.

Compare the amount of work done by the heart in 1 day with the person’s average daily intake of 10000 kJ

(about 2400 kcal) of food energy. One calorie (1 cal) of heat is the amount required to warm 1 g of water by

1ºC, and is equivalent to 4.184 J, while one food calorie (1 kcal) is equivalent to 4184 J.

Considering 70 heartbeats in 1 minute, give the power of the heart.

𝑊 = ∆𝑝∆𝑉 = (16 𝑘𝑃𝑎 − 1 𝑘𝑃𝑎) ∗ (140 𝑚𝐿 − 80 𝑚𝐿) ∗ 10−6𝑚3 = 0.9 𝐽

1 minute 70 heartbeat, 1 day 70*24*60 = 100800 heartbeat

𝑊 = 0.9 𝐽 ∗ 100800 = 90.72 𝑘𝐽

𝑊(𝑐𝑎𝑙) =90.72 𝑘𝐽

4.184 𝑘𝐽= 21.68 𝑘𝑐𝑎𝑙

The ratio of the work to the daily consumption:

21.68 𝑘𝑐𝑎𝑙

2400 𝑘𝑐𝑎𝑙= 9 ∗ 10−3 = 0.9 %

1 heartbeat 0.857 seconds

𝑃 =𝑊

𝑡=

0.9 𝐽

0.857 𝑠= 1.1 𝑊

Mechanics I. 12

ENERGY

Example

Source: College Physics Page 243. Example 7.3

A package is pushed horizontally with a

force of F = 120 N at a distance of d = 0.8 m.

Neglect the friction.

The work done by the force: 𝑊 = 𝐹𝑑𝑐𝑜𝑠𝛼 = 𝐹𝑑. According to Newton’s laws net force results in a change

in the velocity of the package changes from 𝑣0 to 𝑣, consequently the package accelerates. Using

Newton’s 2nd law, the work can be expressed as follows:

𝑊 = 𝐹𝑑 = 𝑚𝑎𝑑

Using the motion equations for constant acceleration 𝑎 =𝑣−𝑣0

𝑡 and 𝑑 =

𝑣+𝑣0

2𝑡, the following equation can

be derived:

𝑊 =1

2𝑚𝑣2 −

1

2𝑚𝑣0

2

The 𝑇𝐾𝐸 =1

2𝑚𝑣2 quantity is defined as the translational kinetic energy:

the energy of an object due to translational motion

mechanical form of energy

The above example illustrates that the work done by the net force gives to a system energy of motion

(translational kinetic energy).

WORK-ENERGY THEOREM

The work of the net forces on a point-like object equals to the change in its translational kinetic

energy:

𝑾 = ∆𝑻𝑲𝑬 =𝟏

𝟐𝒎𝒗𝟐 −

𝟏

𝟐𝒎𝒗𝟎

𝟐

Example

Source: College Physics Page ?. Example ?.

A package of 30 kg is lifted through a height of d = 1 m. The is work done against the gravitational force

𝑊 = 𝐹𝑑 = 𝑚𝑔ℎ

The 𝐺𝑃𝐸 = 𝑚𝑔ℎ is defined as the gravitational potential energy:

the energy of an object due to its position in a gravitational field

mechanical form of energy

The above example illustrates that the work done by the net force gives to a system energy due to the

change its position in the gravitational field (gravitational potential energy).

Mechanics I. 13

CONSERVATIVE FORCES

the work does not depend on the path, only on the starting and end points of the motion: 𝑾𝒄

e.g.: gravitational force

NONCONSERVATIVE FORCES

work depends on the path: 𝑾𝒏𝒄

e.g.: frictional force

changes mechanical energy into thermal energy

LAW OF CONSERVATION OF ENERGY

When only conservative forces act on a system the total mechanical energy of a system is constant:

𝑮𝑷𝑬 + 𝑻𝑲𝑬 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕

Total energy (gravitational potential energy, translational kinetic energy, other forms of energy) is

constant in any process; energy may change in form or be transferred from one system to another,

but the total remains the same:

𝑮𝑷𝑬𝟏 + 𝑻𝑲𝑬𝟏 +𝑾𝒏𝒄 +𝑶𝑬𝟏 = 𝑮𝑷𝑬𝟐 + 𝑻𝑲𝑬𝟐 +𝑶𝑬𝟐

Example

Source: College Physics Page 249. Example 7.6

A 60 kg person jumps onto the floor from a height of 3 m. Give the kinetic energy just before landing. If he

lands stiffly with his knee joints compressing by 0.5 cm, calculate the force on the knee joints.

The person loses his GPE upon fall, while he gains TKE. The kinetic energy of the person upon reaching the

floor is:

𝑊 = ∆𝐾𝐸=−𝑚𝑔ℎ = −60𝑘𝑔 ∗ 𝑔 ∗ (0 𝑚 − 3 𝑚) = 1766 𝐽

The floor stops the person within the distance of the bending of the knee joints, the work done:

𝑊 = −𝐹𝑑

The floor brings his TKE to zero, the work done:

𝑊 = −∆𝐾𝐸 = 𝑚𝑔ℎ

𝑚𝑔ℎ = −𝐹𝑑 → 𝐹 = −𝑚𝑔ℎ

𝑑=

60𝑘𝑔 ∗ 𝑔 ∗ 3𝑚

5 ∗ 10−3𝑚= 353000 𝑁 = 3.53 ∗ 105𝑁

This large force over a short period of time is enough to break bones. A better way of landing is to bend

the legs or rolling on the ground, thereby increasing the time/distance over which the force acts.

As example bending motion of 0.5 m results in a force 100 times smaller.

Mechanics I. 14

ENERGY CONVERSION IN HUMANS

According to the conservation of energy, the chemical

energy stored in food is converted to work, thermal

energy and stored as chemical energy in fatty tissue. The

fraction of each energy form depends on how much we

eat and on the level of our physical activity.

Metabolic rate (MR)

characteristic to the energy conversion rate of a person

the rate at which our body uses energy to sustain life and do different activities.

Basal metabolic rate (BMR): energy conversion rate at rest

ORGAN POWER CONSUMED

AT REST (W) OXYGEN CONSUMPTION

(mL/min) BMR (%)

liver, spleen 23 67 27

brain 16 47 19

skeletal muscle 15 45 18

kidney 9 26 10

heart 6 17 7

other 16 48 19

total 85 250 100

Energy consumption is directly proportional to

oxygen consumption because the digestive process

is basically one of oxidizing food. We can measure

the energy people use during various activities by

measuring their oxygen use. Approximately 20 kJ of

energy are produced for each liter of oxygen

consumed, independent of the type of food. The

amount of oxygen in blood can be measured by a

pulse oximeter.

ACTIVITY POWER CONSUMED AT REST

(W) OXYGEN CONSUMPTION (L/min)

sleeping 83 0.24

walking (5 km/h) 280 0.8

cycling (13 -18 km/h) 400 1.14

shivering 425 1.21

swimming breaststroke 475 1.36

running cross country 740 2.12

sprinting 2415 6.90

Mechanics I. 15

Examples

Source: College Physics Page 269. Example 7.13

If a person who normally requires an average of 10000 kJ (2400 kcal) of food energy per day consumes 11000

kJ per day, he will steadily gain weight. How much bicycling per day is required to work off this extra 1000 kJ?

Consider that 400 w is used when cycling at a moderate rate.

𝑃 =𝑊

𝑡→ 𝑡 =

𝑊

𝑃=

1000 𝑘𝐽

400 𝑊= 2500𝑠 = 42 𝑚𝑖𝑛

Source: College Physics Page 269. Example 7.13

If this person uses more energy than he or she consumes, the person’s body will obtain the needed energy by

metabolizing body fat. Give the amount of fat loss if the person uses 11000 kJ but consumes only 10000 kJ,

assuming the energy content of fat to be 39 kJ/g.

𝑓𝑎𝑡 𝑙𝑜𝑠𝑠 = 1000 𝑘𝐽1 𝑔

39 𝑘𝐽= 25.6 𝑔

Source: College Physics Page ?. Example ?.

Energy that is not utilized for work or heat transfer is converted to the chemical energy of body fat containing

about 39 kJ/g. How many grams of fat will you gain if you eat 10000 kJ (about 2400 kcal) corresponding to

the average daily consumption. Give the amount of fat loss due to 8 h shivering (P = 425 W), or due to sleeping

(P = 83 W).

𝑓𝑎𝑡 𝑔𝑎𝑖𝑛 = 10000 𝑘𝐽1 𝑔

39 𝑘𝐽= 256 𝑔

𝑃 =𝑊

𝑡→ 𝑊 = 𝑃𝑡 = 425 𝑊 ∗ 8 ℎ ∗ 60 ∗ 60 = 12240 𝑘𝐽

𝑓𝑎𝑡 𝑙𝑜𝑠𝑠 = 12240 𝑘𝐽1 𝑔

39 𝑘𝐽= 313.8 𝑔

𝑃 =𝑊

𝑡→ 𝑊 = 𝑃𝑡 = 83 𝑊 ∗ 8 ℎ ∗ 60 ∗ 60 = 2390 𝑘𝐽

𝑓𝑎𝑡 𝑙𝑜𝑠𝑠 = 2390 𝑘𝐽1 𝑔

39 𝑘𝐽= 61.3 𝑔

Mechanics I. 16

APPENDIX 1

SCALAR AND VECTOR QUANTITIES

We measure the mass of an object to be 60 kilograms (kg). In this case, the physical quantity is the mass

and its value – or magnitude – is 60 kg; composed of a number (60) and a unit (kg). Note that without

knowing the exact unit the physical quantity is meaningless! These kind of physical quantities –

composed of a number and a unit – are called scalar quantities. 60 kg contain all the information about

the mass of the object. Let’s see another example. A particle moved 10 meters (m) from a given place.

Where is the particle at the end of its movement? Even if we know the magnitude of the displacement,

we do not know the exact new position of the particle. In this case, the magnitude of the physical quantity

does not contain all the information about the movement of the particle. What else do we have to know?

The direction of the movement in space. These kind of physical quantities – composed of a number, a

unit and a direction – are called vector quantities.

SCALAR AND VECTOR QUANTITIES

quantity

(notation) property how to treat examples

scalar

𝒂

only magnitude, no

direction ordinary algebra

mass, temperature, time,

energy, work, refractive

index, charge, pressure,

volume, area…

vector

�⃗⃗⃗�

both magnitude and

direction vector algebra

displacement, velocity,

acceleration, force, linear

momentum, torque,

angular momentum, …

REPRESENTATION OF VECTORS, OPERATIONS WITH VECTORS

Representation of vector quantities

�⃗� −�⃗�

Mechanics I. 17

�⃗� �⃗⃗�

�⃗� + �⃗⃗� = 𝑐 �⃗� + −�⃗⃗� = �⃗� − �⃗⃗� = 𝑐

�⃗�

−�⃗⃗�

�⃗�

𝑥

𝛼

𝑎𝑥⃗⃗⃗⃗⃗

𝑎𝑦⃗⃗ ⃗⃗⃗

𝑦

Components of a vector

Horizontal component: 𝑎𝑥 = 𝑎 ∙ 𝑐𝑜𝑠𝛼

Vertical component: 𝑎𝑦 = 𝑎 ∙ 𝑠𝑖𝑛𝛼

Magnitude of the resultant (Pythagorean theorem):√𝑎𝑥2 + 𝑎𝑦

2 = 𝑎

Direction of the resultant: 𝑎𝑦

𝑎𝑥= 𝑡𝑎𝑛𝛼

Adding and subtracting vectors

Graphical method (head-to-tail method)

�⃗� + �⃗⃗� = 𝑐 (vector sum, resultant)

�⃗� + �⃗⃗� = �⃗⃗� + �⃗� (commutative law)

�⃗� + �⃗⃗� + 𝑐 = �⃗� + (�⃗⃗� + 𝑐) (associative law)

�⃗� + −�⃗⃗� = �⃗� − �⃗⃗� (vector subtraction)

Mechanics I. 18

Analytical method (see the components of a vector section)

𝑅 = √𝑅𝑥2 + 𝑅𝑦

2

𝑅𝑦

𝑅𝑥= 𝑡𝑎𝑛𝛼

Multiplying vectors

MULTIPLYING VECTORS

Multiplying a vector by a scalar vector

𝑠 ∙ �⃗� = �⃗⃗�

magnitude:

scalar times the magnitude of the vector

direction:

same (positive scalar)

opposite (negative scalar)

force: �⃗� = 𝑚 ∙ �⃗�

linear momentum: 𝑝 = 𝑚 ∙ �⃗�

electric force law: �⃗� = 𝑄 ∙ �⃗⃗�

Multiplying a vector by a vector – scalar product (dot product) scalar

�⃗� ∙ �⃗⃗� = 𝑐

magnitude:

𝑎 ∙ 𝑏 ∙ 𝑐𝑜𝑠𝛼

direction:

no

work: 𝑊 = �⃗� ∙ 𝑠

electric potential: 𝑉 = �⃗⃗� ∙ 𝑠

Multiplying a vector by a vector – vector product (cross product) vector

�⃗� × �⃗⃗� = 𝑐

magnitude:

𝑎 ∙ 𝑏 ∙ 𝑠𝑖𝑛𝛼

direction:

torque: 𝜏 = 𝑟 × �⃗�

angular momentum: 𝑙 = 𝑟 × 𝑝

magnetic force law: �⃗� = 𝑄 ∙ �⃗� × �⃗⃗�

Mechanics I. 19

forefinger

�⃗⃗�

middle finger 𝑐

thumb

�⃗�

right-hand rule

Right-hand rule.

Examples

Source: College Physics Page 171. Example 41.

Two muscles in the back of the leg pull upward on the Achilles tendon (the medial

and lateral heads of the gastrocnemius muscle). Find the magnitude and direction

of the total force on the Achilles tendon.

𝑐𝑜𝑠200 =𝐹𝐴𝐹

→ 𝐹𝐴 = 200 𝑁 ∗ 𝑐𝑜𝑠200 = 188 𝑁

Source: College Physics Page 342. Example 30.

A person working at a drafting board may hold her head as shown in Figure, requiring

muscle action to support the head. The three major acting forces are shown. Calculate

the direction and magnitude of the force supplied by the upper vertebrae (Fv) to hold the

head stationary, assuming that this force acts along a line through the center of mass

as do the weight and muscle force.

Vertical components: 𝐹𝑣 = 50 𝑁 + 60𝑁 ∗ 𝑠𝑖𝑛330 = 82.6 𝑁

Horizontal components: 𝐹ℎ = 60𝑁 ∗ 𝑐𝑜𝑠330 = 50.32 𝑁

Magnitude of the resultant: 𝐹 = √𝐹𝑣2 + 𝐹ℎ

2 = 96.7 𝑁

Direction of the resultant: 𝑡𝑎𝑛𝛼 =𝐹𝑣

𝐹ℎ→ 𝛼 = 58.6𝑜

Mechanics I. 20

Source: College Physics Page 170. Example 33.

Braces are used to apply forces to teeth to realign them. The total force applied to

the tooth by the wire, points straight toward the back of the mouth. What force is

exerted on the tooth if the tension in the wire is T = 25 N? Note that the force applied

to the tooth is smaller than the tension in the wire, but this is necessitated by

practical considerations of how force can be applied in the mouth.

𝑐𝑜𝑠750 =𝐹𝑎2𝑇

→ 𝐹𝑎 = 2 ∗ 25 𝑁 ∗ 𝑐𝑜𝑠750 = 2 ∗ 6.5 𝑁

Source: College Physics Page 342. Example 28.

The upper leg muscle (quadriceps) exerts a force of 1250 N, which is carried by a

tendon over the kneecap (the patella) at the angles shown in Figure. Find the

direction and magnitude of the force exerted by the kneecap on the upper leg

bone (the femur).

Vertical components: 𝐹𝑣 = 1250 𝑁 ∗ 𝑠𝑖𝑛550 − 1250 𝑁 ∗ 𝑠𝑖𝑛750 = −183 𝑁

Horizontal components: 𝐹ℎ = 1250 𝑁 ∗ 𝑐𝑜𝑠550 + 1250 𝑁 ∗ 𝑐𝑜𝑠750 =

1040 𝑁

Magnitude of the resultant: 𝐹 = √𝐹𝑣2 + 𝐹ℎ

2 = 1560 𝑁

Direction of the resultant: 𝑡𝑎𝑛𝛼 =𝐹𝑣

𝐹ℎ→ 𝛼 = 9.98𝑜

CREDIT AND REFERENCE

https://openstax.org/details/college-physics