newtons laws and energy

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Newton’s laws

Although he did not know it at the time, Isaac Newton’s

work in the 17th century signalled an end to the transition

in the way the world was conceived and understood. The

transition was begun by Copernicus and Galileo, but Newton was

able to use mathematics to develop laws and theories that could

account for the motion of the heavens. These showed the universe

to be a mechanism that could readily be understood, one that was

regulated by simple natural laws. The universe taught by Aristotle,

and accepted up until the time of Newton, was one in which objectswere classified into categories and their motion depended on the

category to which they belonged.

Isaac Newton was born in rural England in 1642, the year Galileo

died. He so impressed his mentors that he was made Professor of

Mathematics at Cambridge University at the age of 26. His interests

spanned light and optics, mathematics (he was one of the inventors

of calculus), astronomy and the study of mechanics. His greatest

achievement was the formulation of the law of universal gravitation.

This, along with a complete explanation of the laws that govern

motion, is laid out in his book Philosophiae Naturalis Principia

Mathematica (Mathematical Principles of Natural Philosophy),

which was published in 1687. The Principia is one of the most

influential publications in natural science. Newton’s framework for

understanding the universe remained intact right up to the advent

of Einstein’s relativity more than 200 years later.

Newton died in 1727 a famous man. He remained self-critical and

shy. That he could ‘see so far’ was only because he was ‘able to

stand on the shoulders of giants’. In saying this, he was referring to

the work of Galileo, Copernicus and many others who had paved the

way for his discoveries.

5

CHAPTER

you will have covered material from the study of

movement including:

• vector techniques in two dimensions

• forces in two dimensions

• Newton’s laws of motion

• problems in mechanics including weight and friction.

by the end of this chapter



Chapter 5 Newton’s laws

5.1

14

Force as a vector

The previous chapter developed the concepts and ideas needed to describe

the motion of a moving body. This branch of mechanics is called kinematics.

In this chapter, rather than simply describe the motion, we will consider

the forces that cause the motion to occur. Treating motion in this way falls

within the branch of mechanics called dynamics. In simple terms, a force

can be thought of as a push or a pull, but forces exist in a wide variety ofsituations in our daily lives and are fundamental to the nature of matter

and the structure of the universe. Consider each of the photographs in

Figure 5.1 and identify each force—push or pull—that is acting.

In each of the situations depicted in Figure 5.1, forces are acting. Some

are applied directly to an object and some act on a body without touching

it. Forces that act directly on a body are called contact forces, because the

body will only experience the force while contact is maintained. Forces that

act on a body at a distance are non-contact forces.

Contact forces are the easiest to understand and include the simple

pushes and pulls that are experienced daily in people’s lives. Examples of

these include the forces between colliding billiard balls, the force that you

exert on a light switch to turn it on, and the forces that act between you and

your chair as you sit reading this book. Friction and drag forces are other

contact forces that you should be familiar with.

Non-contact forces occur when the object causing the push or pull is

physically separated from the object that experiences the force. These forces

are said to ‘act at a distance’. Gravitation, magnetic and electric forces areexamples of non-contact forces.

The action of a force is usually recognised through its effect on an object

or body. A force may do one or more of a number of things to the object.

It may change its shape, change its speed or change only the direction of its

motion. The tennis racquet in Figure 5.1a has applied a force to the tennis

ball, and, as a consequence, the speed of the ball changes along with its

direction. The ball also changes shape while the force acts!

Figure5.1 (a) At the moment of impact, both the tennis ball and racquet strings aredistorted by the forces acting at this instant. (b) The rock climber is relying on the frictionalforce between his hands and feet and the rock face. (c) A continual force causes the clay todeform into the required shape. (d) The gravitational force between the Earth and the Moon isresponsible for two high tides each day. (e) The globe is suspended in mid-air because of themagnetic forces of repulsion and attraction.

(c)(b)

(d)

(e)

(a)



Motion144

The amount of force acting can be measured using the SI unit called

the newton, which is given the symbol N. The unit, which will be defined

later in the chapter, honours Sir Isaac Newton (1642–1727), who is still

considered to be one of the most significant physicists to have lived. A force

of one newton, 1 N, is approximately the force you have to exert when

holding a 100 g mass against the downward pull of gravity. In everyday

life this is about the same as holding a small apple. Table 5.1 provides acomparison of the magnitude of some forces.

Table 5.1 A comparison of the magnitude of various forces

Force Magnitude (N)

Force on the electron in a hydrogen atom 10−7

Holding a small apple against gravity 1

Opening a door 10

Pedalling a bicycle 300

Thrust of a Boeing 747 at take-off 106

Gravitational force between the Earth and the Sun 1022

Force: a vector quantityIn Chapter 4, quantities associated with motion were classified as being

either vectors or scalars. Scalar quantities such as time and mass do not

have a direction. Only their size or ‘magnitude’ should be given. Quantities

that require a direction as well as a magnitude are called vectors. Force

is a vector quantity because the direction in which a force acts is always

significant. In this text, vectors are set in bold italics.

If a question only requires the magnitude of a vector, the direction can be

ignored. In this text, italics will be used to show this.

In a diagram, a force is usually shown as an arrow whose length

represents the magnitude of the force and whose direction is indicated by

the arrow.

Consider the case of a soccer player who kicks the ball horizontally with

a force of 95 N towards the east. The horizontal forces acting on the ball can

be illustrated by a vector diagram as shown in Figure 5.3.

If there are two or more forces acting on the same object, these forces can

be shown on the same diagram. If one force is larger, it should be represented

by a longer vector. If, for example, the soccer ball just discussed was sitting

in thick mud so that a frictional force of 20 N towards the west was acting

as it was kicked, this could be represented as shown in Figure 5.4.

The subsequent motion of the soccer ball will be different in the two

situations described above. When there is a large frictional force acting on

the ball, its speed will be significantly reduced. The muddy ground will

act to make the ball travel more slowly as it leaves the boot. To analyse the

horizontal motion of the ball, it is necessary to add all the horizontal forces

FORC… is measured in newtons (N) and is a vector quantity. It requires a

magnitude and a direction to describe it fully.

i

Figure 5.2 The netball will only go throughthe hoop if a force of the right magnitude anddirection is applied. Force is a vector; it can onlybe completely specified if both the direction andmagnitude are given.

Figure 5.3 When drawing force diagrams, it isimportant that the force is shown to be acting atthe correct location. In this example, the force onthe ball acts at the point of contact between theball and the foot.

W E

95 N



14Chapter 5 Newton’s laws

that are acting on it at this instant. The ball is simply treated as a point mass

located at its centre of mass.

If more than one force acts on a body at the same time, the body behaves

as if only one force—the vector sum of all the forces—is acting. The vector

sum of the forces is called the resultant or net force, Σ F (shown as a double-

headed arrow).

Because force is a vector quantity, the addition of a number of forces

must be undertaken with the directions of the individual forces in mind.

Vector addition is shown in Figure 5.5.

95 N+

20 N

=95 N

20 NΣ F = 75 N

Figure5.5 When the forces (95 N acting towards the east and 20 N acting to the west) areadded, the resultant or net force is 75 N towards the east. The ball will move as though thisresultant force is the only force acting on it.

If the forces that are acting are perpendicular (or any other angle) to

each other, the resultant force must still be found by performing a vector

addition. Consider the example of a shopping trolley that is being

simultaneously pushed from behind by one person and pushed from the

side by another. This situation is illustrated in Figure 5.6.

To find the magnitude of the resultant force, Pythagoras’s theorem must

be used:

ΣF = √802 + 602 = √10 000 = 100 N

60 N

80 N

North

80 N + 60 N =

80 N

View from above

Σ F = 100 N 60 N

Person 1

Person 2(a)

(b)

Figure5.6 (a) Two perpendicular forces are acting on the trolley. (b) The vector additionof these two forces gives the resultant force (ΣF ) that is acting on the trolley to be 100 N at127°T. The trolley is treated as a point mass located at its centre of mass.

The N…T FORC… acting on a body experiencing a number of forces actingsimultaneously is given by the vector sum of all the individual forces acting:ΣF = F

1 + F

2 + ... + F

n

i

Figure5.4 The two forces being consideredare acting at different locations and havedifferent strengths. The larger force is shownas a longer vector.

W E

20 N

95 N

Remember, when adding vectors, the

tail of the second vector is placed at

the head of the first. The resultant

vector is from the tail of the first

vector to the head of the second. A full

explanation of one-dimensional and two-

dimensional vector addition is included

in Appendix A.

Physics file



Motion146

To find the direction of the resultant force, trigonometry must be used:

tan θ = 60

80 = 0.75

θ = 37°

This is a direction of 37° south of east, which is equivalent to a bearing

of 127°T.

Hence, the net force acting on the trolley is 100 N at a bearing of 127°T.

Worked example 5.1A

A motorboat is being driven west along the Yarra River. The engine is providing a driving force

of 560 N towards the west. A frictional force of 180 N from the water and a drag force of 60

N from the air are acting towards the east as the boat travels along.

a Draw a force diagram showing the horizontal forces of this situation.

b Determine the resultant force acting on the motorboat.

Solution

a

b For the vector addition, treat the boat as a point mass located at its centre of mass.

The resultant or net force acting on the motorboat is ΣF = 320 N due west.

Worked example 5.1B

While playing at the beach, Sally and Ken kick a stationary beachball simultaneously

with forces of 100 N south and 150 N west respectively. The ball moves as if it were only

subjected to the net force. In what direction will it travel, and what is the magnitude of the

net force on the ball?

Solution

The net force is found by treating the beachball as a point mass and is given by:

ΣF = F Sally + F Ken

Figure 5.7 The golf ball moves in the direction ofthe applied force and is in the direction of the line

joining the centre of the club-head with the centreof the ball. The force will be very large.

60 N

180 N 560 N

W E

F 560 N

= +180 N 60 N

+

560 N

180 N 60 N F = 320 N=

Σ

Σ

100 N +150 N

= Σ F

100 N

150 N

There are two methods for describing

the direction of a vector in a two-

dimensional plane. In each case, the

direction has to be referenced to a

known direction.

A full circle bearing describes north

as ‘zero degrees true’—written as 0°T. In

this convention, all directions are given

as a clockwise angle from north. 90°T is

90° clockwise from north, i.e. due east.

A force acting in a direction 220°T is

acting in a direction of 220° clockwise

from north. This is the method most

commonly used in industry.

An alternative method is to provide

a quadrant bearing , where all angles

are between 0° and 90° and so lie within

one quadrant. The particular quadrant is

identified using two cardinal directions,

the first being either north or south.

In this method, 220°T becomes S40° W,

literally ‘40° west of south’.

220°Tor S40°W

F

N

W E

S

Figure5.8 The direction 220°T lies 220° clockwise from north. This direction can alsobe written as S40°W meaning 40° west ofsouth.

Physics file




ΣF = √1002 + 1502 = 180 N

tan θ = 150

100 = 1.5

θ = 56°

This is a quadrant bearing of 56° west of south, which is equivalent to a true bearing of

236°T. Hence, the net force acting on the beachball is 180 N in the direction 236°T.

Vector components

It is often helpful to divide a force acting in a two-dimensional plane into

two vectors. These two vectors are called the components of the force. This

can be done because the force can be considered to act in each of the two

directions at once. Consider, for example, the pulling force of 45 N acting

on the cart shown in Figure 5.9.

This pulling force is acting through the rope and is known as tension

or a tensile force. The force is acting at an angle of 20° to the horizontal, so

it has some effect in the horizontal direction and some effect in the vertical

direction. The amounts of force acting in each direction are the components

of the force.It is usual to construct a right-angled triangle around the force vector.

The force vector is the hypotenuse of the triangle, and the adjacent and

opposite sides become the components of the force. The horizontal and

vertical components of the pulling force can then be determined using

trigonometry. It is important to remember that there is only one pulling

force acting on the cart, but this force can be treated as two component

forces.

So, the cart will move as though a horizontal force of 42 N pulling the

cart along and a vertical pulling force of 15 N upwards were acting on

it simultaneously. When the components are added together, the original

45 N force is the resultant force.

Is this the most effective way of using a 45 N force to move the cart

forwards? No, it would be slightly more effective if the 45 N force was

acting in the horizontal direction. This would make the cart travel faster,

but it may be impractical or inconvenient to apply the force in this way.

Worked example 5.1C

A stationary hockey ball is struck with a force of 100 N in the direction N30°W. What are the

northerly and westerly components of this force?

Solution

Figure5.9 The pulling force acting on the carthas a component in the horizontal directionand a component in the vertical direction.

Figure5.10 The magnitudes of the vectorcomponents F

h and F

v can be calculated using

trigonometry.

F v = 45 sin 20° = 15 N

F h = 45 cos 20° = 42 N

20°

45 N

30

30

N

S

E

W

westerly component ofthe force = 50 N

northerlycomponent of

the force = 87 Nforce fromthe hockey

stick = 100 N

F = 45 N

20°



Motion148

F W

= 100sin30° = 50 N

F N = 100cos30° = 87 N

The ball moves as though forces of 50 N west and 87 N north were acting on it simultaneously.

Worked example 5.1D

When walking, a person’s foot pushes backwards and downwards at the same time. While

playing basketball, Kate’s foot pushes back along the court with a force of 400 N, and downwith a force of 600 N. What is the actual force applied by Kate’s foot?

Solution

400 N horizontally and 600 N vertically downwards are the components of the force

supplied by Kate’s foot. Therefore, the force she supplies will be F = F horizontal

+ F vertical

and a

vector diagram is needed.

Using Pythagoras’ theorem:

F = √F h

2 + F v

2 = √4002 + 6002 = √520 000 = 721 N

θ = tan−1 600

400 = tan−11.5 = 56°

So Kate supplies a force of 721 N backwards at 56° down from the horizontal.

F v = 600 N down F

F h = 400 N

• A force is a push or a pull. Some forces act on contact

while others can act at a distance.

• Force is a vector quantity whose SI unit is the

newton (N).

• A vector can be represented by a directed line

segment whose length represents the magnitude of

the vector and whose arrowhead gives the direction

of the vector.

• The net force acting on a body that experiences a

number of forces acting simultaneously is given by

the vector sum of all the individual forces acting:

Σ F = F 1 + F

2 + … + F

n

• A vector addition may be calculated using a

sketch vector diagram that can be solved using

trigonometry.

• A force F acting at an angle θ to a given direction

will have components Fcos θ parallel to the reference

direction, and Fsin θ perpendicular to that reference

direction.

Force as a vector5.1 summary




11 Which one or more of the following quantities are

vectors?

A mass B velocity

CC temperature D force

22 Calculate the resultant force in each of the followingvector additions:

a 200 N up and 50 N down

b 65 N west and 25 N east

c 10 N north and 10 N south

d 10 N north and 10 N west

33 If the force you have to exert when holding a small

apple is about 1 N and holding a kilogram of sugar is

10 N, estimate the force required for:

a using a stapler

b kicking a beachball

c lifting your school bagd doing a chin-up exercise.

44 Which one or more of the following directions are

identical?

A 40°T and S40°E B 140°T and S40°E

C 200°T and S20°W D 280°T and N80°W

55 Convert the following into full circle bearings (i.e. °T).

a N60°E b N40°W

c S60°W d SE

e NNE

66 Use the vectors below to determine the forces rep-

resented in the following situations.Scale: 1 cm represents 20 N

77 Use trigonometry if necessary to add the following

forces:

a 3 N east and 4 N west

b 60 N east and 80 N south

88 A small car is pulled by two people using ropes. Eachperson supplies a force of 400 N at an angle of 40°

to the direction in which the car travels. What is the

total force applied to the car?

40°40°

400 N

400 N

99 Resolve the following forces into their perpendicular

components around the north–south line. In part d,

use the horizontal and vertical directions.

a 100 N south 60° east

b 60 N north

c 300 N 160°T

d 3.0 × 105 N 30° upward from the horizontal

110 What are the horizontal and vertical components of

a 300 N force that is applied along a rope at 60° to the

horizontal used to drag a Christmas tree across the

backyard?

Force as a vector5.1 questions

a

F

b

F

c

F

Worked Solutions



Motion150

Newton’s first law of motion5.2

Aristotle and GalileoThe first attempt to explain why bodies move as they do was made more

than 2000 years ago by the Greek philosopher Aristotle. As discussed in

Chapter 4, Aristotle and his followers felt that there was a natural state

for matter and that all matter would always tend towards its natural placewhere it would be at rest. Aristotle’s thesis was based on the everyday

observation that a moving body will always slow down and come to rest

unless a force is continually applied. Try giving this book a (gentle) push

along a table top and see what happens.

Aristotle’s ideas were an attempt to explain the motion of a body as it was

seen, but they do not help to explain why a body moves as it does. It was

not until the early 17th century that Galileo Galilei was able to explain

things more fully. Galileo performed experiments that led him to conclude

that the natural state of a moving body is not at rest. Significantly, Galileo

introduced the idea that friction was a force that, like other forces, could

be added to other forces. A generation later, Newton developed Galileo’s

ideas further to produce what we now call the first law of motion.To understand Newton’s first law, follow the logic of this thought

experiment, similar to one used by Galileo. Consider a steel ball and a

smooth length of track. In Figure 5.11a, the ball is held at one end of an

elevated track; the other end of the track is also elevated. When the ball is

released, it will roll downhill, then along the horizontal section and then up

the elevated section. In reality, it will not quite reach the height it started at,

due to friction.

Now, imagine there were no friction. The ball, in this ideal case, would

slow down as it rolled uphill and finally come to rest when it reached the

height at which it started.

Now consider what would happen if the angle of the elevated sectionwas made smaller as in Figure 5.11b. The ball would now roll further along

the track before coming to rest because the track is not as steep. If we could

again imagine zero friction, the ball would again slow down as it rolls

uphill and reach the same height before stopping, but travelling further in

the process.

What would happen if we made the angle of this elevated section

progressively smaller? Logically, we would expect that, if we had enough

track, the ball would travel even further before stopping.

Now consider Figure 5.11c. Here the end of the track is not elevated at

all. As the ball rolls along, there is nothing to slow it down because it is not

going uphill. If we can ignore friction, what will the ball do as it rolls along

the horizontal track? Galileo reasoned that it would not speed up, norwould it slow down. Ideally, the ball should keep travelling horizontally

with constant speed and never reach its starting height. According to

Galileo, the natural state of a body was to keep doing what it was doing. This

tendency of objects to maintain their original motion is known as inertia.

As the ball travels along the horizontal track, there is no driving force,

nor is there any retarding force acting. The net force on the ball is zero and

so it keeps moving with a constant velocity. This is the breakthrough in

Figure 5.11 Galileo used a thought experimentto derive his law of inertia. This became Newton’sfirst law and stated that the natural state of

bodies was to maintain their original motion.This contradicted Aristotle’s idea that the naturalstate of bodies was at rest.

Ideally you would expect theball to reach this height.

Ball ideally would roll further

and reach this position.

The ball ideally will keep

moving with a constant velocity.

(a)

(b)

(c)

Interactive




understanding that Newton was able to make. Any body will continue

with constant velocity if zero net force (Σ F = 0) acts upon it.

A good example of inertia and Newton’s first law is illustrated by the

air-track. With the air turned on, give a glider a gentle push along the

track. It will travel along the track with a constant velocity as described by

Newton’s first law. There are no driving or retarding forces acting on the

glider, so it simply maintains its original motion. Aristotle’s laws would

not be able to explain the motion of the glider.

The motion of a spacecraft cruising in deep space is another good

example of a body moving with constant velocity as required by Newton’s

first law. As there is no gravitational force, and no air in space to retard its

motion, the spacecraft will continue with constant speed in a straight line.

The absence of air explains why there is no need to make a space probe

aerodynamic in shape.

N…WTON’S FIRST LAW OF MOTION states that a body will either remain at rest orcontinue with constant speed in a straight line (i.e. constant velocity) unless itis acted on by an unbalanced force.

i

At the time of the Roman Empire some

2000 years ago, it cost as much money

to have a bag of wheat moved 100 km

across land as it did to transport it

across the whole Mediterranean Sea. One

of the reasons for this stemmed from

the enormous friction that acted between

the wheel and the axle in the cart of

the day. Some animal fats were used

as crude lubricants, but the effect was

minimal. It has only been during the last

century that engineering has provided a

mechanical solution.

Today’s wheels are connected to the

axle by a wheel bearing. An outer ring

is attached to the wheel, and an inner

ring is attached to the axle. Separating

the rings are a number of small ball

bearings, which are able to roll freely

between the rings. In this way, the area

of contact and the friction between the

wheel and axle is reduced dramatically.

Figure5.13 Ball bearings reduce friction andenable wheels to work very efficiently.

Physics file

Several decades before Newton, Galileo

Galilei had concluded that objects tended

to maintain their state of motion. He

called this tendency inertia , so this

conclusion is also known as Galileo’s

law of inertia. Inertia is not a force; it

simply describes the property of bodies

to continue their motion.

Physics file

Figure 5.14 Two Voyager spacecraft were launched from Cape Canaveral in 1977 withthe mission to investigate the outer planets of the solar system at close hand. Both craftcompleted the mission successfully, passing Saturn in 1981, Uranus in 1986 and Neptunein 1989. Voyager 1 and 2 have now left the solar system and since they have effectivelyzero net force acting on them, they continue to travel away from the Earth with a constantvelocity.

Figure 5.12 An air-track glider moves with a constant velocity because there is zero net forceacting on it. This is an example of inertia.



Motion152

Forces in equilibriumNewton’s first law states that a body will travel with a constant velocity (or

remain at rest) when the vector sum of all the forces acting on it is zero, i.e.

when the net force is zero. When the net force is zero, the forces are said to

be in equilibrium or balance.

• If a body is at rest and zero net force acts on it, it will remain at rest. Thisapplies to any stationary object such as a parked car or a book resting

on a desk, as shown in Figure 5.15. In these cases, the velocity is zero

and it is constant. The forces that are acting are balanced and so the net

force is zero.

• If a body is moving with a constant velocity and zero net force acts on

it, it will continue to move with the same constant velocity. An example

of this is a spacecraft with its engines off travelling through deep space.

If gravitation is ignored, there is nothing to slow the craft down or to

speed it up, and so it will continue with a constant velocity. The net force

acting on the spacecraft is zero, so it will move with a constant velocity.

Similarly, if a bus is travelling along a road with a constant velocity, the

vector sum of the forces acting on the bus must be zero. The drivingforces must balance the retarding forces, i.e. Σ F = 0.

Worked example 5.2A

During a car accident, a passenger travelling without a fastened seatbelt may fly through

the windscreen and land on the road. Using Newton’s first law of motion, explain how this

will occur.

Solution

During the accident, the car is brought to rest suddenly. Any occupants of the car will

continue to travel with the original speed of the car until a force acts to slow them down. Ifthe seatbelt were fastened, this would provide the necessary force to slow the passenger

within the car. In the absence of an opposing force, the passenger continues to move—often

crashing through the windscreen.

(The injuries received as a consequence of not wearing a seatbelt are usually far more

serious than those received if the person were fixed in the car during an accident. This is

why laws require seatbelts to be worn.)

Figure 5.15 The forces acting on this book are

balanced—i.e.ΣF = 0—so the velocity of thebook will not change; the book will continue tostay at rest. That is, the book’s velocity will beconstant at 0 m s−1.

Figure 5.17 (a) There are no forces acting on the spacecraft. The net force is zero and so itcontinues to move with constant velocity. (b) The forces acting on the bus are in equilibrium,so the net force is zero. The bus continues to move with constant velocity.

F T

F g

Σ F = 0

velocity doesn’t

change

no forces acting

deep space(a) v = 50 km h–1 F ground

F gravity

F drag

F driving

forces are balanced

velocity doesn’t change

(b)

F = 0

velocity doesn’t change

F = 0

Friction had a lot to do with Australia’s

Steven Bradbury winning a surprise gold

medal at the 2002 Winter Olympics. His

opponents did not experience enough

friction as they skated around the home

turn. Their inertia, in the absence of any

horizontal forces, sent them crashing

into the side wall. Steven stayed upright,

his skates cutting into the ice and

producing enough friction to allow him

to turn the corner and win the gold

medal.

Figure5.16 Steven Bradbury won the goldmedal as his opponents learned first-handabout inertia and friction (or rather a lack offriction!).

Physics file




Worked example 5.2B

A cyclist keeps her bicycle travelling with a constant velocity of 8.0 m s−1 east on a horizontal

surface by continuing to pedal. A force (due to air resistance) of 60 N acts against the

motion. How large is the driving force that is provided by the rear tyre at the point of contact

with the ground?

v = 8 m s–1

F drag = 60 N

F applied = 60 N

Σ F = F applied

+ F drag

= 0

Solution

If the cyclist is to continue at a constant 8.0 m s −1 east, then the forces that act on the

bicycle must be in equilibrium, i.e. ΣF = 0. This means that the forces due to air resistance

are exactly balanced by the pedalling force. A force of 60 N east must be produced at the

rear wheel.

(The cyclist will actually have to produce more than 60 N as the gearing of the bike is

designed to increase speed, not reduce the force that has to be applied.)

Prac 21

Figure5.18 In these examples, the forces acting on the objects are in equilibrium—the net force is zero. The body will either remain at rest, like thepicture hanging on the wal l, or continue with constant velocity, like the aircraft. A more complex situation involves the groundsman pushing the heavyroller with constant velocity. The horizontal component of the force he applies along the handle exactly balances the frictional force that opposes themotion of the roller in the horizontal direction. The vertical component of the applied force acts downwards, and adds to the weight of the roller, butthese two downward forces are balanced by an upward force provided by the ground.

lift

weight, F g

F g

drag

F thrust

F thrust

F drag

F lift

F 2

F 1

F 1

F g

F g

applied

force

weight of rollerfriction

upwards

force from

ground on

roller

applied push

upward force

from ground

weight of roller

friction

F g + F 1 + F 2 = 0 F thrust + F g + F drag + F lift = 0

F 2

SPARKlab



Motion154

Galileo Galilei was born into an academic family in Pisa, Italy,

in 1564. Galileo made significant contributions to physics,

mathematics and the scientific method through intellectual

rigour and the quality of his experimental design. But more

than this, Galileo helped to change the way in which the

universe was understood.Galileo’s most significant contributions were in astronomy.

Through his development of the refracting telescope he

discovered sunspots, lunar mountains and valleys, the four

largest moons of Jupiter (now called the Galilean moons)

and the phases of Venus. In mechanics, he demonstrated that

projectiles moved with a parabolic path and that different

masses fall at the same rate (the law of falling bodies).

These developments were most important because

they changed the framework within which mechanics was

understood. This framework had been in place since Aristotle

had constructed it in the 4th century BCE. By the 16th century,

the work of the Greek philosophers had become entrenched,

and it was widely supported in the universities. It was also

supported at a political level. In Italy at that time, governmentwas controlled by the Catholic Church. Today one would

think that Galileo would have been praised by his peers for

making such progress, but so ingrained and supported was

the Aristotelian view that Galileo actually lost his job as a

professor of mathematics in Pisa in 1592.

Galileo was not without supporters, though, and he was

able to move from Pisa to Padua where he continued teaching

mathematics. At Padua, Galileo began to use measurements

from carefully constructed experiments to strengthen his

ideas. He entered into vigorous debate in which his ideas

(founded as they were on observation) were pitted against

the philosophy of the past and the politics of the day. The

most divisive debate involved the motion of the planets.

The ancient Greek view, formalised by Ptolemy in the 2ndcentury AD, was that the Earth was at the centre of the

solar system and that all the planets, the Moon and the Sun

were in orbit around it. This view was taught by the Church

and was also supported by common sense. As such, it was

accepted as the establishment view. In 1630 Galileo published

a book in which he debated the Ptolemaic view and the new

Sun-centred model proposed by Copernicus. On the basis of

his own observations, Galileo supported the Copernican view

of the universe. However, despite the book having been passed

by the censors of the day, Galileo was summoned to Rome

to face the Inquisition for heresy. The finding went against

Galileo, and all copies of his book had to be burned and he was

sentenced to permanent house arrest for the term of his life.

Galileo died in 1642 in a village near Florence. Hehad become an influential thinker across Europe and the

scientific revolution he had helped start accelerated in

the freer Protestant countries in northern Europe. For its

part, the Catholic Church under Pope John Paul II began an

investigation in 1979 into Galileo’s trial, and in 1992 a papal

commission reversed the Church’s condemnation of him.

Galileo Galilei—revolutionary

Physics in action

Figure5.19 Galileo Galilei (1564–1642) was a short, active man withred hair. Galileo made significant contributions to our understanding ofthe forces that act on moving bodies. In his book thePrincipia , Newtonwas quick to acknowledge his debt to Galileo’s genius.This portrait was drawn 8 years before Galileo’s trial.

• Aristotle theorised that the natural state of matter

was to be at rest in its natural place.

• Galileo performed experiments and from these

developed the idea of inertia.

• Newton developed Galileo’s ideas further and

devised the first law of motion, stated as ‘A body

will either remain at rest or continue with constant

velocity unless it is acted on by a non-zero net force

(or an unbalanced force).’

• Where the net force on a body is zero, i.e. Σ = 0,

the forces are said to be balanced and are in

equilibrium.

Newton’s first law of motion5.2 summary




11 In just a few sentences, distinguish between the

understandings held by Aristotle and Newton about

the natural state of matter. Describe an experiment

that might help support each of these views.

22 A billiard ball is rolling freely across a smoothhorizontal surface. Ignore drag and frictional forces

when answering these questions.

a Which of the following force diagrams shows the

horizontal forces acting on the ball according to

the theories of Aristotle?

b Which of the following force diagrams is correct

for the ball according to the theories of Newton?

c Which force diagram correctly describes this

situation?

33 If a person is standing up in a moving bus that stops

suddenly, the person will seem to fall forwards.

Has a force acted to push the person forwards? UseNewton’s first law of motion to explain what is

happening.

44 What horizontal force has to be applied to a wheelie

bin if it is to be wheeled to the street on a horizontal

path against a retarding force of 20 N at a constant

1.5 m s−1?

55 When flying at constant speed at a constant altitude,

a light aircraft has a weight of 50 kN down, and the

thrust produced by the engines is 12 kN to the east.

What is the lift force required by the wings of the

plane, and how large is the drag force that is acting?

66 A young boy is using a horizontal rope to pull his

go-kart at a constant velocity. A frictional force of

25 N also acts on the go-kart.

a What force must the boy apply to the rope?

b The boy’s father then attaches a longer rope to thekart because the short rope is uncomfortable to

use. The rope now makes an angle of 30° to the

horizontal. What is the horizontal component of

the force that the boy needs to apply in order to

move the kart with constant velocity?

c What is the tension force acting along the rope

that must be supplied by the boy?

77 Passengers on commercial flights are required to be

seated and have their seatbelts done up when their

plane is coming in to land. What would happen to

a person who was standing in the aisle as the plane

travelled along the runway during landing?

88 Consider the following situations, and name the

force that causes each object not to move in a straight

line.

a The Earth moves in a circle around the Sun with

constant speed.

b An electron orbits the nucleus with constant

speed.

c A cyclist turns a corner at constant speed.

d An athlete swings a hammer in a circle with

constant speed.

99 A magician performs a trick in which a cloth is pulled

quickly from under a glass filled with water without

the glass falling over or the water spilling out.

a Explain the physics principles underlying this

trick.

b Does using a full glass make the trick easier or

more difficult? Explain.

110 Which of these objects would find it most difficult

to come to a stop: a cyclist travelling at 50 km h −1, a

car travelling at 50 km h−1 or a fully laden semitrailer

travelling at 50 km h−1? Explain.

Newton’s first law of motion5.2 questions

F F

F F

A B

C D

Worked Solutions



Motion156

Newton’s second law of motion5.3

Newton’s first law of motion states that when all the forces on a body

are balanced, the body can only remain at rest or continue with constant

velocity. Newton’s second law of motion deals with situations in which a

body is acted on by a non-zero net force, i.e. Σ F ≠ 0; in other words, when

the forces are unbalanced.

When there is a non-zero net force acting on a body, the body willaccelerate in the direction of the net force. Newton explained that the rate

of this acceleration will depend on both the size of the net force and the

mass of the body. Experiments show that the acceleration produced is

directly proportional to the size of the net force acting:

∝ Σ F

Experiments also show that the acceleration produced by a given net

force depends on the mass of the body. We know that a greater mass has

a greater inertia, so it will be more difficult to accelerate. Not surprisingly,

experiments reveal that the acceleration produced by a particular force is

inversely proportional to the mass of the body:

∝ 1

m

If the two relationships are combined, we get:

∝ Σ F × 1

m

or

∝ Σ F

m

The relationship can be converted into an equality by including a constant

of proportionality, so:

= k × Σ F

m

By definition, 1 newton is the force needed to accelerate a mass of 1 kg at

1 m s−2. In the SI system of units, this makes the constant k equal to 1. The

relationship is therefore simplified to Σ F = m , a mathematical statement of

Newton’s second law of motion.

The SI unit, the newton (N), will be required for force when the mass

of the accelerating body is given in kilograms (kg) and its acceleration is

provided in metres per second squared (m s−2

).Σ F = m is a vector equation in which the direction of the acceleration is

in the same direction as the net force. If only one force acts, the acceleration

will be in the direction of that force.

Worked example 5.3A

Determine the size of the force required to accelerate an 80 kg athlete from rest to 12 m s−1

in a westerly direction in 5.0 s.

N…WTON’S S…COND LAW OF MOTION states that the acceleration of a body, a, isdirectly proportional to the net force acting on it, ΣF , and inversely proportionalto its mass,m:

ΣF = ma

i

Figure 5.20 This sprinter is about to leavethe starting blocks. The starting blocks stophis foot from slipping backwards, increasingthe size of the forward force acting on him andincreasing his forward acceleration.

From Newton’s second law, it can be

seen that the unit of a newton (N) is

equivalent to the product of the mass

unit (kg) and the acceleration unit

(m s−2). In other words: N = kg m s−2.

When writing the value of a force,

either unit is correct; but newton is the

SI unit and is obviously more convenient

to use!

Physics file




Solution

First, determine the acceleration of the athlete:

v = u + at

a = v − u

t

= 12 − 0

5.0a = 2.4 m s−2 west

The net force can now be found using Newton’s second law:

ΣF = ma

= 80 × 2.4

= 190 N west

If more than one force acts on a body, the acceleration will be in the direction

of the net force, i.e. the vector sum of all of the forces.

Worked example 5.3B

A swimmer whose mass is 75 kg applies a force of 50 N as he starts a lap. The water opposes

his efforts to accelerate with a drag force of 20 N. What is his initial acceleration?

Solution

The net force on the swimmer in the horizontal direction will be:

ΣF = F applied

+ F drag

A vector addition gives ΣF = 30 N forwards.

So,

a = ΣF

m =

30

75

= 0.40 m s−2 in the direction of the applied force

(It is worth noting that the drag applied by the water will increase with the swimmer’s

speed.)

Worked example 5.3C

A 150 g hockey ball is simultaneously struck by two hockey sticks. If the sticks supply a

force of 15 N north and 20 N east respectively, determine the acceleration of the ball, and

the direction in which it will travel.

Solution

Remember to work in kilograms. Calculate the net force acting on the ball by performing a

vector addition:

ΣF = F 1 + F

2

ΣF = √F 1

2 + F 2

2

= √152 + 202 = √225 + 400 = √625 = 25 N

a = ΣF

m =

25

0.15

= 170 m s−2

Looking at the vector diagram showing the addition of the forces, we can see that θ will be

given by:

tan θ = F

2

F 1

= 20

15 = 1.33

so θ = tan−1 1.33 = 53°

The ball will travel in the direction N53°E or (053ºT).

Prac 22

50 N

20 N

30 N

+

F applied

F drag

Σ F

F 1 = 15 N north

F 2 = 20 N east

F 2

F 1

N

S

W EΣ F

θ



Motion158

Mass and weight

Mass of a body

To this point, the idea of the mass of an object has been taken for granted.

However, the concept of a body’s mass is rather subtle and, importantly

in physics, the mass of a body is a fundamentally different quantity from

its weight—even though people (even physics teachers) tend to use theseexpressions interchangeably in everyday life.

In earlier science courses, mass may have been defined as ‘the amount

of matter in an object’. To understand what mass really is, this description

says very little. The international standard for the kilogram is not very

helpful either. Since the time of the French Revolution (late 1700s), the

kilogram has been defined in terms of an amount of a standard material. At

first, 1 litre of water at 4°C was used to define the kilogram. More recently

an international mass standard has been introduced. This is a 1 kg cylinder

of platinum–iridium alloy that is kept in Paris. Copies are made from the

standard and sent around the world.

Newton’s second law can help to provide a better understanding ofmass through the effect of a force on a massive body. Think about a mass

resting on a frictionless surface. If a force is applied to the mass in the

horizontal direction, an acceleration is produced that is given by =Σ F

m.

The greater the mass, the smaller will be the acceleration. If the mass is

reduced, the acceleration will increase. Here the mass can be seen as the

property of the body resisting the force. Mass is the closest quantity in

physics to the concept of inertia.

If the above experiment is repeated at another location, the same net

force acting on the body will give the same acceleration regardless of where

the experiment is performed. This is because—on Earth, on the Moon, in

space—the mass of the body remains the same. Mass is a property of the body, and is not affected by its environment. In fact, for any situation at this

level in physics, the mass of a body will be a constant value. As discussed

in the adjacent Physics file, Albert Einstein was able to show in his theory

of relativity that the mass of an object does change as its speed changes.

The (inertial) MASS of a body is its ability to resist acceleration when the bodyis acted on by a net force. Mass is a constant property of the body.

i

Mass is usually considered to be an

unchanging property of an object. This is

true in Newtonian mechanics where the

speed with which an object is considered

to travel matches everyday experience.

However, at very high speeds, Newton’s

laws of motion do not apply, and the

theory of relativity must be used. In

1905, Albert Einstein showed that a

body with a rest mass m0 (i.e. mass

when stationary) will experience an

increase in mass as it gets faster. This

increase is usually undetectable except

when the object nears the speed of light.

At these very high speeds, the mass

will become greater and greater, tending

to infinity as the speed approaches the

speed of light.

Figure5.21 Scientists working at theAustralian synchrotron in Clayton have to takeaccount of these relativistic effects on theelectrons they accelerate to extreme speeds.

Table 5.2 gives the mass of a 1 kg

block if it were to travel at speeds of

0.1c (10% of the speed of light or3 × 107 m s−1), 0.8c and 0.99c .

Table 5.2 The mass of a 1 kg blockat different speeds

Speed Mass

0.1c 1.0050 kg (i.e. 5 g increase)

0.8c 1.6667 kg (667 g increase)

0.99c 7.1 kg (over 700% increase)

Relativistic mass increase provides

the reason why no object can travel at

the speed of light. To do so would require

an infinite quantity of energy, since the

mass of the body would itself be infinite.

Only objects with no rest mass (such as

light ‘particles’) can travel at the speed

of light.

Physics file

Figure 5.22 Regardless of the external conditions, the inertial qualities of a mass remain thesame. A net force of 1 N will always produce an acceleration of 1 m s−2 for a 1 kg mass. In thisway, mass can be understood as the resistance to a force. The greater the mass of the body,the smaller the acceleration caused by the force.

1 kg

on Earth in deep space

a = 1 m s–2

a = 1 m s–2

F = 1 N

F = 1 N

1 kg




Weight of a body

In the late 1500s, Galileo was able to show that all objects dropped near

the surface of the Earth accelerate at the same rate, g, towards the centre of

the Earth. The force that produces this acceleration is the force of gravity.

In physics, the force on a body due to gravity is called the weight of a body,

F g or W .

Consider a pumpkin of mass 10 kg and a banana of mass 0.10 kg that are

dropped together from several metres above the surface of the Earth.

The pumpkin and the banana will fall with a uniform acceleration that

is equal to 9.8 m s−2. The acceleration of a freely falling object (i.e. one on

which the only force that is acting is gravity) does not depend on the mass

of the object.

If we took the pumpkin and the banana to the Moon and dropped them,

they would fall with an acceleration of just 1.6 m s −2. Gravity is weaker on

the Moon because it is much less massive than Earth.

The acceleration of a freely falling object due to gravity is known as g.i

Prac 23

Figure5.23 If air resistance is ignored, the pumpkin and banana will fall side by side with anacceleration of 9.8 m s−2 towards the Earth.

Figure5.24 The pumpkin and banana will fall side by side with a uniform acceleration of1.6 m s−2 towards the Moon.

a = 9.8 m s–2

a = 9.8 m s–2

a = 1.6 m s–2 a = 1.6 m s

–2

SPARKlab



Motion160

We will now use Newton’s second law to analyse the motion of the

pumpkin as it falls to the Earth. The only force acting on the pumpkin is the

force of gravity or weight, F g. Hence:

Σ F = F g

m = F g

The acceleration of the pumpkin is 9.8 m s−2 or g, so F g = mg.

The acceleration of a mass due to gravity is numerically identical to the

gravitational field strength, g. These two quantities have different names

and different units but are numerically equal. It can be shown that 1 m s−2

is equal to 1 N kg−1.

As a consequence of this, the weight of a body will change as it is placed

in different gravitational fields. On the Earth a 10 kg pumpkin will have a

weight of 10 × 9.8 = 98 N downwards. On the Moon, the gravitational field

strength is lower at 1.6 N kg−1, and so the pumpkin will be easier to lift as

its weight is now only 10 × 1.6 = 16 N. In deep space, far from any stars or

planets, where g = 0, the pumpkin would be truly weightless, although its

mass would still be 10 kg.

Why do heavy and light objects fall with equal acceleration?

Although Galileo was able to show that heavy and light objects fell atthe same rate, he was not able to explain why. Newton, however, after

stating his laws of motion, was able to show why this happens. We can

use Newton’s second law to analyse the motion of the pumpkin and the

banana as they fall towards Earth.

The W…IGHT of a body, W or F g, is defined as the force of attraction on a body

due to gravity:W = F

g = m g

where m is the mass of the body (kg) g is the acceleration due to gravity (m s−2)

g is also known as the GRAVITATIONAL FI…LD STR…NGTH. The unit of thegravitational field strength is newton/kg or N kg−1.

i

Figure 5.26 The force dragging the pumpkin to the ground is much greater than the forcethat is acting on the banana. However, the mass and inertia of the pumpkin is much greaterthan the mass and inertia of the banana. The acceleration that results in both cases isidentical: 9.8 m s–2 down.

Figure 5.25 The pumpkin is in free-fall. Theonly force acting on it is gravity, F

g, and it

accelerates at 9.8 m s−2 towards the ground.

a = 9.8 m s–2

F g

F g = 98 N F g

= 0.98 N

10 kg

0.10 kg




The force of gravity acting on the 10 kg pumpkin is:

W = F g = mg = 10 × 9.8 = 98 N down

This is the only force acting on the pumpkin so the net force, Σ F , is also

98 N down.

The acceleration of the pumpkin can be calculated:

a =

Σ F

m =

98

10 = 9.8 m s−2

down

The force of gravity acting on the 0.10 kg banana is:

W = F g = mg = 0.10 × 9.8 = 0.98 N down

This is the only force acting on the banana so the net force, Σ F , is also 0.98

N down.

The acceleration of the banana can be calculated:

a = Σ F

m =

0.98

0.10 = 9.8 m s−2 down

The pumpkin has a large force dragging it towards the ground, but this

large force is acting on a large mass, which has more inertia. The banana

has a small force acting on it, but this small force is moving a small mass.

The acceleration produced in each case is exactly the same: 9.8 m s−2 down.

Worked example 5.3D

A 1.5 kg trolley cart is connected by a cord to a 2.5 kg mass as shown. The cord is placed

over a pulley and allowed to fall under the influence of gravity.

a Assuming that the cart can move over the table unhindered by friction, determine the

acceleration of the cart.

b If a frictional force of 8.5 N acts against the cart, what is the magnitude of the acceleration

now?

Solution

a The cart and mass experience a net force equal to the weight of the falling mass. SoΣF = F

g = m g = 2.5 × 9.8 = 24.5 N down. This force has to accelerate not only the cart

but also the falling mass, and so the total mass to be accelerated is 1.5 + 2.5 = 4.0 kg.

a = ΣF

m

=

24.5

4.0 = 6.1 m s−2

to the rightb In analysing the forces that now act on the cart, the net force is:

ΣF = 24.5 − 8.5 = 16.0 N to the right,

and a = ΣF

m =

16.0

4.0 = 4.0 m s−2 to the right.

Figure5.27 The weight of this boulder is theforce it experiences due to gravity given byF

g = m g. This is approximately 2.5 × 105 N directed

to the centre of the Earth. The mass of the boulderis approximately 25 000 kg. If the boulder weretaken to outer space where the gravitational fieldstrength is zero, the boulder would still have thesame mass but no weight.

1.5 kg

2.5 kg

8.5 N

24.5 N



Motion162

Galileo was able to show more than 400 years ago that the

mass of a body does not affect the rate at which it falls

towards the ground. However, our common experience is that

not all objects behave in this way. A light object, such as a

feather or a balloon, does not accelerate at 9.8 m s−2 as it falls.

It drifts slowly to the ground, far slower than other droppedobjects. Parachutists and skydivers also eventually fall with a

constant speed. Why is this so?

Skydivers, BASE-jumpers and air-surfers are able to use

the force of air resistance to their advantage. As a jumper first

steps off, the forces acting on him are drag (air resistance),

a, and gravity,

g . Since his speed is low, the drag force is

small (Figure 5.28a). There is a large net force downwards,

so he experiences a large downwards acceleration of just less

than 9.8 m s−2, causing him to speed up. This causes the drag

force to increase because he is colliding harder with the air

molecules. In fact, the drag force increases in proportion to

the square of the speed:a ∝ v 2. This results in a smaller net

force downwards (Figure 5.28b). His downwards acceleration

is therefore reduced. It is important to remember that he isstill speeding up, but at a reduced rate.

As his speed continues to increase, so too does the

magnitude of the drag force. Eventually, the drag force

becomes as large as the weight force (Figure 5.28c). When this

happens, the net force is zero and the jumper will fall with a

constant velocity. Since the velocity is now constant, the drag

force will also remain constant and the motion of the jumperwill not change (Figure 5.28d). This velocity is commonly

known as the terminal velocity .

For skydivers, the terminal velocity is usually around

200 km h−1. Opening the parachute greatly increases the air

resistance force that is acting, resulting in a lower terminal

velocity. This is typically around 70 km h−1.

Terminal velocity

Physics in action

Figure5.28 As the jumper falls, the force of gravity does not change,but the drag force increases as he travels faster. Eventually these twoforces will be in equilibrium and the jumper will fall with a constant orterminal velocity.

Σ F

Σ F

F a

F g

F a

F g

v

v v

v

F a

F g

F a

F g

Σ F = 0

Σ F = 0

(a)

(b)

(c)

(d)

Figure5.29 Skydivers can change their speed by changing their body profile. If they assume a tuck position they will fall faster and if theyspreadeagle they will fall slower. This enables them to meet and form spectacular patterns as they fall.




• Newton’s second law of motion states that the

acceleration a body experiences is directly pro-

portional to the net force acting on it, and inversely

proportional to its mass:

Σ = m where m is measured in kilograms (kg), is

measured in metres per second squared (m s−2), and

Σ in newtons (N).

• The mass of a body can be considered to be its

ability to resist a force. Mass is a constant property

of the body and is not affected by its environment or

location.

• The weight of a body W or F g is defined as the forceof attraction on a body due to gravity. This will be

given by W = F g = mg where m is the mass of the body

and g is the strength of the gravitational field.

Newton’s second law of motion5.3 summary

Use g = 9.8 m s−2 when answering these questions.

11 State whether the forces are balanced (B) or

unbalanced (U) for each of these situations.

a a netball falling towards the groundb a stationary bus

c a tram travelling at a constant speed of 50 km h–1

d a cyclist slowing down at a traffic light

22 During a tennis serve, a ball of mass 0.060 kg is

accelerated to 30 m s–1 from rest in just 7.0 ms.

a Calculate the average acceleration of the ball.

b What is the average net force acting on the ball

during the serve?

33 Use Newton’s laws to explain why a 1.0 kg shot-put

can be thrown further than a 1.5 kg shotput.

44 In a game of soccer, the ball is simultaneously kicked

by two players who impart horizontal forces of

100 N east and 125 N south on the ball. Determine:

a the net force acting on the ball

b the direction in which the ball will travel

c the acceleration of the ball if its mass is 750 g.

55 When travelling at 100 km h−1 along a horizontal

road, a car has to overcome a drag force due to air

resistance of 800 N. If the car has a mass of 900 kg,

determine the average driving force that the motor

needs to provide if the car is to accelerate at 2.0 m s−2.

66 Mary is paddling a canoe. The paddles are providinga constant driving force of 45 N south and the drag

forces total 25 N north. The mass of the canoe is 15 kg

and Mary has a mass of 50 kg.

a What is Mary’s mass?

b Calculate Mary’s weight.

c Find the net horizontal force acting on the canoe.

d Calculate the magnitude of the acceleration of the

canoe.

77 On the surface of the Earth, a geological hammer hasa mass of 1.5 kg. Determine its mass and weight on

Mars where g = 3.6 m s−2.

88 What is the average braking force required of a

1200 kg car in order for it to come to rest from

60 km h−1 in a distance of 25 m?

99 Consider a 70 kg parachutist leaping from an aircraft

and taking the time to reach terminal velocity before

activating the parachute. Draw a sketch graph of

the net force against time for the parachutist in the

period from the start of the jump until terminal

velocity has been reached. Explain your reasoning. 110 A 0.50 kg metal block is attached by a piece of string

to a dynamics cart as shown below. The block is

allowed to fall from rest, dragging the cart along.

The mass of the cart is 2.5 kg.

2.5 kg

0.50 kg

a If friction is ignored, what is the acceleration of

the block as it falls?

b How fast will the block be travelling after 0.50 s?

c If a frictional force of 4.3 N acts on the cart, what is

its acceleration?

Newton’s second law of motion5.3 questions

Worked Solutions



Motion164

Newton’s third law of motion5.4

Newton’s first two laws of motion describe the motion of a body resulting

from the forces that act on that body. The third law is easily stated, and seems

to be widely known by students, but is often misunderstood and misused!

It is a very important law in physics, and assists our understanding of the

origin and nature of forces.

Newton realised that all forces exist in pairs and that each force in thepair acts on a different object. Consider the example of a hammer gently

tapping a nail. Both the hammer and nail experience forces during this. The

nail experiences a small downwards force as the hammer hits it and this

moves it a small distance into the wood. The hammer experiences a small

upwards force as it hits the nail causing the hammer to stop. These forces

are known as an action–reaction pair and are shown in Figure 5.31a.

Now consider what will happen if the hammer is smashed into the nail.

The nail now experiences a large downwards force as the hammer smashesinto it and this moves it a larger distance into the wood. The hammer itself

experiences a large upwards force as it hits the nail, again causing the

hammer to stop. You should notice that the forces acting on the hammer

and nail are both larger, as shown in Figure 5.31b. This is what Newton

realised. If the hammer exerted a downwards force of 25 N on the nail, the

nail would exert an upwards force of 25 N on the hammer.

It is important to recognise that the action force and the reaction force in

Newton’s third law act on different objects and so should never be added

together. Figure 5.32 shows some examples of action–reaction forces.

It is also important to understand that even though action–reaction

forces are always equal in size, the effect of these forces may be very

different. A good example of this is the collision between the bus and the

car shown in 5.32c. Because of the car’s small mass, the force acting on

N…WTON’S THIRD LAW OF MOTION states that for every action force (object A onB), there is an equal and opposite reaction force (object B on A):

F(A on B) = −F(B on A)When body A exerts a force F on body B, body B will exert an equal and opposite

force –F on body A.

i

Figure 5.30 A hammer hitting a nail is agood example of an action–reaction pair andNewton’s third law.

Figure 5.31 (a) As the hammer gently taps the nail, both the hammer and nail experiencesmall forces. (b) When the hammer smashes into the nail, both the hammer and nailexperience large forces.

force thatnail exertson hammer

force thathammer exertson nail

force thatnail exertson hammer

force thathammer exertson nail

(b)(a)




the car will cause the car to undergo a large acceleration backwards. The

occupants may be seriously injured as a result of this. The force acting onthe bus is equal in size, but is acting on a much larger mass. The bus will

have a relatively small acceleration as result and the occupants will not be

as seriously affected.

Worked example 5.4A

In each of the following diagrams, one of the forces is given.

i Describe each given force using the phrase ‘force that exerts on ’.

ii Describe the reaction pair to the given force using the phrase ‘force that

exerts on ’.

iii Draw each reaction force on the diagram, carefully showing its size and location.

Figure5.32 Some action–reaction force pairs. Notice that these can be contact or non-contact forces. (a) The girl exerts a downwards force on the floor and the floor exerts an equalupwards force on the girl. (b) The foot exerts a forwards force on the ball and the ball exertsan equal size backwards force on the foot. (c) The bus exerts a backwards force on the carand the car exerts an equal size forwards force on the bus. (d) The Earth exerts a downwardsgravitational force on the brick and the brick exerts an equal size upwards force on the Earth.

Figure5.33 The soccer ball and the player’shead exert equal forces on each other during thiscollision, but only the player will experience pain!

force that foot

exerts on ball

force that floor

exerts on girl

force that girl

exerts on floor

force that bus

exerts on car

gravitational force

that Earth exertson brick ( F g )

force that car

exerts on bus

force that ballexerts on foot

gravitational force

that brick exertson Earth

(a) (b)

(c) (d)

(a) (b) (c)



Motion166

Solution

a i force that bat exerts on ball

ii force that ball exerts on bat

iii see diagram at right

b i force that ball exerts on floor

ii force that floor exerts on ball


c iii gravitational force that Earth exerts on astronaut

ii gravitational force that astronaut exerts on Earth


Motion explainedNewton’s third law also explains how we are able to move around. In fact,

the third law is needed to explain all locomotion. Consider walking. Your

leg pushes backwards with each step. This is an action force acting on the

ground. As shown in Figure 5.34, a component of the force acts downwards

and another component pushes backwards horizontally along the surface

of the Earth. The force is transmitted because there is friction between your

shoe and the Earth’s surface. In response, the ground then pushes forwards

on your leg. This forwards component of the reaction force enables you

to move forwards. In other words, it is the ground pushing forwards on

your leg that moves you forwards. It is important to remember that in

Newton’s second law, ∑ F = m , the net force ∑ F is the sum of the forces

acting on the body. This does not include forces that are exerted by the body

on other objects. When you push back on the ground, this force is acting on

the ground and may affect the ground’s motion. If the ground is firm, this

effect is usually not noticed, but if you run along a sandy beach, the sand is

clearly pushed back by your feet.

The act of walking relies on there being some friction between your shoe

and the ground. Without it, there is no grip and it is impossible to supplythe action force to the ground. Consequently, the ground cannot supply the

reaction force needed to enable forward motion. Walking on smooth ice is

a good example of this, and so mountaineers will use crampons (basically

a rack of nails) attached to the soles of their boots in order to gain purchase

in icy conditions.

The situation outlined above is fundamental to all motion.

Figure 5.34 Walking relies on an action and

reaction force pair in which the foot will push downand backwards with an action force. In response, theground will push upwards and forwards. The forwardcomponent of the reaction force is actually friction.This is responsible for the body moving forward as awhole, while the back foot remains at rest.

θ

θ

θ

F up

F forwards

F (reaction)

F (reaction)

F (action)




Table 5.3 All motion can be explained in terms of action and reactionforce pairs

Motion Action force Reaction force

Swimming Hand pushes back on water Water pushes forwards on hand

Jumping Legs push down on Earth Earth pushes up on legs

Bicycle/car Tyre pushes back on ground Ground pushes forwards on tyre

Jet aircraft and

rockets

Hot gas is forced backwards out

of engine

Gases push craft forwards

Skydiving Force of gravitation on the

skydiver from Earth

Force of gravitation on Earth

from skydiver

Worked example 5.4B

A front-wheel drive car of mass 1.2 tonne accelerates from traffic lights at 2.5 m s –2.

a Discuss and identify the horizontal force that enables the car to accelerate forwards.

b Draw this force on the diagram at right. Label it force A.

c Carefully draw the reaction pair to this force as described in Newton’s third law. Label it

force B.

d Discuss and identify the reaction force that you have drawn.

Solution

a This is the forwards force that the road exerts on the front tyre and could be called a

frictional force.

b c

d This is the backwards force that the front tyre exerts on the road. If the road surface was

ice, both of the forces in (b) and (c) would be very small and the car would not be able

to drive forwards.

The normal forceWhen an object, say a rubbish bin, is allowed to fall under the influence of

gravity, it is easy to see the effect of the force of gravity. As shown in Figure

5.35a, the only force acting is the weight, so the net force is the weight, and

the bin therefore accelerates at g.

When the bin is at rest on a table, the force of gravity ( F g = W = mg) is

still acting. Since the bin is at rest, there must be another force (equal inmagnitude and opposite in direction) acting to balance the weight. This

upwards force is provided by the table. Because of the weight of the bin,

the table is deformed a little, and being elastic, it will push upwards. The

elastic force it provides is perpendicular to its surface and is called a normal

reaction force F N or N , (often abbreviated to the normal force).

The NORMAL FORC… is a reaction force supplied by a surface at 90° to its plane.i

1200 kg

force A force B



Motion168

This means that there are two forces that act on the bin which will

completely balance each other so that the net force is zero. In Figure 5.35b,

the bin is empty so its weight is small and the normal force is also small.

However, when the bin fills up, its weight increases and so too does the

normal force.

The normal force and the weight force in these examples are equal and

opposite. However, this does not mean that they are an action–reaction pair

as described by Newton’s third law! The weight force and the normal force

act on the same body (the bin) so they cannot be an action–reaction force

pair. Remember that in Newton’s third law, one force acts on one object andthe other force acts on the other object. Let us identify the reaction pair to

each of the forces shown in Figure 5.36.

In Figure 5.36a the action force shown is the force of gravity F g on the

bin. This is the attractive force that the Earth exerts on the bin. The reaction

force, therefore, is the attractive force that the bin exerts on the Earth. This is

shown as F G. In Figure 5.36b, the action force shown is the normal force on

the bin, F N. This is the force that the table exerts on the bin. So the reaction

force is the force that the bin exerts on the table. This is shown as F T.

Worked example 5.4C

An 8.0 kg computer rests on a table.

a Identify the forces that act on the computer.

b Determine the magnitude and direction of the force that the computer exerts on thetable.

c If a 3.0 kg monitor is placed on the computer box, determine the new normal force acting

on the computer.

Solution

a The weight of the computer will beF g = m g = 8.0× 9.8 = 78 N down, so that if the net force

on the computer is zero, the normal force supplied by the table must be 78 N upwards:

ΣF = F g + F

N = 0.

Figure 5.35 (a) When the bin is in mid-air,

there is an unbalanced force acting on it so itaccelerates towards the ground. (b) When theempty bin is resting on the table, there is a smallupwards force from the table acting on it. Thebin remains at rest, so the forces are balanced.(c) The forces acting on the full bin are alsobalanced. The weight of the bin is greater, so thenormal force exerted by the table is larger thanfor the empty bin.

Figure 5.36 (a) The reaction force pair tothe weight of the bin is the gravitational forceof attraction that the bin exerts on the Earth.(b) The reaction force pair to the normal forceon the bin is the force that the bin exerts on thetable. The force pairs are equal and oppositebut they do not cancel out because they arenot acting on the same object.

F N

F N F g

F g

F g

(a) (b) (b)

F N =

F G =

F g =

F T =

gravitational force

Earth exerts on bin

force table

exerts on bin

gravitational force bin

exerts on Earth

force bin exerts

on table

(b)(a)

F N

F g

Σ F = F g

+ F N = 0




b Since the force that the table exerts on the computer has been found to be 78 N up, then,

according to Newton’s third law, the force that the computer exerts on the table must be

78 N down.

c If a 3.0 kg monitor is placed on top of the computer, the table must supply a further

3.0 × 9.8 = 29 N, bringing the total normal force to 107 N. (The computer will also have

to provide a normal force of 29 N upwards to balance the weight of the monitor.)

The inclined planeThe Guinness Book of Records identifies the steepest road in the world as

being at an angle of 20° to the horizontal. It is located in Dunedin, New

Zealand. Living on such a road requires the residents to ensure that the

handbrake in their car is always in good repair! To determine the force

required by the handbrake of a car parked on this steep road, the physics of

forces acting on a body on an inclined plane must be used.

Start by thinking of a body at rest on a horizontal surface. Two forces act

on the body: the weight of the body F g and the normal force F

N supplied

by the surface. The weight force always acts downwards and is given by

F g = mg. The normal force is supplied by the surface and will vary dependingon the situation, but it will always act upwards and perpendicular to the

surface. This means that the net force on the body will be the sum of these

two forces, and in this case it has to be zero since the body does not move.

If the surface is tilted so that it makes an angle to the horizontal, the

weight force remains the same: F g = mg. However, the normal force

continues to act at right angles to the surface and will change in magnitude,

getting smaller as the angle increases. If there is no friction between the

body and the surface, the two forces will not balance and a non-zero net

force will be directed down the incline as shown in Figure 5.37.

From the vector diagram of the forces:

Σ F = F g + F

N = F

g sin θ = mg gsinθ

From Newton’s second law, the net force is:Σ F = m

so, m = mg gsin θ

or = g gsin θ

This means that the acceleration down an incline is a function of the angle

of the incline alone, and not the mass of the body. Ignoring any friction, a

car rolling down the steep street in Dunedin will accelerate at = g gsinθ

= 9.8 × sin 20° = 3.4 m s−2—quite a rate!

You may have observed that the mass

of an object can be used in two different

contexts. First, mass is a measure of

the ability of an object to resist being

accelerated by a force. This mass can be

determined from Newton’s second law

and is known as inertial mass . Second,

mass can give an indication of the

degree to which an object experiences a

gravitational force when in the presence

of a gravitational field. This mass is

known as gravitational mass . Some

very accurate experiments have shown

that the inertial and gravitational

masses are equal, although there is no

theoretical reason why this should be

the case.

Physics file

Figure5.37 (a) Where the surface isperpendicular to the weight force, the normal forcewill act directly upwards and cancel the weightforce. (b) On an inclined plane,F

N is at an angle to

F g and as long as no friction acts, there will be a net

force down the incline. The body will accelerate.

Σ F = F g + F N

= 0

Σ F =

F g +

F N

Σ F

θ

θ

F N

F N

F N

F N

F g

F g

F g

F g

body remains

at rest

(a) (b)

Prac 24 SPARKlab



Motion170

80 m

ice patch

35°

35°

F g F N

F g

F N

Σ F

Worked example 5.4D

Driver error allows a 5 tonne truck to roll down a steep ramp inclined at 30° to the horizontal.

As it is a high-technology vehicle, there is negligible friction between the wheels of the truck

and the wheel bearings. Find the acceleration of the truck if the acceleration due to gravity

is taken as 9.8 m s−2.

Solution

ΣF = F g + F

N

From the vector diagram:

ΣF = F g sin θSo, ma = m g sin θ

a = g sin 30°

= 9.8 × sin 30°

= 4.9 m s−2 down the ramp

If friction exists between a body on an inclined plane and the surface, its

direction will be along the incline but against the motion. If a frictional force

is great enough to balance the sum of the normal force and the weight of

the body, the net force is zero and the body will either travel with a constant

velocity or remain stationary. Worked example 5.4E illustrates this.

Worked example 5.4E

Kristie is a 60 kg skier. At the start of a ski slope that is at 35° to the horizontal, she crouches

into a tuck. The surface is very icy, so there is no friction between her skis and the ice. Ignore

air resistance when answering these questions.

a If Kristie starts from rest, what is her speed after travelling a distance of 80 m on the ice?

b The snow conditions change at the end of the ice patch so that Kristie continues down

the slope with a constant velocity. What is the force due to friction that must be acting

between Kristie’s skis and the snow?

Solution

a The net force on Kristie will be ΣF = F g + F

N. This is a vector addition. From the vector

diagram:

ΣF = F g sin θSo, a = g sin 35°

= 9.8 sin 35°

= 5.6 m s−2

In February 2003, a train driver pulled

into Broadmeadows station and went

for a toilet break. Unfortunately, he

forgot to put on the handbrake. When

he returned, the train was rolling

away from the platform, heading for

Jacana. It is a slight downhill incline

from Broadmeadows to Jacana and the

train simply rolled off down the hill,

accelerating gradually. After Glenroy,

the incline of the track is greater and

the train’s acceleration increased. It

is estimated that it reached speeds of

around 100 km h –1 at times. Fortunately,

no-one was injured as it hurtled through

level crossings and stations on its way

into the city. At Spencer Street station,

it crashed into a stationary V/Line train

at about 60 km h−1. The express trip

from Broadmeadows to Spencer Street

took about 16 minutes.

Jacana

Glenroy

5 km

Oak ParkPascoe Vale

Strathmore

Moonee Ponds

Glenbervie

Broadmeadows

Essendon

Ascot Vale

Newmarket

Kensington

North Melbourne

Spencer Street Station

0

Figure5.38 An empty three-carriagetrain took 16 minutes to roll downhill fromBroadmeadows to Spencer Street (nowSouthern Cross) station. The track was like along inclined plane and the train acceleratedalong it after the driver forgot to put thehandbrake on!

Physics file

30

30

F = F g

+ F N

F N

F N

F g

F g




If this acceleration continues over 80 m, Kristie’s final speed would be:

v 2 = u2 + 2ax

so v 2 = 0 + 2 × 5.6 × 80 = 900

v = 30 m s−1 (110 km h−1)

b Kristie is travelling with a constant velocity, so ΣF = 0, i.e. the force of friction F f would

balance the component of the weight force parallel to the incline.

So F f = F g sin 35 = 340 N up the incline.

TensionAnother force that is experienced in everyday life is the tension force that

is found in stretched ropes, wires, cables and rubber bands. If you stretch

a rubber band, it will exert a restoring force on both your hands. This force

is known as a tensile force and is present in any material that has been

stretched.

Consider the situation in which a person hangs from a cable that is tied

to a beam as shown in Figure 5.39a. We will assume that the mass of the

cable is negligible.

At the top end of the cable, the tension force F T pulls down on the beam.

At the other end, the cable exerts an upwards force F T on the person, holding

them in mid-air. In other words, the same size tension force acts at both

ends of the cable, but in opposite directions. If a second identical person

also hung from the cable, the tension acting at both ends of the cable would

double and the cable would become tauter (and more likely to snap!)

If the mass of the person in Figure 5.39a is known, the size of the tension

can be determined. If the mass of the person is 50 kg, then the forces acting

on them are, as shown in Figure 5.39b, an upwards tension force and the

downwards pull of gravity of 490 N. If the person is at rest, the forces acting

are balanced and so the upwards tensile force acting from the cable must

also be 490 N.

Calculations involving tension are illustrated in the following example.

A naughty monkey of mass 15 kg has escaped backstage in a circus. Nearby

is a rope threaded through an ideal (frictionless and massless) pulley.

Attached to one end of the rope is a 10 kg bag of sand. The monkey climbs

a ladder and jumps onto the free end of the rope.

Figure5.39 (a) The stretched cable exerts an upwards force on the person and an equal sizedownwards force on the beam. (b) The forces acting on the person are balanced, so the tensionforce is equal in magnitude to the weight force (490 N).

tension T

tension T

T = 490 N

g = 490 N

= 50 kgΣ

= 0

(a) (b)



Motion172

The system of the rope and the monkey is now subjected to a net force of:

Σ F = 15 × 9.8 − 10 × 9.8 = 49 N down on the side of the monkey.

As a consequence, both objects and the rope will accelerate at:

a = Σ F

Σm =

49

25 = 2.0 m s−2

While all of this is occurring, the rope is under tension. To find the

amount, we look at the forces acting on one of the masses (Figure 5.40c).

Take the monkey: the net force on the monkey will be Σ F = F g + F

T.

So, F T = Σ F − F

g

= 15 × 2.0 down − 15 × 9.8 down

= 30 N down + 147 N up = 117 N up

To check, find the tension acting on the sandbag.

Again, Σ F = F g + F

T

So, F T =

Σ F

− F g

= 10

× 2.0 up

− 10

× 9.8 down = 20 N up + 98 N up = 118 N up

Note that the small difference between the two results is due to rounding

error. The tension is equal on both sides of the pulley, regardless of how the

masses are arranged (provided the pulley is frictionless).

Intuitively, you might have thought that the tension would have been

(15 + 10) × 9.8 = 245 N since the two weights are pulling in opposite

directions. This is not the case because the system is allowed to accelerate,

causing a reduction in tension.

Figure 5.40 The monkey has a greater weight than the sandbag, and so the rope will accelerate in the direction of the monkey. The tension in therope is found by considering the forces that act on each weight.

98 N

10 kg

15 kg

a

F g =147 N F

g = 98 N

Σ F = 49 N

Σ F = 15 × 2.0Σ F = 10 × 2.0

F T = 117 N F T = 118 N

= 30 N= 20 N

(b)(a) (c)

147 N




Friction is a force that opposes movement. Suppose you want

to push your textbook along the table. This simple experiment

can reveal a significant amount of information about the

nature of friction. As you start to push the book, you find that,

at first, the book does not move. You then increase the force

that you apply, and suddenly, at a certain critical value, the

book starts to move relatively freely.

The maximum frictional force resists the onset of sliding.

This force is called the static friction force,s. Once the book

has begun to slide, a much lower force thans is needed to

keep the book moving. This force is called the kinetic friction

force, and is represented byk.

This phenomenon can be understood when it is realised

that even the smoothest surfaces are quite jagged at the

microscopic level. When the book is resting on the table, the

jagged points of its bottom surface have settled into the valleys

of the surface of the table, and this helps to resist attempts to

try to slide the book. Once the book is moving, the surfaces do

not have any time to settle into each other, and so less force isrequired keep it moving.

Another fact that helps to explain friction arises from the

forces of attraction between the atoms and molecules of the

two different surfaces that are in contact. These produce weak

bonding between the particles within each material; before

one surface can move across the other, these bonds must

be broken. This extra effort adds to the static friction force.

Once there is relative motion between the surfaces, the bonds

cannot re-form.

In everyday life, there are situations in which friction is

desirable (e.g. walking) and others in which it is a definite

problem. Consider the moving parts within the engine of a

car. Friction can rob an engine of its fuel economy and cause

it to wear out. Special oils and lubricants are introduced in

order to prevent moving metal surfaces from touching. If the

moving surfaces actually moved over each other they would

quickly wear, producing metal filings that could damage the

engine. Instead, both metal surfaces are separated by a thin

layer of oil. The oils are chosen on the basis of their viscosity

(thickness). For example, low viscosity oils can be used in

the engine while heavier oils are needed in the gear box and

differential of the car where greater forces are applied to the

moving parts.

At other times, we want friction to act. When driving to

the snowfields, if there is any ice on the road, drivers are

required to fit chains to their cars. When driving over a patch

of ice, the chain will break through the ice, and the car is

again able to grip the road. Similarly, friction is definitely

required within the car’s brakes when the driver wants to

slow down. In fact modern brake-pads are specially designed

to maximise the friction between the pads and the brake drum

or disk.

When a car is braking in a controlled fashion, the brake-

pads grip a disk that is attached to the wheel of the car. The

retarding force, applied through friction, slows the disk and

hence the car will come to rest. If the brakes are applied

too strongly they may grab the disk, locking up the wheels

in the process. The car then slides over the road, with two

undesirable consequences. First, the car usually takes about

20% longer to come to rest. This is because the car is relying

on the kinetic frictional force between the tyres and the road

to stop. As was seen when pushing the book over the desk,

this force is less than the static friction force. The other

consequence is that the car has lost its grip with the road,

and so the driver can no longer steer the car. Most cars now

employ anti-lock braking systems (ABS) to overcome the

possibility of skidding. This is achieved by using feedback

systems that automatically reduce the pressure applied by the

brake-pads regardless of the pressure applied by the driver to

the brake pedal.

Frictional forces

Physics in action

Figure5.41 To get things moving, the static friction between anobject and the surface must be overcome. This requires a largerforce than that needed to maintain constant velocity.

maximumstatic friction

kinetic friction

Applied force

Time

overcoming friction constant velocity

motion

F s

F s

F k

F k

P H Y S IC S

1 1

P H Y S IC S

1 1

Figure 5.42 This magnetic levitation train in China rides 1 cm abovethe track, so the frictional forces are negligible. The train is propelled bya magnetic force to a cruising speed of about 430 km h−1.



Motion174

• Newton’s third law of motion recognises that forces

exist in pairs and states that for every action force,

there is an equal and opposite reaction force:

F (A on B) = − F (B on A)

• Action and reaction forces act on different objectsand so should never be added together.

• Whenever a force acts against a fixed surface, the

surface provides a normal force, F N or N , at right

angles to the surface. The size of the normal force

depends on the orientation of the surface to the

contact force.

• All locomotion is made possible through the

existence of action and reaction force pairs.

• On smooth inclined planes at an angle of θ to the

horizontal, objects will move with an acceleration of

= g sin θ.• Materials that have been stretched, such as ropes,

cables and rubber bands, exert tensile forces on the

objects to which they are attached. These forces are

equal and opposite.

Newton’s third law of motion5.4 summary

11 Determine the action and reaction forces involved

when:

a a tennis ball is hit with a racquet

b a pine cone falls from the top of a tree towards the

ground

c a pine cone lands on the ground

d the Earth orbits the Sun.

22 A 70 kg fisherman is quietly fishing in a 40 kg dinghy

at rest on a still lake when, suddenly, he is attacked

by a swarm of wasps. To escape, he leaps into

the water and exerts a horizontal force of 140 N on

the boat.

a What force does the boat exert on the fisherman?

b With what acceleration will the boat move

initially?

c If the force on the fisherman lasted for 0.50 s,

determine the initial speed attained by both the

man and the boat.

33 A 100 kg astronaut (including the space suit)

becomes untethered during a space walk and drifts

to a distance of 10 m from the mother ship. To get

back to the ship, he throws his 2.5 kg tool kit away

with an acceleration of 8.0 m s−2 that acts over 0.50 s.

a How does throwing the tool kit away help theastronaut in this situation?

b How large is the force that acts on the tool kit and

the astronaut?

c With what speed will the astronaut drift to the

mother ship?

d How long will it take for the astronaut to reach the

ship?

44 A 2.0 kg bowl strikes the stationary ‘jack’, which

has a mass of 1.0 kg, during a game of bowls. It is a

head-on collision, and the acceleration of the jack is

found to be 25 m s−2 north. What is the acceleration

of the bowl?

55 a A ball rolls down an incline as shown in diagram

(a). Which one of the following best describes the

speed and acceleration of the ball?

(a) (b)

A The speed and acceleration both increase.

B The speed increases and the acceleration is

constant.

C The speed is constant and the acceleration is

zero.

D The speed and acceleration are both constant.

b A ball rolls down the slope shown in diagram(b). Which one of the following best describes its

speed and acceleration?

A Its speed and acceleration both increase.

B Its speed and acceleration both decrease.

C Its speed increases and its acceleration

decreases.

D Its speed decreases and its acceleration

increases.

Newton’s third law of motion5.4 questions




66 During the Winter Olympics, a 65 kg competitor in

the women’s luge has to accelerate down a course

that is inclined at 50° to the horizontal.

a Name the forces acting on the competitor.

b Ignoring friction (because it’s an icy slope), deter-mine the magnitude and direction of the forces

that act.

c Determine the magnitude of the net force on the

competitor.

d What acceleration will the competitor experience?

77 A cyclist is coasting down a hill that is inclined at

15° to the horizontal. The mass of the cyclist and her

bike is 110 kg, and for the purposes of the problem,

no air resistance or other forces are acting. After

accelerating to the speed limit, she applies the brakes

a little. What braking force is needed for her to be

able to travel with a constant velocity down the hill?

88 Discuss and compare the size of the normal force

that acts on the skater as he travels down the half-

pipe from A to B to C as shown.

A

B

Cx

x

x

99 Two students, James and Tania, are discussing the

forces acting on a lunchbox that is sitting on the

laboratory bench. James states that a weight force

and a normal force are acting on the lunchbox and

that since these forces are equal in magnitude butopposite in direction, they comprise a Newton’s

third law action–reaction pair. Tania disagrees saying

that these forces are not an action–reaction pair. Who

is correct and why?

110 A rope is allowed to move freely over a ‘frictionless’

pulley backstage at a theatre. A 30 kg sandbag,

which is at rest on the ground, is attached at one

end. A 50 kg work-experience

student, standing on a ladder,

grabs onto the other end of the

rope to lower himself.

a When the student steps off

the ladder, what is the net

external force on the rope?

b With what acceleration will

the system move?

c What is the tension in the

rope?

1 A boxer receives a punch to the head during a training session.

His opponent is wearing boxing gloves. Which of the following is

correct?

A The force that the glove exerts to the head is greater than the

force that the head exerts on the glove.B The force that the glove exerts to the head is less than the

force that the head exerts on the glove.

C The force that the glove exerts to the head is always equal to

the force that the head exerts on the glove.

D None of the above is correct.

2 When pushing a shopping trolley along a horizontal path, James

has to continue to provide a force of 30 N just to maintain his

speed. If the trolley (and shopping) has a mass of 35 kg, what is

the total horizontal force that he will have to provide to accelerate

the cart at 0.50 m s−2?

3 A force of 25 N is applied to a 6.0 kg ten-pin bowling ball for 1.2 s.

If the ball was initially at rest, what is its final speed?

4 A parachutist of mass 60 kg is falling vertically towards the

ground with a constant speed of 100 km h–1.

a Calculate the size of the drag force acting on the parachutistat this speed.

b The parachutist pulls the ripcord and the chute opens,

slowing the parachutist to 20 km h –1. How does the drag

force compare to the weight force during this period?

c During the final stage of the skydive, the parachutist

continues towards the ground at 20 km h –1. How does the

drag force compare to the weight force during this period?

Chapter reviewNewton’s laws

30 kg

50 kg

Continued on next page

Worked Solutions



Motion176

5 Jane has a mass of 55 kg. She steps into a lift which goes up

to the second floor. The lift accelerates upward at 2.0 m s−2 for

2.5 s, then travels with constant speed.

a What is the maximum speed that the lift attains as it travels

between floors?

b What is Jane’s weight:

i when the lift is stationary?

ii when the lift is accelerating upwards?

6 a What is the mass of an 85 kg astronaut on the surface of

Earth where g = 9.8 m s−2?

b What is the mass of an 85 kg astronaut on the surface of the

Moon where g = 1.6 m s−2?

c What is the weight of an 85 kg astronaut on the surface of

Mars where g = 3.6 m s−2?

7 The series of photographs shows a stack of smooth blocks in a

tall pile. One of the blocks in the pile is struck by a hammer andthe blocks above it fall onto the block below, and the pile remains

standing. Explain this in terms of Newton’s laws of motion.

8 A force of 120 N is used to push a 20 kg shopping trolley along

the line of its handle—at 20° down from the horizontal. This is

enough to cause the trolley to travel with constant velocity to the

north along a horizontal path.

a Determine the horizontal and vertical components of the

force applied to the trolley.

b What is the value of the frictional force acting against the

trolley?

c How large is the normal force that is supplied by the ground

on which the trolley is pushed?

d Why is it often easier to pull rather than push a trolley?

9 A 100 g glider is at rest on a horizontal air track, and a force is

applied to it as shown in the following graph. What will be its

speed at the end of the time interval?

10 The following diagrams show force vectors on a puck travelling

at constant speed across an air table in a games arcade. The

puck experiences no friction as it moves across its cushion of

air. Which diagram A–D correctly shows the forces that act on

the puck?

Applied

force(N)

0.5

0.2

021

t (s)

A

B

C

D

F g

F N

F drag

F

F g

F N

F

F g

F N

F g

wton’s laws (continued)




11 Two shopping trolleys with masses 30 kg and 50 kg stand

together. A force of 120 N is applied to the 30 kg trolley.

a With what acceleration will the trolleys move?

b Calculate the size of the contact force that the 30 kg trolley

exerts on the 50 kg trolley.

12 A young girl of mass 40 kg leaps horizontally off her stationary

10 kg skateboard. Assuming that no frictional forces are involved,

determine the following ratios:

a horizontal force on the girlhorizontal force on the skateboard

b acceleration of the girl

acceleration of the skateboard

c final velocity of the girl

final velocity of the skateboard

13 A rope has a breaking tension of 100 N. How can a full bucket

of mass 12 kg be lowered using the rope, without the rope

breaking?

14 The force diagram below shows the forces acting on a full

water tank.

F N

F g

a Are these forces an action–reaction pair as described by

Newton’s third law?

b Justify your answer to part a.

15 A car begins to roll down a steep road that has a grade of 1 in 5

(i.e. a 1 m drop for every 5 m in length). If retarding forces are

ignored, determine the speed of the car in km h−1 after it has

travelled a distance of 100 m if it begins its journey at rest.

16 A small boy’s racing set includes an inclined track along which a

car accelerates at1

2 g (i.e. 4.9 m s−2). At what angle is the track

to the horizontal?

17 When skiing down an incline, Eddie found that there was a

frictional force of 250 N acting up the incline of the mountainside

due to slushy snow. The slope was at 45° to the horizontal, so if

Eddie had a mass of 70 kg, what was his acceleration?

18 On a sketch, draw vectors to indicate the forces that act on a

tennis ball:

a at the instant it is struck

b an instant after it has been struck.

19 Two masses, 5.0 kg and 10.0 kg, are suspended from the ends

of a rope that passes over a frictionless pulley. The masses

are released and allowed to accelerate under the influence of

gravity. What is the acceleration of the system, and what is the

tension in the rope?

20 Tim and Julia are discussing a couple of physics problems over

dinner.

a First, they discuss a collision between a marble and a billiard

ball. Tim argues that since the billiard ball is much heavier

than the marble, it will exert a larger force on the marble

than the marble exerts on it. Julia thinks that the marble

and billiard ball will exert equal forces on each other as theycollide. Who is correct? Explain.

b Then they discuss a basketball as it bounces on a concrete

floor. Tim claims that the ball must exert a smaller force on

the floor than the floor exerts on it, otherwise the ball would

not rebound. Julia thinks that the ball and the floor will exert

forces on each other that are equal in magnitude. Who is

correct this time? Explain.

120 N30 kg 50 kg

Chapter QuizWorked Solutions



Momentum, energy,work and power6

CHAPTER

Is skydiving on your list of things to do in your future? Base-

jumping? Mountain boarding? Are you a person who would love to

experience the exhilaration of taking that leap out of a plane, or do

you question why someone would choose to jump out of a perfectly safe

aeroplane in mid-flight? When an aeroplane is climbing to the required

height for the thrill-seekers, we say that the aeroplane’s engines are

doing work against gravity. When the parachutist takes the jump, we

say that gravity is doing work on the parachutist. The search for the

ultimate extreme-sport thrill is often about ‘taking on gravity’. Whether

it is snowboarding, BASE-jumping or bungee jumping, the participant

is experimenting with the conversion of gravitational potential energy

into kinetic energy.

Throughout this chapter you will be able to see the common threadof energy conversion that is present in so many of our activities. In

our everyday lives we try to harness and transform various forms

of energy as efficiently and cleverly as possible. We burn up our own

personal energy stores as we climb up the steps to the classroom. We

make sure the tennis ball hits the ‘sweet spot’ on our racquet when

we hit it back to our opponent. In more thrill-seeking adventures we

also make the most of our understanding of energy transformations.

Converting lots of gravitational potential energy into kinetic energy can

be an extreme experience and great fun, as shown in the photograph.

Don’t be deceived though, the laws of physics cannot be switched off!

you will have covered material from the study of

movement including:

• momentum and impulse

• work done as a change in energy

• Hooke’s law

• kinetic, gravitational and elastic potential energy

• energy transfers

• power.

by the end of this chapter



Chapter 6 Momentum, energy, work and power 17

6.1 The relationship between momentum and force

Consider a collision between two footballers on the football field. From

Newton’s second law, each force can be expressed as:

Σnet

= m

and using the relationship for acceleration:

= −

∆t Σ =

m( − )

∆twhere a is the acceleration during the collision (m s−2)

∆t is the time of contact (s)

u is the velocity of either one of the footballers before the collision (m s−1)

v is the velocity of the footballer after the collision (m s−1).

Rearranging:

Σ ∆t = m( − )

or:

Σ ∆t = m∆

This relationship introduces two important ideas.

• The product of net force and the contact time is referred to as impulse,

. The idea of impulse is commonly applied to objects during collisionswhen the time of contact is small. This concept will be further explained

later.

• The product of the mass of an object and its velocity is referred to as

momentum:

=

where is momentum (kg m s−1)

m is the mass of the object (kg)

is the velocity of the object (m s−1).

Momentum

Momentum can be thought of as the tendency of an object to keep movingwith the same speed in the same direction. It is a property of any moving

object. As it is the product of a scalar quantity (mass) and a vector quantity

(velocity), momentum is a vector quantity. The direction of the momentum

of an object is the same as the direction of the velocity of that object. The

unit for momentum is kg m s−1, which is readily determined from the

product of the units for mass and velocity.

Momentum often indicates the difficulty a moving object has in

stopping. A fast-moving car has more momentum than a slower car of

the same mass; equally so, an elephant will have more momentum than

a person travelling at the same speed (just as a greater force is needed to

cause the same acceleration). The more momentum an object gains as its

velocity increases, the more it has to lose to stop and the greater the effect

it will have if involved in a collision. A football player is more likely to be

knocked over if tackled by a heavy follower than a light rover, since the

product p = mv will be larger for the heavy follower.

Although he used different language, Newton understood this idea, and

his second law of motion can be stated in terms of momentum.

Figure6.1 When two footballers collide, theyexert an equal and opposite force on eachother. The effect this force will have on thevelocity of each footballer can be investigatedusing the concept of momentum.

Figure6.2 The enormous mass of a large shipendows it with very large momentum despiteits relatively slow speed. After turning off itsengines the ship can continue against theresistance of the water for more than 4 km ifno other braking is applied.

Interactive



Motion180

That is:

Σ = ∆ p

∆t

where Σ is the average net force applied to the object during the collision,

in newtons (N)

∆p is the change in momentum during contact for a time ∆t.

An unbalanced net force is required to change the momentum of an

object—to increase it, decrease it or change its direction. This force might

result from a collision or an interaction with another object. The change in

momentum (∆p) of the object will be given by:

Worked example 6.1A

A footballer trying to take a mark collided with a goal post and came to rest. The footballer

has a mass of 80 kg and was travelling at 8.2 m s −1 at the time of the collision.

a What was the change in momentum during the collision for the footballer?

b Estimate the average force the footballer experienced in this collision.

Solution

a Prior to the collision the footballer’s momentum was given by:

p = mv

= 80 × 8.2

= 656 kg m s−1

towards the pole After the collision the momentum was zero since the footballer stopped moving. So:

∆ p = 0 − 656 = −656 kg m s−1 towards the pole

or:

∆ p = 6.6 × 102 kg m s−1 away from the pole

b The negative value for the change in momentum indicates that the direction of the

momentum, and hence the force applied to the footballer, is opposite to the direction in

which the footballer was travelling. The time that the footballer took to stop has not been

given but a reasonable estimate of the force can be made by estimating the stopping

time. Keeping to magnitudes of 10 for easy working, it would be reasonable to assume

that the stopping time in this sort of collision would be less than 1 s (10 0) and greater

than 0.01 s (10−2). Something in the order of 0.1–0.5 s (10−1) would make sense on

the basis of observations of similar situations. Then using:

F = ∆ p

∆t

F = 656

10−1 = 6560 ≈ 7 × 103 N away from the pole, i.e. a retarding force.

The change in momentum of a body is proportional to the net force applied to it:∆ p ∝ ΣF

i

Change in momentum = final momentum − initial momentum∆ p = p

f − p

i

i



18Chapter 6 Momentum, energy, work and power

ImpulseThink about what it feels like to fall onto a concrete floor. Even from a small

height it can hurt. A fall from the same height onto a tumbling mat is barely

felt. Your speed is the same, your mass hasn’t changed and gravity is still

providing the same acceleration. So what is different about the fall onto the

mat that reduces the force you experience?

Remember that, according to Newton’s second law of motion, the

velocity of an object only changes when a net force is applied to that object.

A larger net force will be more effective in creating a change in the velocity

of the object. The faster the change occurs (i.e. a smaller time interval ∆t),

the greater the net force that is needed to produce that change. Landing on

a concrete floor changes the velocity very quickly as you are brought to an

abrupt stop. When landing on a tumble mat the change occurs over a much

greater time. The force needed to produce the change is smaller.

Another illustration of this could be a tennis player striking a ball with

a racquet. At the instant the ball comes in contact with the racquet the

applied force will be small. As the strings distort and the ball compresses,

the force will increase until the ball has been stopped. The force will thendecrease as the ball accelerates away from the racquet. A graph of force

against time will look like that in Figure 6.3.

The impulse affecting the ball at any time will be the product of applied

force and time, i.e. = av

∆t. The total impulse during the time the ball is in

contact will be = av

× t, whereav

is the average force applied during the

collision and t is the total time the ball is in contact with the racquet. This

is equivalent to the total area under the force–time graph. The total impulse

for any collision can be found in this way.

The concept of impulse is appropriate when dealing with forces during

any collision since it links force and contact time, for example a person

hitting the ground, as described above, or a ball being hit by a bat or

racquet. If applied to situations where contact is over an extended time, the

average net force involved is used since the forces are generally changing

(as the ball deforms for example). The average net applied force can be

found directly from the formula for impulse. The instantaneous applied

force at any particular time during the collision must be determined from a

graph of the force against time.

The relationship between impulse and

momentumFrom the derivation earlier in the chapter, the impulse is also equal to the

change of momentum for an object. Previously we had:

= m∆

∆t

The IMPULS… affecting an object during a collision is the product of the netaverage applied force and the time of contact and is equivalent to the areaunder a force–time graph.

i

Figure6.3 (a) When a tennis player hits a ball, anunbalanced force is applied to the ball, producinga change in its momentum; hence an impulse isapplied to the ball. The magnitude of the force willchange over time. (b) The impulse can be foundfrom the area under the force–time graph sincearea = x -axis × y -axis = F

av × ∆t = impulse.

(a)

t (s)0 0.150.120.090.060.03

Impulse = F av × t

= area under graph

F

( N )

(b)



Motion182

Multiplying by the time interval ∆t:

av

∆t = m∆

or:

= av

∆t = ∆p

The units for impulse (N s) and the units for momentum (kg m s−1) have

each been introduced separately. Since we have just seen that impulse is

equal to the change in momentum:

1 N s = 1 kg m s−1

Worked example 6.1B

A tennis racquet applies a force to a tennis ball for a period of 0.15 s, bringing the ball

(momentarily) to a halt. The tennis ball has a mass of 58 g and was originally travelling

towards the racquet at 55 m s−1.

a Find the change in momentum as the ball is momentarily brought to a halt by the

racquet.

b Find the magnitude of the impulse during this part of the collision.

c Find the average force applied during the time it takes to stop the ball.

Solution

a Initial momentum: pi = mu = 0.058 × 55= 3.19 kg m s−1

Final momentum: pf = mv = 0.058 × 0 = 0 kg m s−1

Change in momentum:∆ p = 0 − 3.19 = −3.19 kg m s−1 in the direction of travel,

i.e. ∼3.2 kg m s−1 in the opposite direction.

b Impulse = change in momentum:

I = 3.19 ≈ 3.2 N s

c Using I = F av∆t

then F av

= I

∆t

so F av = 3.19

0.15 = 21.27 N ≈ 21 N in the opposite direction to the ball’s travel.

Try this simple experiment. Grab a can of fruit or similar

relatively soft non-corrugated steel can. Place your finger flat

on a bench top and, carefully avoiding the can’s seam, bring

the side of the can crashing down on your finger. (We take no

responsibility for you using the wrong part of the can!)

Were you actually game to try it? If you did, how much did it

hurt? Not nearly as much as you expected, right? Why?

Bringing a rigid hammer down on your finger in similar

circumstances would have caused considerable damage to the

finger. Yet the can crumpled in around your finger and, even

though it had a similar mass to the hammer and travelled at

a reasonable velocity, it caused no damage to the finger and

little pain. This observation can be explained with the concept

of impulse.

By assuming a mass of 500 g for both hammer and can

and an impact speed of 20 m s−1, the magnitude of the change

in momentum, and hence impulse, can be estimated.

Forces during collisions

Physics in action

Figure6.4 A simple example of the effect on the applied force of thestopping time during a collision can be achieved with nothing morecomplicated than a hammer, a can of fruit and your finger. (a) A rigidobject, such as the hammer, will stop quickly. The applied force will belarge. (b) A can will experience the same change in momentum, but,having a simple crumple zone, will stop more slowly, thus reducingthe applied force to a tolerable amount.

(a) (b)




I = ∆ p = m∆v

= 0.5 × 20 = 10 N s

The hammer, being rigid, will quickly come to a stop. This

time can be estimated at about 0.1 s, so:

F =

I

∆t =

10

0.1 = 100 N

—a considerable force that could be expected to do some

damage to the finger!

The can is able to compress and so the stopping time will

be somewhat longer, say around 0.5 s. The average force

will be:

F = I

∆t =

10

0.5 = 20 N

Increasing the stopping time by five times has reduced

the average force applied to the finger to one-fifth that

applied by a rigid object to something which, while perhaps

not totally pain-free, is quite tolerable and will do no real

damage. The applied force is inversely proportional to the

stopping time. Increase the stopping time and the applied

force is decreased. Try it on a friend and see if you can provethis bit of physics to them!

This simple idea is the basis upon which the absorbency

systems of sports shoes, crash helmets, airbags and crumple

zones of cars, and other safety devices are designed.

Walking, running and sports shoe design

As athletes walk or run, they experience action–reaction

forces due to gravity, the surface of the track and the air

around them. These forces have been investigated in some

detail in the previous chapter.

The force the ground exerts on a runner creates a change

in momentum as the runner’s feet strike the ground. This

force can be quite large and cause considerable damage to the

runner’s ankles, shins, knees and hips as it is transmitted up the bones of the leg. Jogging in bare feet can increase the

forces experienced to nearly three times that applied when

simply standing still. Table 6.1 lists the relative size of some

forces associated with common movements in sport.

An understanding of the forces generated and the

elastic properties of materials are used by designers in the

development of athletics tracks, playing surfaces and sports

shoes. Elastic materials can reduce the forces developed

between foot and track by increasing the stopping time. Based

on an understanding of impulse, sports shoes are designed

with soles that include gels, air and cushioning grids, which

extend the stopping time and thus reduce the force applied

to the runner’s body. Sophisticated modern sports shoes,

properly fitted to suit the wearer, have substantially reducedthe size and effect of forces on the runner, with consequential

benefits for the runner’s knees and hips.

The running surface can also be designed to minimise

the forces and produce fewer injuries. Cushioned surfaces

reduce the impact considerably. Grass is actually quite

effective but can sometimes be too spongy. The extra time

spent rebounding from the surface slows the runner down.

The response must be quick if good running times are to be

achieved. As a result artificial surfaces such as polyurethane,

‘AstroTurf’ and ‘Rebound Ace’ have become popular.

They offer good cushioning but are more responsive, allowing

faster take-offs than grass.

Vehicle safety designDesigning a successful car is a complex task. A vehicle must

be reliable, economical, powerful, visually appealing, secure

and safe. Public perception of the relative importance of these

issues varies. Magazines and newspapers concentrate on

appearance, price and performance. The introduction of air-

bag technology into most cars has altered the focus towards

safety. Vehicle safety is primarily about crash avoidance.

Research shows potential accidents are avoided 99% of

the time. The success of accident avoidance is primarily

attributable to accident avoidance systems such as antilock

brakes. When a collision does happen, passive safety features

come into operation, for example the air bag. Understanding

the theory behind accidents involves an understanding

primarily of impulse and force.

Table 6.1 Relative size of forces associated withsome common movements in sport

Movement Footwear Ratio of normal toweight force

Standing still Barefoot or shoes 1.0

Walking Barefoot 1.6

Jogging Barefoot 2.9

Jogging Running shoes 2.2

Sprinting Barefoot 3.8

Fast bowling Cricket spikes 4.1

Long jump take-off Athletic spikes 7.8

Figure6.5 The forces developed between track and foot can doconsiderable damage to a runner’s body unless they are reduced byincreasing the stopping time through cushioning of the foot. Moderntrack shoes incorporate sophisticated design principles to increasestopping time and thus decrease the forces generated.




Motion184

The air bag

Seat belts save lives. They also cause injuries. Strangely, the

number of people surviving an accident but with serious

injury has increased since the introduction of safety belts.

Previously many of today’s survivors would have died

instantly in the accident. A further safety device is required

to minimise these injuries.

The air bag in a car is designed to inflate within a few

milliseconds of a collision to reduce secondary injuries

during the collision. It is designed to inflate only when the

vehicle experiences an 18–20 km h−1 or greater impact with

a solid object. The required deceleration must be high or

accidental nudges with another car would cause the air bag

to inflate. The car’s computer control makes a decision in a

few milliseconds to detonate the gas cylinders that inflate

the air bag. The propellant detonates and inflates the air bag

while the driver collapses towards the dashboard. As the body

lunges forwards into the air bag, the bag deflates, allowing

the body to slow in a longer time as it moves towards the

dashboard. Injury is thus minimised.

Calculating exactly when the air bag should inflate, and

for how long, is a difficult task. Many cars have been crash

tested and the results painstakingly analysed. High-speed film

demonstrates precisely why the air bag is so effective. During

a collision the arms, legs and heads of the occupants are

restrained only by the joints and muscles. Enormous forces

are involved because of the large deceleration. The shoulders

and hips can, in most cases, sustain the large forces for the

short duration. However, the neck is the weak link. Victims

of road accidents regularly receive neck and spinal injuries.

An air bag reduces the enormous forces the neck must

withstand by extending the duration of the collision, a direct

application of the concept of impulse.

The extent of injuries during a collision is not only

dependent on the size of the force but also the duration

and deflection resulting from the applied force. An increase

in localised pressure will result in a greater compressionor deflection of the skull. The air bag reduces the

localised pressure by increasing the contact surface area

and decreasing the force. The effect can be seen by the

relationship:

= A

where is the pressure (N m−2)

is the force (N)

A is the contact area (m2).

An air bag has a contact surface area of about 0.2 m2

compared with 0.05 m2 for a seat belt. This reduces injuries

caused by seat belts, such as bruising and broken ribs and

collar bones, since it increases the stopping time. It alsosupports the head and chest, preventing high neck loads

caused by the seat belt restraining the upper torso. Most

importantly, it prevents the high forces caused by contact of

the head with the steering wheel. The air bag ensures that

the main thrust of the expansion is directed outwards instead

of towards the driver. The deflation rate, governed by the size

of the holes in the rear of the air bag, provides the optimum

deceleration of the head for a large range of impact speeds.

The air bag is not the answer to all safety concerns

associated with a collision, but it is one of many safety

features that form a chain of defence in a collision.

Forces during collisions (continued)

A p p l i e d

f o r c e ( N

)

Time (ms)0 20 40 60 80 100 120 140

no air bag

with air bag

(a)

(b)

Figure6.6 (a) The air bag is one of a number of passive safetyfeatures incorporated into the design of modern cars. It extendsthe stopping time, significantly reducing the forces on the headand neck during a collision. It also distributes the force required todecelerate the mass of the driver or passenger over a larger areathan a seat belt. (b) The deflation rate of the bag is governed by thesize of holes in the rear of the air bag, and is designed to providethe optimum deceleration of the head for a large range of impacts.




• The momentum of a moving object is the product of

its mass and its velocity:

p = mv

where p is in kg m s−1

m is in kgv is in m s−1

Momentum is a vector quantity.

• The change in momentum (∆p) = final momentum

(pf ) − initial momentum (p

i).

• Impulse is the product of the net force during a

collision and the time interval ∆t during which the

force acts: I = F av

∆t. It can also be found from the

area under a force–time graph and is measured in

newton seconds (N s). Impulse is equal to the change

in momentum, ∆p, caused by the action of the net

applied force: F

av∆t = ∆p

• Extending the time over which a collision occurs will

decrease the average net force applied since F av

∝ 1

∆t.

This is the principle behind many safety designs.

The relationship between momentum and force6.1 summary

Use g = 9.8 m s−2 where required.

11 What is the momentum, in kg m s−1, of a 20 kg cart

travelling at:

a 5.0 m s−1?

b 5.0 cm s−1?

c 5.0 km h−1?

22 The velocity of an object of mass 8.0 kg increases

from an initial 3.0 m s−1 to 8.0 m s−1 when a force acts

on it for 5.0 s.

a What is the initial momentum?

b What is the momentum after the action of the

force?c How much momentum is the object gaining each

second when the force is acting?

d What impulse does the object experience?

e What is the magnitude of the force?

33 Which object has the greater momentum—a

medicine ball of mass 4.5 kg travelling at 3.5 m s−1 or

one of mass 2.5 kg travelling at 6.8 m s−1?

44 Calculate the momentum of an object:

a of mass 4.5 kg and velocity 9.1 m s−1

b of mass 250 g and velocity 3.5 km h−1

c that has fallen freely from rest for 15 s and has amass of 3.4 kg

d that experiences a net force of magnitude 45 N, if

the net force is applied for 3.5 s.

55 A tennis ball may leave the racquet of a top player

with a speed of 61 m s−1 when served. If the mass of

the ball is 65 g and it is actually in contact with theracquet for 0.032 s:

a what momentum does the ball have on leaving

the racquet?

b what is the average force applied by the racquet

on the ball?

66 A 200 g cricket ball (at rest) is struck by a cricket bat.

The ball and bat are in contact for 0.05 s, during which

time the ball is accelerated to a speed of 45 m s−1.

a What is the magnitude of the impulse the ball

experiences?

b

What is the net average force acting on the ballduring the contact time?

c What is the net average force acting on the bat

during the contact time?

77 The following graph shows the net vertical force

generated as an athlete’s foot strikes an asphalt

running track.

F

o r c e ( N )

1400

1200

1000

800

600

400

200

0 10 20 30 40 50 60 70

Time (ms)

a Estimate the maximum force acting on the

athlete’s foot during the contact time.

The relationship between momentum and force6.1 questions




Motion186

b Estimate the total impulse during the contact

time.

88 A 25 g arrow buries its head 2 cm into a target on

striking it. The arrow was travelling at 50 m s−1 just

before impact.

a What change in momentum does the arrow

experience as it comes to rest?

b What is the impulse experienced by the arrow?

c What is the average force that acts on the arrow

during the period of deceleration after it hits the

target?

99 Crash helmets are designed to reduce the force of

impact on the head during a collision.

a Explain how their design reduces the net force on

the head.

b Would a rigid ‘shell’ be as successful? Explain.

110 Describe, with the aid of diagrams, a simple collision

involving one moving object and one fixed in

position. Estimate, by making reasonable estimates

of the magnitudes of the mass and velocity of the

moving object, the net force acting on the objects

during the collision.

The relationship between momentum and force (continued)

Worked Solutions




Conservation of momentum6.2

The most significant feature of momentum is that it is conserved. This

means that the total momentum in any complete system will be constant.

For this reason momentum is very useful in investigating the forces

experienced by two colliding objects—as long as they are unaffected by

outside forces. The law of conservation of momentum, as it is known, is

derived from Newton’s third law.From Newton’s third law, the force applied by each object in a collision

will be of the same magnitude but opposite in direction:

1 = −

2

From Newton’s second law, Σ = ma, so the forces could be expressed as:

m1a

1 = −m

2a

2

and using a = v − u

∆t we get:

m1

v1 − u

1

∆t = −m

2

v2 − u

2

∆t

where a1 and a

2 are the respective accelerations of the two objects during

the collision in m s−2

∆t is the time of contact (s) u

1 and u

2 are the velocities of the objects prior to collision (m s−1)

v1 and v

2 are the velocities after collision (m s−1)

Since the time that each object is in contact with the other will be the same,

∆t will cancel out:

m1(v

1 − u

1) = −m

2(v

2 − u

2)

or:

m1u

1 + m

2u

2 = m

1v

1 + m

2v

2

In other words, the total momentum before colliding is the same as the total

momentum after the collision.

It is most important to realise that momentum is only conserved in an

isolated system; that is, a system in which no external forces affect the

objects involved. The only forces involved are the action–reaction forces on

the objects in the collision. Consider two skaters coming together on a near-frictionless ice rink. In this near-ideal situation it is realistic to apply the

law of conservation of momentum. The only significant horizontal forces

between the two skaters are those of the action–reaction pair as the two

skaters collide.

If the skaters were to skate through a puddle of water as they come

together then friction would become noticeable. This force is an external

force since it is not acting between the two skaters. The interaction between

TH… LAW OF CONS…RVATION OF MOM…NTUM states that, in any collision orinteraction between two or more objects in an isolated system, the totalmomentum of the system will remain constant; that is, the total initialmomentum will equal the total final momentum:

Σ pi = Σ p

f

orm

1u

1 + m

2u

2 = m

1v

1 + m

2v

2

i

The principle of conservation of

momentum was responsible for the

interpretation of investigations that led

to the discovery of the neutron. Neutral

in charge, the neutron could not beinvestigated through the interactions

of charged particles that had led to the

discovery of the proton and electron. In

1932 Chadwick found that in collisions

between alpha particles and the element

beryllium, the principle of conservation

of momentum only held true if it could

be assumed that there was an additional

particle within the atom, which had

close to the same mass as a proton

but no electric charge. Subsequent

investigations confirmed his experiments

and led to the naming of this particle as

the neutron.

Physics file

If you release an inflated rubber

balloon with its neck open, it will fly

off around the room. In the diagram

below, the momentum of the air to the

left is moving the balloon to the right.

Momentum is conserved.

air balloon

This is the principle upon which

rockets and jet engines are based.

Both rockets and jet engines employ

a high-velocity stream of hot gases

that are vented after the combustion

of a fuel–air mixture. The hot exhaust

gases have a very large momentum as

a result of the high velocities involved,

and can accelerate rockets and jets to

high velocities as they acquire an equalmomentum in the opposite direction.

Rockets destined for space carry their

own oxygen supply, while jet engines use

the surrounding air supply.

Physics file



Motion188

a skater and the puddle would constitute a separate isolated system (as

would that between puddle and ice, ice and ground etc.). Momentum

would still be conserved within this other separate system. It is virtually

impossible to find a perfectly isolated system here on Earth because of the

presence of gravitational, frictional and air-resistance forces. Only where

any external forces are insignificant in comparison to the collision forces is

it reasonable to apply the law of conservation of momentum.Also important is that in any collision involving the ground, Earth

itself must be part of the system. Theoretically, any calculation based on

conservation of momentum should include the Earth as one of the objects,

momentum only being conserved when all the objects in the system are

considered. In practice, the very large mass of the Earth in relation to the

other objects involved means that there is a negligible change in the Earth’s

velocity and it can be ignored in most calculations. Of course, a collision

with a fast-moving asteroid would be another matter!

Worked example 6.2A

Skater 1 in Figure 6.7, with mass 80 kg, was skating in a straight line with a velocity of

6.0 m s−1 while the skater 2, of mass 70 kg, was skating in the opposite direction, also witha speed of 6.0 m s−1.

a The two skaters collide and skater 1 comes to rest. Assuming that friction can be ignored,

what will happen to the skater 2 after the collision?

b What would happen if the two skaters had hung on to each other and stayed together

after the collision?

Solution

a From conservation of momentum:

Σ pi = Σ p

f

or m1u

1 + m

2u

2 = m

1v

1 + m

2v

2

and m1

= 80 kg, m2

= 70 kg

As both velocity and momentum are vector quantities, a positive direction should be

established and taken into consideration. Adopting the direction of motion of skater 1

as the positive direction:

u1 = 6.0 m s−1, u

2 = −6.0 m s−1, v

1 = 0 m s−1, v

2 = ?

Substituting into equation:

80 × 6.0 + 70× (−6.0) = 80 × 0 + 70× v 2

and v 2 = +0.86 m s−1.

The 70 kg skater bounces back in the opposite direction with a speed of 0.86 m s−1.

b Treating the two skaters as one mass after the collision:

80 × 6.0 + 70× (−6.0) = (80 + 70) × v 2

and now v 2 = +0.4 m s−1.

A different outcome after the collision results in a different velocity for each skater.There is no unique answer when applying the idea of conservation of momentum. The

final velocity of any object depends on what happens to all the objects involved in the

collision.

The law of conservation of momentum can be extended to any number

of colliding objects. The total initial momentum is found by calculating the

vector sum of the initial momentum of every object involved. The total final

momentum will then also be the vector sum of each separate momentum

involved. Separation into two or more parts after the ‘collision’ (interaction

is a better word since it does not have to be destructive), for example the

firing of a bullet, can also be dealt with in the same manner.

Figure 6.7 When two skaters collide on a near-frictionless skating rink they exert equal andopposite forces on each other. The total momentumof the two skaters before the collision will equalthe total momentum of the two skaters after thecollision. Because no other large horizontal forcesare involved other than those in the collision, thiscan be considered as an isolated system.

Figure 6.8 Newton’s cradle, or Newton’s ballsto some, is an instructive ‘executive toy’ basedon the principle of conservation of momentumextended over a number of objects.

6.0 m s–1 6.0 m s–1

Prac 25




A car is designed to keep its occupants safe. Unfortunately,

however, very little can be done to protect a pedestrian from

the onslaught of a 1400 kg car travelling at 60 km h −1. Bull

bars in residential areas are currently under review because

of the enormous damage they inflict on a pedestrian. The

effect on the pedestrian will depend on the person’s height and

mass, the height of the front of the oncoming vehicle, and the

speed, mass and shape of the vehicle.

Consider the following possibilities:

• a pedestrian being struck by a truck moving at 30 km h−1

• a pedestrian being struck by a car moving at 30 km h−1

• a pedestrian being struck by a cyclist moving at 30 km h−1.

The injuries to the pedestrian are due largely to the

change in the pedestrian’s momentum. Being hit by the cyclist

will obviously result in the least injury to the pedestrian

because of the lower momentum of the cycle and its rider. The

mass of the other vehicles is such that they have a far larger

momentum to impart to the pedestrian.

Consider the following situation. A 1400 kg car istravelling at 60 km h−1 when it strikes a stationary 70 kg

pedestrian. The pedestrian lands on the bonnet of the car and

travels with the car until it finally comes to a halt. Assuming

that frictional forces are minimal:

total momentum before the collision

= total momentum after the collision

Before the collision:

car

= m

= 1400 × (60 × 1000

3600)

= 2.3 × 104 kg m s−1

pedestrian

= 0 (pedestrian is stationary)

After the collision:

total

= 2.3 × 104 kg m s−1

= (mcar

+ mpedestrian

)

so

1470 = 2.3 × 104 kg m s−1

and

total

≈ 16 m s−1 or 57 km h−1

This means that the pedestrian accelerates from rest to

a speed of 57 km h−1 in the short duration of the collision. A

similar collision between the pedestrian and a cyclist travelling

at 30 km h−1 would result in a final speed of 5.2 m s−1 (19

km h−1). The speed of the car changes very little. The speed of

the cyclist is almost halved.

Antilock brakes, excellent road handling and reducedspeed limits in some areas reduce the likelihood of a vehicle

striking a pedestrian. Unfortunately accidents can still happen.

There are essentially two possibilities that can occur

when a pedestrian is struck by a car.

1. The pedestrian bounces off the front of the car

and is projected through the air. This type of motion tends

to happen when the vehicle is travelling relatively slowly.

The pedestrian is rapidly accelerated forwards to near

the velocity of the vehicle. Injuries occur to the pedestrian

when the car strikes and again when they land on the

ground. Leg and hip injuries are general in this form of

pedestrian–vehicle collision.

Head injuries result from the pedestrian colliding with

the road. Very little of a car’s design will alter the severity

of head injuries received.

Vehicles can be designed with a low, energy-absorbing

bumper bar to reduce knee and hip damage. If the

pedestrian’s knee strikes the bumper bar, knee damage

is very likely. Knees do not heal as well as broken legs. A

lower, energy-absorbing bar is, for this reason, preferable.

22. When a vehicle is moving very fast at the point of impact,

the pedestrian’s inertia acts against rapid acceleration.

The pedestrian does not initially move forward with the

same velocity as the vehicle. If the pedestrian remains in

approximately the same place, he or she will go either over

or under the car. The pedestrian will be either run over or

run under (i.e. the car goes under the pedestrian).

Being run over usually results in serious injury or

fatality. Massive head injuries occur as the pedestrian’shead strikes the ground. The relative height of the vehicle’s

bumper bar and the height of the pedestrian determines

whether they will be run over or under. Most bumper bars

are below adult waist level. Small children, however, have

much more chance of being run over as the height of the

bumper bar is relatively much higher.

If the pedestrian is run under, ‘passive’ safety features

of modern car design come into play. Removal of protruding

hood ornaments is essential since they can easily penetrate

the body of a person and cause enormous injuries. The

bonnet of a car acts as a good impact absorber, particularly

in comparison with the hard surface of the road. Bull bars,

however, can block the path of the pedestrian, making it

more likely that they be run over. Further, they have littleimpact-absorbing ability. It is, therefore, logical to ban bull

bars in residential areas.

Collisions and pedestrians

Physics in action

Figure 6.9 This Nissan car has a pop-up bonnet that has beendesigned to result in less damage to a pedestrian in a collision,by making space between the bonnet and the engine.



Motion190

• The law of conservation of momentum states that

in any collision or interaction between two or more

objects in an isolated system the total momentum

of the system will remain constant. The total initial

momentum will equal the total final momentum:

Σpi = Σp

f

• Conservation of momentum can be extended to

any number of colliding objects within an isolated

system.

Conservation of momentum6.2 summary

11 A white billiard ball of mass 100 g travelling at

2.0 m s−1 across a low-friction billiard table has a

head-on collision with a black ball of the same mass

initially at rest. The white ball stops while the black

ball moves off. What is the velocity of the black ball?

22 A girl with mass 50 kg running at 5 m s−1 jumps onto

a 4 kg skateboard travelling in the same direction at

1.0 m s−1. What is their new common velocity?

33 A man of mass 70 kg steps forward out of a boat and

onto the nearby river bank with a velocity, when he

leaves the boat, of 2.5 m s−1 relative to the ground.

The boat has a mass of 400 kg and was initially at

rest. With what velocity relative to the ground does

the boat begin to move?

44 A railway car of mass 2 tonnes moving along a

horizontal track at 2 m s−1 runs into a stationary train

and is coupled to it. After the collision the train and

car move off at a slow 0.3 m s−1. What is the mass of

the train alone?

55 A trolley of mass 4.0 kg and moving at 4.5 m s−1

collides with, and sticks to, a stationary trolley of

mass 2.0 kg. Their combined speed in m s−1 after the

collision is:

A 2.0 B 3.0 C 4.5 D 9.0

66 A superhero stops a truck simply by blocking it with

his outreached arm.

aa Is this consistent with the law of conservation of

momentum? Explain.

b Using reasonable estimates for the initial

speed and mass of the truck and the superhero,

demonstrate what will happen. Use appropriate

physics concepts.

77 A car of mass 1100 kg has a head-on collision with

a large four-wheel drive vehicle of mass 2200 kg,immediately after which both vehicles are stationary.

The four-wheel drive vehicle was travelling at

50 km h−1 prior to the collision in an area where the

speed limit was 70 km h−1. Was the car breaking the

speed limit?

88 A 100 g apple is balanced on the head of young

master Tell. William, the boy’s father, fires an arrow

with a mass of 80 g at the apple. It reaches the apple

with a velocity of 35 m s−1. The arrow passes right

through the apple and goes on with a velocity of

25 m s−1

. With what speed will the apple fly off the boy’s head? (Assume there is no friction between

apple and head.)

99 A space shuttle of mass 10 000 kg, initially at rest,

burns 5.0 kg of fuel and oxygen in its rockets to

produce exhaust gases ejected at a velocity of

6000 m s−1. Calculate the velocity that this exchange

will give to the space shuttle.

110 A small research rocket of mass 250 kg is launched

vertically as part of a weather study. It sends out

50 kg of burnt fuel and exhaust gases with a velocity

of 180 m s−1

in a 2 s initial acceleration period.a What is the velocity of the rocket after this initial

acceleration?

b What upward force does this apply to the rocket?

c What is the net upward acceleration acting on the

rocket? (Use g = 10 m s−2 if required.)

Conservation of momentum6.2 questions

Worked Solutions




Work 6.3

Aristotle, Galileo and Newton each made significant contributions to our

developing understanding of the relationships between the forces that are

applied to objects and their resultant behaviour. Aristotle and those who

followed him were often locked into philosophical views involving, for

example, the natural resting places of objects. Unlike his predecessors,

Galileo, over 400 years ago, based his proposed theories largely on theobservations that he made. Although he could not be completely free from

the influences of his era, observational scientific experimentation had

been born through him. The implications of this change in approach were

immeasurable.

In this chapter our study of the interactions between forces and objects

focuses on the resultant displacement that objects experience, rather than

the resulting velocities and accelerations discussed earlier in our study of

Newton’s laws. This leads to the examination of the concepts of work and

energy. The notion of energy was not developed until a relatively short time

ago, and was only fully understood in the early 1800s. Today, the concept

has become one of the most fundamental in science. We will see that in

physics an object is said to have energy if it can cause particular changes to

occur. Energy is a conserved quantity and is useful not only in the study of

motion, but in all areas of the physical sciences. Before discussing energy, it

is necessary to first examine the concept of work.

In common usage the term ‘work’ has a variety of meanings. Most

convey the idea of something being done. At the end of a long, tiring day

we might say that we have done a lot of work. This could also be said

because the person feels that their reserves of energy have been used up.

Imagine lifting a heavy book up onto a high shelf. The heavier the book,

the more force must be applied to overcome its weight. The higher the

shelf, the greater the displacement over which the force must be applied.

A very heavy book lifted to a high shelf will require a considerably greatereffort than moving a few pieces of paper from floor to table. Thus there are

two features that constitute the amount of work done: the amount of effort

required and the displacement involved.

In physics work is done on an object by the action of a force or forces.

The object is often referred to as the load. Many interactions are complex

and there is often more than one force present. As work can only be done

in the presence of a force, it is imperative that any time the work done in a

particular situation is being discussed, the relevant force, forces or net force

should be clearly stated. For clarity, the item upon which the work is done,

the load, should also be specified. Clearly specified examples of work are:

• the work done by gravity on a diver as she falls

• the work done by arm muscles on a schoolbag lifted to your shoulder

• the work done by the heart muscle on a volume of blood during a

contraction

• the work done by the net force acting on a cyclist climbing a hill.

Always being clear about the particular forces and objects examined will

prevent considerable confusion in this area of study.

For work to be done on a body, the energy of the body must change. Thus

the work done is measured in joules, which is also the unit of energy.Figure6.10 In each situation involving work,a load can clearly be specified.



Motion192

The different forms of energy are discussed later in this chapter. A test to

decide whether work has been done on a particular object involves deter-

mining whether the object’s energy has altered. If its energy is unaltered,

no net work has been done on it, even though forces clearly may have been

acting.

Work done by a constant forceIf the net force acting on an object in a particular situation has a constant

value, or if it is appropriate to utilise an average force value, then:

From the definition of work, it can be seen that if a person pushes a load

a horizontal distance of 5 m by exerting a horizontal force of 30 N on the

load, then the person does 150 J of work on the load. This is straightforward.

A displacement in the direction of the force is achieved so work is done. If

an applied force does not produce any displacement of the object (x = 0),

then we say that no work is done on the object.

Situations also occur in which a constant force acts at an angle θ to the

direction of motion. A force acting at an angle will be less effective than the

force acting solely in the direction of the displacement. The component of

the force in the direction of the displacement, F cos θ, is used in calculating

the work done in the required direction.

The net WORK done on an object is defined as the product of the net force onthe object and its displacement in the direction of the net force.When the force and displacement are in the same direction, the work done bythe stated force is given by:

W = Fx where W is the work done by the stated force in joules (J) F is the magnitude of the stated force in newtons (N)

x is the magnitude of the displacement in metres (m)Work is the area under a force–displacement graph.

i

One JOUL… of work is done on an object when the application of a net force of1 newton moves an object through a distance of 1 metre in the direction of thenet force.

i

The symbol W is a little over-used in this

area of physics. In the area of motion

and mechanics it can stand for work or

the abbreviation of watt. Be careful to

read the context when you come across

the symbol!

Physics file

The convention for naming units in

physics is to use small letters when

writing the unit in full (e.g. joule,

newton, metre). A capital letter is used

for the symbol only when the unit is

named in recognition of a scientist’s

contributions, otherwise the symbol is

lower case (e.g. J for joule, N for newton,

but m for metre).

Physics file

Figure 6.12 During the fall the force due togravity does work on the person and producesa displacement.

Figure 6.11 (a) No work is done on the crate since its energy is not altered. (b) Theenergy of the crate is changing, so work is being done on the crate.

I’m doing no work

on the crate since

x = 0

W = Fx = 0

I’m doing nowork on the

crate since

x = 0

I’m doing

work!

crate speeding up

work is doneon the load

W = Fx

(a) (b)

The unit for work, the joule (J), is used

for all forms of energy in honour of

James Prescott Joule, an English brewer

and physicist, who pioneered work on

energy in the 19th century.

Physics file




Work and frictionIf an object is forced to move across a surface by the application of a force,

its motion may be slowed by friction. In this case the applied force is doing

work on the object and the frictional force can be considered to be doing

‘negative’ work on the object. In Figure 6.14 an applied force of 300 N across

a displacement of 5 m does 1500 J of work on the object. If a 100 N frictional

force occurs, we can state that work done by the frictional force is:

−100 × 5 = −500 J

Hence the net work done on the object is 1000 J. An alternative approachwould involve first calculating the net force, ΣF, on the object to be 200 N.

The net work done on the load (by the net force) is therefore:

ΣF × x = 200 × 5 = 1000 J

If work is done against a frictional force as a load continues to move,

then some of the energy expended by the person pushing is converted into

heat and sound energy, and transferred to the ground and the load. As

the surfaces slide past one another friction would cause them to heat up

slightly and make some noise. Keep in mind that on a frictionless surface

the load would accelerate, increasing its energy.

Figure 6.15 shows a situation in which the size of the frictional force is

not large enough to prevent motion, but it is large enough to balance theapplied force. As a result the object moves at a steady speed. Although the

person is doing work on the object, this is opposed by friction and the net

work on the object is zero. This is consistent with our earlier discussion, which

stated that if the energy of the object is not altered, then no net work has

been done on the object.

W = Fx cosθ

where θ is the angle between the applied force and the direction of motion.i

Figure6.14 The object slides across adisplacement of 5 m. Due to friction the net work

done on the object is less than the work done by theperson on the object.

Figure6.13 (a) If a force is applied in the direction of motion of the cart, then the force is at its most effective in movingthe cart. (b) When the force is applied at an angle θ to the direction of motion of the cart, the force is less effective. Thecomponent of the force in the direction of the displacement, F cos θ, is used to calculate the work. (c) When the angle atwhich the force is acting is increased to a right angle (θ = 90°), then the component of the force in the direction of theintended displacement is zero and it does no work on the cart—provided of course that it doesn’t lift the cart, in which case

work would also be done against gravity.

Figure6.15 Due to friction the net work done on

the object is zero since the object has no increasein kinetic energy.

Direction of motion

F

F F

θ

F cos θ

Direction of motion Direction of motion

(a) (b) (b)

frictional

force F f= 100 N

Σ F = 200 N

F applied

= 300 N

Net work

done on crate

= 1000 J

frictional

force F f= 100 N

F applied

= 100 N

steady speed

object’s energy is unchanged



Motion194

Worked example 6.3A

Calculate the work done against gravity by an athlete of mass 60 kg competing in the Empire

State Building Run-up illustrated in Figure 6.16.

Use g = 9.8 m s−2. Assume the athlete climbs at a constant speed.

Solution

Only the weight force needs to be considered in this example as the work in the vertical

direction is all that is required.

m = 60 kg, g = 9.8 m s−2, x = ∆h = 320 m

Force applied = weight = mg = 60 × 9.8 = 588 NW = Fx

= 588 × 320

= 188 160 N m = 1.9 × 105 J

Worked example 6.3B

The girl in Figure 6.13 pulls the cart by applying a force of 50 N at an angle of 30° to the

horizontal. Assuming a force due to friction of 10 N is also acting on the wheels of the cart,

calculate the net work done on the cart if the cart is moved 10 m along the ground in a

straight line.

Solution

F applied

= 50 N, θ = 30°, F f = 10 N, x = 10 m

ΣF = F applied

× cos 30° − F f

= 50 × 0.866 − 10

= 33.3 N

Now W = ΣFx

= 33.3 × 10 = 330 J

Figure 6.16 The Empire State Building Run-upis one of a number of races to the tops of tallbuildings around the world. The current recordfor the 320 m (vertical) race is 9 minutes and33 seconds.




Upward force does no work A more difficult idea to comprehend is that in the apparent absence of

friction, a force can be exerted on an object yet do no work on it. For

example, when a person carries an armload of books horizontally the

upward force does no work on the books since the direction of the applied

force (i.e. up) is at right angles to the displacement (i.e. horizontal). An

examination of the definition of work—W = Fx cos θ—confirms this finding

since the value of θ is 90° and, hence, the value of cos θ is 0.

Similarly if a person is holding a heavy item, such as a TV, stationary,

they may be exerting great effort. However, since the upward force applied

to support the object does not produce any vertical (nor indeed horizontal)

displacement, x = 0 and there is no work done by this upward applied force

on the object.

Force–displacement graphsA graphical approach can also be used to understand the action of a force

and the work expended in the direction of motion. This is particularly

useful in situations in which the force is changing with displacement. The

area under a graph of force against displacement always represents the work

produced by the force, even in situations when the force is changing, such

as during a collision. The area can be shown to be equivalent to work as

follows. From Figure 6.17:

the area enclosed by the graph = Fav

× x

Work = Fav

× x

When the force is changing, a good estimate of the area can be found by

dividing the area into small squares and counting the number or by dividing

it into thin segments. The segments can be considered to be rectangles with

an area equal to the work for that small part of the displacement. The total

work will be the sum of the areas of all the separate rectangles.

Whenever the net force is perpendicular to the direction of motion no (net) workis done on the object.

i

Figure 6.17 The area under a force–displacementgraph is equivalent to the work done by a forceacting in the direction of the displacement. Wherethe net applied force is changing, the area can befound by counting squares or by dividing the areainto segments. The area of each segment thenequals the work done by a constant force duringthat small displacement and the total area willrepresent the total work.

Units for area = N m = N m

= J

i.e. area = work done

F o r c e

( N )

Displacement (m)

The area under a force–displacement

graph can also be found by using

calculus if the equation of the graph

is known. In most instances a good

estimate by counting squares or

segments is sufficient.

Physics file



Motion196

Worked example 6.3C

The force–displacement graph on the left represents the work done on the sole of a sports

shoe as it compresses against the surface of a rigid track. The displacement shown

represents the amount of compression the sole undergoes. Find the work done on the shoe

by the compressive forces.

SolutionThis is a simple case of working out the area represented by each square and then counting

the total number of squares to find the total work done. Be careful to consider the scale of

each axis in your working.

Area of one square = 10 N × 0.001 m = 0.01 J

Total number of squares (part squares can be added to give whole squares) = 33

Work = 33× 0.01 = 0.33 J

Impulse and work The concepts of impulse and work seem quite similar and, when solving

problems, can easily be confused. Actually, problems focusing on forces in

collisions may be solved using either concept, but it should be understood

that each is derived from a different idea. Impulse comes from an

understanding of the action of a force on an object over time and is equal to

the change in momentum the force produces. Work is related to the action

of a force on an object as it moves the object, or part of it, through some

displacement. This equals the change in the object’s energy, ∆E.

Summarising:

• Impulse is equal to F × ∆t, is equivalent to ∆ p, has the units newton

seconds (N s), and can be determined from the area under a force–time

graph.

• Work is equal to F × x, is equivalent to ∆E, has the units joules (J), and can

be determined from the area under a force–displacement graph.

• When a force does work on an object, a change occurs

in the displacement and energy of the object.

• The work done on an object, W in joules (J), is the

product of the net applied force and its displacement

in the direction of the force:

W = Fx

• The work done by a force acting at an angle to thedisplacement is given by Fx cos θ where θ is the

angle between the force and the direction of the

displacement. When the force is at right angles to

the direction of the displacement, no work is done in

that direction.

• The area under a force–displacement graph is

equivalent to the work done. The area under the

graph for a variable force can be found by counting

squares or narrow segments.

Work6.3 summary

F o r c e ( N )

Displacement (m)

100

90

80

7060

50

40

30

20

10

0 . 0

0 1

0 . 0

0 2

0 . 0

0 3

0 . 0

0 4

0 . 0

0 5

0 . 0

0 6

0 . 0

0 7

0 . 0

0 8




Where appropriate use g = 9.8 m s−2.

11 How much work is done on an object of 4.5 kg when

it is lifted vertically at a constant speed through a

displacement of 6.0 m?

22 A bushwalker climbs a hill 250 m high. If her mass is

50 kg and her pack has an additional mass of 10 kg,

calculate the work she needs to do in climbing to the

top of the hill.

33 Is the quantity you calculated in Question 2 the only

work that the bushwalker has done? Explain.

44 The work done by a force is:

ii calculated by multiplying the force by the

distance moved

iiii measured in joules

iiiiii not affected by the angle at which the force acts.

Which statement/s is/are correct?

A i, ii, iii

B i, ii

C ii, iii

D ii

E iii

55 A removalist is loading five boxes onto a truck.

Each has a mass of 10 kg and a height of 30 cm. The

tray of the truck is 1.5 m above the ground and the

removalist is placing each box on top of the previous

one.

a How much work does the removalist do in liftingthe first box onto the truck tray?

b How much energy has the removalist used in

lifting this first box?

c What is the total work done on the boxes in lifting

all the boxes onto the truck as described?

66 If a lift of mass 500 kg is raised through a height of

15 m by an electric motor:

ii the weight of the lift is 4900 N

iiii the useful work done on the lift is 73 500 J

iiiiii the useful work done is the only energy used by

the motor.Which statement/s is/are correct?

A i, ii, iii

B i, ii

C ii, iii

D ii

E iii

77 The diagram shows the position of a student’s arm as

the weight of a sandbag is measured using a spring

balance. The balance is held still to take the reading.

What net work is done on the sandbag while the

measurement is being made?

sandbag

springbalance

88 A weightlifter raises a 100 kg mass 2.4 m above the

ground in a weightlifting competition. After holding

it for 3.0 s he places it back on the ground.

a How much work has been done by the weightlifter

in raising the mass?b How much additional work is done during the

3.0 s he holds it steady?

99 A rope that is at 35° to the horizontal is used to pull a

10.0 kg crate across a rough floor. The crate is initially

at rest and is dragged for a distance of 4.00 m. The

tension in the rope is 60.0 N and the frictional force

opposing the motion is 10.0 N.

a Draw a diagram illustrating the direction of all

relevant forces.

b Calculate the work done on the crate by the

tension in the rope.c Find the total work done on the crate.

d Determine the energy lost from the system as heat

and sound due to the frictional force.

110 The graph represents the size of a variable force, F ,

as a rubber band is stretched from a resting length of

5 cm to 25 cm. Estimate the total work done on the

rubber band by the force.

F o r c e ( N )

Extension (cm)10 20

10

5

Work6.3 questions

Worked Solutions



Motion198

Mechanical energy6.4

Although the concept of energy is quite abstract, each of us, from an early

age, will have begun to develop an understanding of its meaning. We are

increasingly aware of our reliance upon the energy resources that allow our

vehicles and computers to run, that keep our homes warm and fuel our

bodies. Energy can take on many forms.

In this section we look at the forms of energy specifically related tomotion. Mechanical energy is defined as the energy that a body possesses

due to its position or motion. Kinetic energy, gravitational potential energy

and elastic potential energy are all forms of mechanical energy. Recall our

earlier assertion that work is done when a force is applied that results in

the displacement of an object in the direction of the applied force. When

work is done the energy of an object will change. We will analyse situations

that result in a change in the kinetic and/or gravitational potential and/or

elastic potential energy of an object.

A hockey puck gains energy when hit because work has been done by the

stick on the puck. The amount of work done on the puck equals the puck’s

change in kinetic energy. A tennis ball at the point of impact is compressed

against the tennis racquet. It has gained elastic potential energy. Work has

been done in compressing the tennis ball. However, the idea of work may

be applied to many forms of energy. The common thread is that, regardless

of the form of energy, whenever work is done there is a change in energy

from one form to another. In order for any energy transformation to occur,

say from motion to heat, work must be done.

We observe many different forms of energy each day. We have come to

take for granted the availability of light, heat, sound and electrical energy

whenever we require it. We rely upon the chemical potential energies that

are available when petrol, diesel and LPG are burnt to run our vehicles, and

food to fuel our bodies. Whenever work is done, energy is expended.

Some comparative energy transformations are included in Table 6.2.

Table 6.2 Comparison of various energy transformations

Energy use Amount of energy

Household in 1 day 150 MJ

Fan heater in 1 hour 8.6 MJ

Adult food intake in 1 day 12 MJ

Making 1 Big Mac 2.1 MJ

Climbing a flight of stairs 5 kJ

Lifting 10 kg to a height of 2 m 200 J

Kinetic energyAn object in motion has the ability to do work and therefore is said to

possess energy. This energy carried by a moving object is called kinetic

energy (from the Greek word ‘kinesis’, literally meaning ‘motion’).

…N…RGY is the ability to do work.i

Figure 6.18 Mechanical energyexists in many forms.

Interactive




If a moving object of mass, m, and initial velocity, u, experiences a

constant net force, , for time, t, then a uniform acceleration results. The

velocity will increase to a final value, v. Work will have been done during

the time the force is applied. Since work is equivalent to the change in

kinetic energy of the object, there should be a relationship linking the two

quantities. This can be found from the definition for work when the net

applied force is in the direction of the displacement:

W = ΣFx

Now substituting Newton’s second law = ma we get:

W = max . . . . . . (i)

Using one of the earlier equations of motion: v2 = u2 + 2ax

and rearranging: x = v2 − u2

2aSubstitute this for x in equation (i): W = ma ×

v2 − u2

2aRearranging gives: W =

1

2mv2 −

1

2mu2

but W = ∆E

If it is accepted that the work done results in a change in kinetic energy,

then an object of mass m with a speed v has kinetic energy equal to1

2mv2.

Like all forms of energy, kinetic energy is a scalar quantity and is

measured in joules (J). There is no direction associated with it. The

kinetic energy of an object depends solely on its mass and velocity. The

approximate kinetic energy of various moving objects is given in Table 6.3.

Table 6.3 Kinetic energy of moving objects

Object Mass (kg)Average speed(m s–1)

… k (J)

Earth in orbit 6 × 1024 3 × 104 2.7 × 1033

Orbiting satellite 100 8 × 103 3 × 109

Large car 1400 28 5.5 × 105

Netball player 60 8 1900

Footballer 90 8 2900

The KIN…TIC …N…RGY , … k, of a body of massm and speed v is:

… k = 1

2mv 2

i

Figure6.19 The kinetic energy of any objectdepends on its mass and the square of its speed.Doubling the velocity will increase the kineticenergy by a factor of four.

The derivation described for kinetic

energy is actually that for translational

kinetic energy, the movement of a

body along a path. A body can also

have rotational kinetic energy if it is

spinning, as does the Earth. A different

relationship is required to calculate the

kinetic energy of rotation.

Physics file



Motion200

Electron in a TV tube 9 × 10−31 7 × 107 2.2 × 10−15

Note that the relationship between work and energy, which was

discussed earlier, has now been quantified for kinetic energy changes.

When an applied force results in the change in kinetic energy of an object, the

work done in joules (J) can be calculated using: W = ∆…

k

= … k(final)

− … k(initial)

= 1

2mv 2 − 1

2mu2

where m is the mass of the object in kilograms (kg) v is the final speed of the object in metres per second (m s−1) u is the initial speed of the object in metres per second (m s−1)As the mass of the object is generally unaltered, often this can be simplified to:

W = ∆… k = 1

2m(v 2 –u2)

i

Therefore, if an object undergoes a known change in kinetic energy

during an interaction, the work done on the object by the net force is known.

Hence the average net force exerted on the object during this interaction

can be calculated by assuming that ∆Ek = W = F

avx.

Worked example 6.4A

Calculate the kinetic energy of an athlete of mass 60 kg running at a speed of 8.0 m s −1.

Solution

m = 60 kg, v = 8.0 m s−1

Using …k =

1

2mv 2:

…k =

1

2 ×

60×

8.0

2

≈ 1900 J.

Worked example 6.4B

Blood is pumped by the heart into the aorta at an average speed of 0.15 m s −1. If 100 g of

blood is pumped by each beat of an adult human’s heart find:

a the amount of work done by the heart during each contraction

b the energy used by the heart each day in pumping blood through the aorta (use an

adult’s average resting rate of 70 beats per minute). Assume that there are no other

energy losses.

Solution

a The work done by the heart is equal to the kinetic energy the blood gains as it is pumped

into the aorta.

m = 0.10 kg, v = 0.15 m s−1, u = 0 m s−1

Using W = ∆…k =

1

2 m(v 2 − u2)

W = 1

2 × 0.10 × (0.152 − 02)

W = 1.125 × 10−3 J = 1.1 mJ

b If there are 70 beats each minute then the amount of energy transferred:

…k per minute = 1.125 × 10−3 × 70 = 0.07875 J per minute

…k per day = 0.07875 × 60 min per hour × 24-hour day

…k = 113.4 J per day ≈ 110 J per day

Interactive




Potential energyAn object can have energy not only because of its motion, but also as a result

of its shape or position. This is called potential energy. A gymnast, crouched

ready to jump, has potential energy. During the jump, work is being done

by the force exerted by the gymnast, and potential energy is converted

into kinetic energy from the stores of chemical energy in the muscles of the

gymnast’s body.

There are many different forms of potential energy: chemical, grav-

itational, elastic etc. Potential energy is a stored energy giving the body

potential to do work or produce a force that creates motion. In this

particular study we are mainly concerned with gravitational and elastic

potential energy which, for the present, we will denote U .

Gravitational potential energy

An athlete at the top of a high-jump has gravitational potential energy

because of his position. As he falls, work is done (Figure 6.20). Recall that in

this case the work done is given by:Work done = ΣFx

The force acting on the body is simply the force due to gravity also called

the person’s weight:

Weight = mg

The displacement that occurs is in a vertical direction and can be

described as a change in height, ∆h. Replacing F and x with these equivalent

terms gives:

W = mg∆h

Similarly, the work done in raising the athlete against a gravitational field

is stored as gravitational potential energy; hence, the athlete has a change

in potential energy:∆U

g = mg∆h

Figure6.20 The energy gained or lost due to achange in height within a gravitational field is called

gravitational potential energy. An increase in heightwill require the transformation of energy fromother sources. A decrease will usually increase thekinetic energy of the body.

Figure6.21 This photograph of a pole-vaulterillustrates that elastic potential energy is stored inthe pole. This energy is largely converted to kineticenergy and then the gravitational potential energyof the athlete.



Motion202

The ∆U g of a body depends only on the vertical height of the object above

some reference point, in this case the ground. It does not depend on the

path taken since it is based on the direction of the gravitational field. It is

the work done against or by the force of gravity that leads to changes in

gravitational potential energy. Similarly, the work it can do when falling

does not depend on whether the object falls vertically or by some other

path, but only on the vertical change in height, ∆h.

The reference level from which the height is measured does not matter

as long as the same reference level is used throughout a given problem

solution. It is only changes in potential energy that are important. Forexample, the height of the high jumper is best referenced to the ground

she jumped from, and commonly it is her centre of gravity that is analysed.

The height of a luggage locker in an aircraft makes a lot more sense when

referenced to the floor of the aircraft than it would referenced to the ground.

The need for considering a change in height in comparison to a reference

level is also made apparent when considering a person standing at ground

level. If the person is standing beside a hole and his centre of gravity is

considered, he will have gravitational potential energy with reference to

the bottom of the hole. It is also quite justifiable to suggest that even with

reference to the ground he has gravitational potential energy; he could

fall over! Gravity would do work on him and his gravitational potential

energy would change. The change in height would be with reference to the

person’s centre of mass.

Worked example 6.4C

A ranger with a mass of 60 kg, checking the surface of Uluru for erosion, walks along a path

that takes her past points A, B and C.

C (0 m)

A (260 m)B (389 m)

a What is her gravitational potential energy at points B and C relative to A?b What is the change in the ranger’s potential energy as she walks from B to C?

c If the ranger was to walk from B to C via A would it alter your answer to part b? Explain.

Solution

a In this question heights are being referenced to point A. The person would have had zero

gravitational potential energy at A using this reference.

m = 60 kg, g = 9.8 m s−2, hA = 260 m, h

B = 389 m, h

C = 0 m

Potential energy change from A to B:

∆Ug = mg(h

B − h

A)

= 60 × 9.8 × (389 − 260)

= 7.6 × 104 J

The change in gravitational potential energy is due to the work done against agravitational field and is given by:

∆Ug = mg∆h

where ∆Ug is the change in gravitational potential energy measured in joules (J)

m is the mass of the body (kg) ∆h is the change in height (m)

g is the acceleration due to gravity (m s–2)

iThe relationship for gravitational

potential energy used here is only

appropriate when the weight force due

to gravity is constant. This will only be

the case when the change in height is

relatively small. As the distance from

the Earth’s surface changes so will

the strength of the gravitational field

according to the relationship

g ∝ 1

r 2

where r is the distance (in metres)

from the centre of the Earth to the

body’s position. The area under a force–

displacement graph can then be used to

find the change in potential energy due

to a change in position and the varying

weight force.

For the purposes of this study only

relatively small changes in height close

to the Earth’s surface will be considered

for which the weight force can be

considered constant.

Physics file




Potential energy change from A to C:

∆Ug = mg(h

C − h

A)

= 60 × 9.8 × (0 − 260)

= −1.5 × 105 J

b There is no need to calculate ∆Ug from B to C separately as the difference between the

two previous results can be used.

Potential energy change from B to C:∆U

g = ∆U

g (A to C) − ∆U

g (A to B)

= −1.5 × 105 − 7.6× 104

= −2.3 × 105 J

c It makes no difference what path is taken to achieve the change in height. Potential

energy change throughout this example is being determined relative to an initial height.

In general, if an object is originally at a height h0, then the change in potential energy as

it moves to a different height, h, is:

∆Ug = mgh − mgh

0

= mg(h − h0)

In general terms, the change in potential energy of an object when it is moved between

two heights is equal to the work needed to take it from one point to another.

Elastic materials and elastic potential energyThe third aspect of mechanical energy that we will study is elastic potential

energy. Like gravitational potential energy it occurs in situations where

energy can be considered to be stored temporarily so that, when this energy

is released, work may be done on an object. Elastic potential energy is stored

when a spring is stretched, a rubber ball is squeezed, air is compressed in a

tyre or a bungee jumper’s rope is extended during a fall. Since each object

possesses energy due to its position or motion, these all suit our earlier

definition of mechanical energy. We will see that when some materials are

manipulated we can think of this as work being done to store energy. This

energy is often released or utilised via work being done on another object.Materials that have the ability to store elastic potential energy when

work is done on them and then release this energy are called elastic materials.

Metal springs are common examples, but also realise that many materials

are at least partially elastic. If their shape is manipulated, items such as

our skin, metal hair clips and wooden rulers all have the ability to restore

themselves to their original shape once released—within limits of course!

Materials that do not return mechanical energy when their shape is

distorted are referred to as plastic materials. Plasticine is an example of a

very plastic material.

Ideal springs obey Hooke’s lawSprings are very useful items in our everyday life due to the consistent way

in which many of them respond to forces and store energy. When a spring

is stretched or compressed by an applied force we say that elastic potential

energy is being stored. In order to store this energy work must be done

on the spring. Recall that in section 6.3 we have discussed that if a force of

a constant value is applied to an object (and a displacement occurs in the

direction of that force) then the quantity of work done can be calculated

using W = Fx. This formula can therefore be used when a set force, F, has

been applied to a spring and a given compression or extension, ∆x, occurs.



Motion204

However, we are usually interested in examining how a spring will behave

in a range of conditions.

Consider the situation in which a spring is stretched by the application

of a steadily increasing force. As the force increases, the extension of the

spring, ∆x, can be graphed against the applied force, F. You can imagine

hanging a spring vertically and gradually adding more and more weight

to it so that it stretches. Many items, such as well-designed springs, will (atleast for a small load) extend in proportion to the applied force. For example,

if a 10 newton force produced an extension of 6 cm, then a 20 newton force

would produce an extension of 12 cm. These items are called ideal springs.

The resulting graph of applied force versus extension would be linear as in

Figure 6.22.

Note that the gradient of this graph tells us the force, in newton, required

to produce each unit of extension. The gradient of the graph is called the

spring constant, k , measured in N m−1. The gradient therefore indicates the

stiffness of the spring, and for an ideal spring this gradient has a set value

(i.e. the F vs ∆x graph is a straight line). A very stiff spring that is difficult

to stretch would have a steep gradient; that is, a large value of k . Although

k is usually called the spring constant, it is sometimes called the stiffness

constant or force constant of a spring. A spring constant of k = 1500 N m−1

indicates that for every metre that the spring is stretched or compressed, a

force of 1500 N is required. This does not necessarily mean that the spring

can be stretched by 1 m, but it tells us that the force and the change in

length are in this proportion.

The relationship between the applied force and the subsequent extension

or compression of an ideal spring is known as Hooke’s law. Since for ideal

springs F ∝ ∆x, we can say F = k ∆x. However, as we are often interested

in using the energy stored by stretched or compressed springs we tend to

refer to the force that the distorted spring is able to exert (rather than the

force that was applied to it). Newton’s third law tells us that an extendedor compressed spring in equilibrium is able to exert a restorative force

equal in size but opposite in direction to the force that is being applied to it.

Therefore Hooke’s law is often written in the form shown below.

Calculating elastic potential energy

Work must be done in order to store elastic potential energy in any elastic

material. Essentially the energy is stored within the atomic bonds of the

material as it is compressed or stretched. The amount of elastic potential

energy stored is given by the area under the force–extension graph for the

item.

For materials that obey Hooke’s law (such as the material shown in

Figure 6.22), an expression can be derived for the area under the F–x graph.

HOOK…’S LAW states that the force applied bya spring is directly proportional,but opposite in direction, to the spring’s extension or compression. That is:

F = −k∆ x

where F is the force applied by the ideal spring (N) k is the spring constant (N m−1) (also called force constant or

stiffness constant) ∆ x is the amount of extension or compression of the ideal spring (m)

i

Prac 26

Figure 6.22 Ideal materials obey Hooke’s law:F ∝ k∆ x.

A p p l i e d

f o r c e ( N )

Extension (m)




Work done = area under F–x graph

= area of a triangle

= 1

2 × base × height

= 1

2 × ∆x × F

But since F = k ∆x:

Work done = 1

2 × ∆x × k ∆x

W = 1

2k ∆x2

= the elastic potential energy stored during the extension/

compression

Although many materials (at least for a small load) extend in proportion

to the applied force, many materials have force–extension graphs more like

that shown in Figure 6.23. For these materials the area under the F–∆x graph

must be used to determine the elastic potential energy stored.

Worked example 6.4D

Three different springs A, B and C are exposed to a range of forces and the subsequent

extension measured. The data collected for each spring has been graphed in the F –∆ x graph

at right.

a Justify the statement that spring B is the only ideal spring shown.

b Calculate the stiffness constant of spring B.

c Calculate the work done in extending spring B by 25 mm. Assume the process of storing

energy is 100% efficient.

d Estimate the work done in extending spring C by 25 mm.

e If all springs are extended such that∆ x = 40 mm, which spring will have stored the most

elastic potential energy? Justify your choice.

Solution

a Ideal springs produce an extension that is consistently proportional to the applied

force. Since spring B has an F –∆ x graph which is a straight line emerging from the

origin, it is behaving ideally and obeying Hooke’s law until an extension of ∼30 mmis reached.

Spring C does not obey Hooke’s law, that is, F is not directly proportional to ∆ x, since the

graph is not a straight line.

On close inspection it can be seen that the initial application of a small force did not

produce any extension in spring A. (This is a common behaviour of real springs where

a certain minimum amount of force must be applied before any extension will occur).

This means that spring A has not obeyed Hooke’s law and therefore is not an ideal

spring.

The elastic (or spring) potential energy,Us, stored in an item is always given

by the area below the force–extension graph for the item. The unit of Us is the

joule, J.In the case of an ideal material that obeys Hooke’s law, the elastic potential

energy is given by the expression:U

s = 1

2k∆ x 2

where Us is the elastic potential energy stored during compression or

extension (J) k is the stiffness constant of the material (N m−1) ∆ x is the extension or compression (m)

i

Figure6.23 Elastic potential energy is a formof mechanical energy. Work is done as elasticpotential energy is stored, as indicated by thearea under theF –∆ x graph.

Extension (m)

area under graph

= work done

F o r c e

a p p l i e d

( N )

A

B

C

10 20 30 40 50 60

20

40

60

80

100

120

F

a p p l i e d ( N )

x (mm)

Take care! The two forms of potential

energy, elastic (or spring) potential

energy , U s, and gravitational potential

energy, U g , have been introduced. Take

extra care when analysing situations

like pole-vaulting, since at some stages

in this event both forms of potential

energy are present at the same time!

Physics file



Motion206

b k = gradient of F –∆ x graph = rise/run = 90/0.030 = 3.0 × 103 N m−1

c Since spring B obeys Hooke’s law, the equation Us =

1

2k∆ x2 can be applied.

W = ∆Us = U

s[final] − U

s[initial]

As there is initially zero energy stored:

W = 1

2k∆ x2

= 1

2 × 3.0 × 103 × 0.0252

= 0.94 Jd Spring C does not obey Hooke’s law, so the work done must be calculated using the area

under the F –∆ x graph:

1 square of area is equal to (0.005 × 10) or 0.05 joules.

There are approximately 7.5 squares of area.

Therefore, W ≈ 7.5 × 0.05 ≈ 0.38 J

e Elastic potential energy is given by the area under the F –∆ x graph. At an extension of

40 mm spring A will have the greatest area under the graph, i.e. it will have stored the

most elastic potential energy.

By the mid-19th century, several scientists had begun to

write of the heating process as an energy change from work

(mechanical energy) to heat. It was eventually realised that all

forms of energy were equivalent and that when a particular

form of energy seemed to disappear, the process was always

associated with the appearance of the same amount of energy

in other forms. This led to the development of the principle

of conservation of energy . At this same time, James Joule

conducted a series of experiments fundamental to our present

understanding of heat.

Joule noticed that stirring water could cause a rise in

temperature. He designed a way of measuring the relationship

between the energy used in stirring the water and the change in

temperature. A metal paddle wheel was rotated by falling masses

and this churned water around in an insulated can. The amount

of work done was calculated by multiplying the weight of the

falling masses by the distance they fell. The heat generated was

calculated from the mass of the water and the temperature rise.

Joule found that exactly the same quantity of heat was always

produced by exactly the same amount of work. Heat was simply

another form of energy, and 4.18 joules of work was equivalent

to 1 calorie of heat.

Joule’s work led to some unusual conclusions for his day. He

stated that as a container of cold water is stirred, the mechanical

energy is being transformed into thermal energy, heating the

water. Theoretically, this means that a cup of water stirred long

enough and fast enough will boil—a novel, if laborious, way

of making a cup of coffee. Of course, the rate at which we can

normally add energy by stirring is less than the transfer of

energy to the surrounding environment. For the cup of water

to boil it would need to be very well insulated.

As a result of Joule’s investigations and other experiments of

the time, we now interpret the process of heating or cooling as a

transfer of energy. When heat ‘flows’ from a hot object to a cold

one, energy is being transferred from the hot to the cold.

James Joule

Physics in action

Figure6.24 James Prescott Joule.




Figure6.25 Joule’s original apparatus forinvestigating the mechanical work equivalent

of heat energy. The falling weights causedthe paddle to turn. The friction between thewheel and the water created heat energy inthe water. For the first time, heat energy couldbe measured and related to other forms ofenergy.

falling masses

water

paddle

pulley

Figure6.26 As a result of the considerable amount of mechanical work being doneon the water of a waterfall, the temperature of the water at the bottom of the falls isusually 1°C or 2°C higher than at the top.

• Energy is the ability to do work. Whenever work

is done, energy is transformed from one form toanother.

• Kinetic energy is the energy a body has because of

its motion. Ek =

1

2mv2 where E

k is the kinetic energy in

joules (J), m is the mass in kilograms (kg) and v is the

speed in metres per second (m s−1).

• Potential energy is stored energy with the potential

to allow work to be done. It may take many forms

including chemical, elastic and gravitational.

• Gravitational potential energy is the energy a body

has because of its position within a gravitational

field: ∆U g = mg∆h where U

g is the gravitational

potential energy in joules (J), m is the mass in

kilograms (kg) and ∆h is the change in height from a

reference height in metres (m).• Ideal materials extend or compress in proportion to

the applied force; that is, they obey Hooke’s law:

F = −k ∆x

• The elastic potential energy, U s, stored in an item is

given by the area below the force–extension graph

for that item, or U s =

1

2kx2 for an ideal spring that

obeys Hooke’s law.

• When work is done to store energy, one or more of

the following may be applied:

W = ∆U g or W = ∆E

k or W = ∆U

s

or W =

area under F–∆

x graph

Mechanical energy6.4 summary



Motion208

Where appropriate use g = 9.8 m s−2.

11 Calculate the kinetic energy of a:

a 1.0 kg mechanics trolley with a velocity of 2.5 m s−1

b 5.0 g bullet travelling with a velocity of 400 m s−1

c 1200 kg car travelling at 75 km h−1.

22 Calculate the gravitational potential energy relativeto the ground when a:

a mass of 1.0 kg is 5 m above the ground

b bird of mass 105 g is 400 m above the ground

c 1200 kg car has travelled a vertical height of 10 m

up a slope.

33 A 100 g rubber ball falls from a height of 2.5 m

onto the ground and rebounds to a height of 1.8 m.

What is the gravitational potential energy of the ball

relative to the ground at its:

a original position?

b final position?

c final position relative to its original position?

44 Which object has the greatest amount of energy?

A a spring with a spring constant k = 40 000 N m−1

compressed by 5.0 cm

B a cricket ball of mass 150 g stuck on the roof of a

grandstand 14 m above the ground

C a cricket ball of mass 150 g travelling at 10 m s−1 at

a height of 10 m above the ground

55 What net braking force must be applied to stop a

car within a straight-line distance of 50 m, if the car

has a mass of 900 kg and was initially travelling at a

velocity of 100 km h−1?

66 A small steel ball with a mass of 80 g is released from

a resting height of 1.25 m above a rigid metal plate.

Calculate the:

a change in gravitational potential energy

b kinetic energy of the ball just before impact

c velocity of the ball just before impact.

The following information relates to questions 7 and 8.

The force–extension graphs for three different springs

are shown below.

A

B

C

0.05 0.10 0.15 0.25

100

200

300

F

a p p l i e d

( N )

x (m)

7 Calculate the spring constant for each spring and

determine which is the stiffest spring.

88 Each spring has a force of 100 N applied to it.Calculate the elastic potential energy stored by each

spring.

99 A piece of gymnasium equipment is operated by

compressing a spring that has a spring constant of

2500 N m−1. How much elastic potential energy is

stored in the spring when it is compressed by:

a 5.00 cm?

b 10.0 cm?

c 15.0 cm?

110 The gymnasium equipment described in Question

9 is adjusted so that its spring constant is now3000 N m−1. If 12.5 J of energy is now stored in

the spring, by what distance has the spring been

compressed?

Mechanical energy6.4 questions

Worked Solutions




Energy transformation and power6.5

Besides the mechanical energy discussed in section 6.4, other forms of

energy exist, for example nuclear, heat, electrical, chemical and sound

energy. Atomic theory has led to each of these other forms being understood

as either kinetic energy or potential energy at the molecular level. Energy

stored in food or fuel and oxygen can be considered as potential energy

stored as a result of the electrical forces in the molecules.Despite the apparently different nature of the various forms of energy,

any energy can be transformed from one form to another. The connecting

factor is that all forms can do work on a body and therefore can be measured

and compared in this way. A stone dropped from some height loses

gravitational potential energy as its height decreases; at the same time its

kinetic energy will increase as its speed increases.

Transformation of energyEnergy transfers or transformations enable people and machines to do work,

and processes and changes to occur. Elastic potential energy stored in a

diving board must be transformed into the kinetic energy of the diver atthe pool. Contracting a muscle converts chemical potential energy stored in

the muscle to the kinetic energy of a person’s motion. In each example, the

transformation of energy means work is being done.

In many cases a transformation of energy produces an unwanted

consequence—a substantial amount of the energy is ‘lost’ as heat energy.

Of a typical adult’s daily food intake of about 12 MJ at least 80% is converted

into heat energy during normal activity. Such transfers can be depicted by

an energy-conversion flow diagram.

h

gravitationalpotential energy

kinetic energy

sound heat

heat

workdone

workdone

workdone

workdone

(b)(a)

Figure6.27 Whenever work is done, energy is transformed from one form to another.(a) As a body falls, gravitational potential energy is transformed to kinetic energy and heat,from the friction with the air. Once the body lands, further energy transformations will takeplace. (b) An energy-conversion flow diagram can be useful in visualising the transformationsthat take place.

A simple, although infinitely unlikely, example is shown in Figure

6.27. As the body falls to the ground there will be a number of energy

transformations. An energy flow diagram illustrates these changes.

Work is done whenever energy is transformed from one form to another.i

Interactive



Motion210

While the body falls, work will be done on the body by the gravitational

field, and gravitational potential energy becomes kinetic energy, the energy

of movement. There will also be some energy converted into heat by the

action of air resistance. When the body hits the ground, the kinetic energy

is converted into elastic potential energy by the compression of the body,

and to other forms, particularly heat but also to sound and kinetic energy.

Each transformation requires a force to do work on the body.

The efficiency of energy transformationsThe percentage of energy that is transformed to a useful form by a device is

known as the efficiency of that device. All practical energy transformations

‘lose’ some energy as heat. The effectiveness of a transfer from one energy

form to another is expressed as:

efficiency (%) = useful energy transferred × 100

total energy supplied =

useful output × 100

total input

Table 6.4 Efficiencies of some common energy transfers

Device Desired energy transfer Efficiency (%)

Large electric motor Electric to kinetic 90

Gas heater or boiler Chemical to heat in water 75

Steam turbine Heat to kinetic 45

High-efficiency solar cell Radiation to electric 25

Coal-fired electric generator Chemical to electric 30

Compact low-energy

fluorescent lightElectric to light 25

Human body Chemical to kinetic 25

Car engine Chemical to kinetic 25

Open fireplace Chemical to heat 15

Filament lamp Electric to light 5

In all the energy transformations included in Table 6.4, the energy lost

in the transfer process is mainly converted into heat. Most losses are caused

by the inefficiencies involved in the process of converting heat into motion.

In the real world, energy must be constantly provided for a device to

continue operating. A device operating at 45% efficiency is converting 45%

of the supplied energy into the new form required. The other 55% is lost to

the surroundings, mainly as heat but also some as sound.

Figure 6.28 In each of these situations belowan energy transformation is taking place. Can youidentify the forms involved in each transformation?




Conservation of energy

No matter what energy transformation occurs overall, no energy is gainedor lost in the process. It is a fundamental law of nature that energy is

conserved.

Consider the example of a diver as depicted in Figure 6.31. When

the diver is at the top, his or her gravitational potential energy will be

at a maximum. As the diver free falls the gravitational potential energy

decreases but the kinetic energy will increase with the increased velocity.

Some of the energy—a small amount—will be converted into heat due to

contact with the air. The moment before the diver reaches the reference

level all the gravitational potential energy has been converted into other

energy forms, mostly kinetic energy. The total at this point will be exactly

Air, like water, is a fluid and therefore there is a force between

the particles of the air and the surface of any object moving

through it. This force is called air resistance or drag . Drag is

due to the air particles that can be thought of as obstacles in

the path of a moving object. There will also be frictional forces

as air particles slide past the object.

At low speeds the effect of air resistance is slight.

However, it has been found that air resistance is proportional

to the square of the velocity (i.e. F a ∝ v 2). A doubling of speed

will increase air resistance approximately four times. At the

racing speeds that Olympic cyclists reach of 50 km h−1 or

more, 90% of the cyclist’s energy is required just to push the

bicycle and rider through the surrounding air. The remaining

10% is needed to overcome frictional forces between the

wheels and the ground.

Air resistance is affected by the frontal area of cross-

section and the shape of the bicycle. Designers have tried

to reduce the frontal area of racing bikes and their riders

by dropping the handle bars and raising the position of thepedals. This allows the rider to race bent forward, reducing the

area presented to the air.

Streamlining the bike helps still further. Some shapes

move through fluids more easily than others. Streamlined

bicycles, such as those used by the Australian Cycling Team

in international competition, have low-profile frames with

relatively smaller front wheels. Brake and gear cables are

run through the frame rather than left loose to create drag.

Moulded three-spoke and solid-disk wheels help still further.

The riders’ clothing and helmets have also been

streamlined. Cyclists wear pointed shoes, streamlined helmetsand skin-tight one-piece lycra bodysuits. Together these reduce

the air resistance on a rider by as much as 10% at higher

speeds. At race time riders shave their legs to reduce energy

losses just that little bit more.

Air resistance in sport

Physics in action

Figure6.29 The efficiency of a cyclist is affected by the velocityof the bike. As the velocity increases, the drag or air resistancecan use as much as 90% of the energy the cyclist’s input.

0 5 9.0 13.5

20

10

Velocity (m s−1 )

D r a g ( N )

Figure 6.30 Technological advances in bike design, pioneered byRMIT University, have led to efficient designs such as that used byAustralian riders in recent international competition.



Motion212

equal to the potential energy at the top. The total energy in the system

remains the same. Energy has been conserved.

This applies to any situation involving energy transfer or transformations

in an isolated system. In this particular case the sum of the gravitational

potential energy and the kinetic energy at any point is called the total

mechanical energy.

Here, the total mechanical energy remains constant. As an object falls,

gravitational potential energy decreases but kinetic energy increases to

compensate, so that the total remains constant. At any point during an

object’s free fall:

total mechanical energy = 1

2mv2 + mgh

There are many examples of this conservation of energy. In athletics,

the pole-vaulters and high-jumpers base their techniques on this principle.

Throwing a ball in the air is another example. When the ball leaves the

hand, its kinetic energy is at a maximum. As it rises, its velocity decreases,

reducing the kinetic energy, and its potential energy increases by the same

amount. At any point Ek + E

p will equal the initial kinetic energy. At the top

of the throw, the ball will have a vertical velocity of zero and, in a vertical

direction, the energy will be totally gravitational potential energy (any

horizontal motion will be represented by a remaining amount of Ek). The

transformation will reverse as the ball falls. Gravitational potential energy

will decrease as the ball returns towards its original height and, with itsspeed increasing, the kinetic energy will increase once more.

A more complex example is provided by the interactions as a gymnast

repeatedly bounces on a trampoline. Figure 6.32 is a series of frames from a

video of a gymnast carrying out a routine on a trampoline. Kinetic energy

and gravitational potential energy changes are shown in the graph below

the frames. Despite the complexity of the motion, the total energy of the

gymnast remains the same during each airborne phase, as illustrated in the

graph. On landing on the bed of the trampoline, the energy is transferred

to elastic potential energy within the trampoline and both kinetic and

gravitational potential energy fall. On take-off some of this energy will be

permanently transferred to the trampoline and its surrounds, thus lowering

the total available to the gymnast. This is represented by the reduced total

energy for each successive jump. Were the gymnast to flex his legs then

additional energy would be added to the mechanical energy available and

this total could be maintained or even increased until the gymnast finally

ran out of available energy himself.

The TOTAL …N…RGY in an isolated system is neither increased nor decreased byany transformation. Energy can be transformed from one kind to another, butthe total amount stays the same.

i

Kinetic energy + potential energy = total mechanical energyi

Figure 6.31 A diver loses gravitational potentialenergy but gains kinetic energy during the fall.

All E p

12—E p

12—E k

All E k

Prac 27 SPARKlab




Worked example 6.5A

Calculate the initial velocity required for a high jumper to pass over a high bar. Assume

the jumper’s centre of gravity rises through a height of 1.5 m and passes over the barwith a horizontal velocity of 1.2 m s −1, and that all of his initial horizontal kinetic energy is

transferred into Ug and …

k.

Use g = 9.8 m s−2.

Solution

As the total mechanical energy is assumed to be conserved after landing, the initial horizontal

kinetic energy equals total mechanical energy at the peak height:1

2mu2 = …

k + U

g (at peak height)

= 1

2mv 2 + mg∆h

The mass cancels out, giving an expression independent of the mass of the athlete. The same

speed at take-off will be required for a light person as a heavy one. (If this doesn’t seem tomake sense remember that all objects fall at the same rate regardless of their mass.)1

2u2 =

1

2v 2 + g∆h

Substituting the values from the question:1

2u2 =

1

2 × 1.22 + 9.8 × 1.5

1

2u2 = 15.42

and u = √2 × 15.42 = 5.6 m s−1.

In reality the take-off speed will need to be a little greater since there will be some losses

to friction.

Figure6.32 During each airborne stage of a gymnast’s trampoline routine (indicated onthe graph by shading), mechanical energy is conserved. The graph shows the relationshipbetween total energy and gravitational potential energy and kinetic energy. Each time thegymnast lands, energy is transferred to the trampoline. The energy returning from thesprings after each landing allows the routine to continue.

E n e r g y

( J )

20 40 60 80 100 120 140 160 180 200 220

Number of frames

on bed on bedflight flight

gravitational

potential energy

plus kinetic energy

gravitational

potential

energy

kinetic

energy

2043

64

81102

122125

147

159174

185

194

207218

flight

1000

2000

3000

4000



Motion214

Worked example 6.5B

A climber abseiling down a cliff uses friction between the climbing rope and specialised

metal fittings to slow down. If a climber of mass 75 kg abseiling down a cliff of height 45 m

reaches a velocity of 3.2 m s−1 by the time the ground is reached, calculate the average

frictional force applied. Use g = 9.8 m s−2.

Solutionm = 75 kg, u = 0 m s−1, v = 3.2 m s−1, h = 45 m

Gravitational potential energy at the top of the cliff:

…p = mgh = 75 × 9.8 × 45= 33 075 J

Kinetic energy at ground level:

…k =

1

2mv 2 =

1

2 × 75 × 3.22 = 384 J

Total energy transformed to forms other than gravitational potential and kinetic energy:

∆… = …p − …

k = 33 075 − 384 = 32 691 J

This change in energy will be equivalent to the work done by the frictional force; that is:

Work = ∆… = F f × x

and:

F f =

∆…

x =

32 691

45 =

726 N≈

730 N

PowerWhy is it that running up a flight of stairs can leave you more tired than

walking up if both require the same amount of energy to overcome the

force of gravity?

The answer lies in the rate at which the energy is used. When horses

were first replaced by steam engines, the engine was rated by how fast it

could perform a given task compared with a horse. An engine that could

complete a task in the same time as one horse was given a rating of one

horsepower. More formally, power is defined as the rate at which energy is

transformed or the rate at which work is done.

Determining the power developed is fairly straightforward when

mechanical work is done; but consider a situation in which a person pushes

a lawnmower, say, at constant speed. Here, there is no increase in kinetic

energy, but energy is being transformed to overcome the frictional forces

acting against the lawnmower.

POW…R = work donetime taken

= energy transformed

time takenor

P = W

∆t =

∆…

∆t

where P is the power developed in watts (W) resulting from an energytransformation ∆… occurring in time ∆t. ∆… is measured in joules (J), time ismeasured in seconds (s).

i

The British Imperial unit for power is

the horsepower, hp, dating from the

time of the Industrial Revolution when

the performance of steam engines was

compared with that of the horses they

were replacing. 1 hp = 746 W. The SI

unit for power honours the inventor of

the steam engine, James Watt.

Physics file




This is useful when finding the power required to produce a constant

speed against a frictional or gravitational force.

The rate of energy use is as much a limiting factor of the work a person

can do as the total energy required. A person may be able to walk or climb a

long distance before having to stop because all available energy is used. The

same person will fall over exhausted after a much shorter time if the same

journey is attempted at a run. Power is the limiting factor, the rate at which

a person’s body can transform chemical energy into mechanical energy.

Few humans can maintain one horsepower, about 750 W, for any lengthof time. Table 6.5 includes comparative figures for the power developed in

various activities and devices.

Table 6.5 Average power ratings for various human activities andmachines

Activity or machine Power rating (W)

Sleeping adult 100

Walking adult 300

Cycling (not racing) 500

Standard light globe 60

Television 200

Fast-boil kettle 2400

Family car 150 000

Worked example 6.5C

The fastest woman to scale the Rialto building stairs in the Great Rialto Stair Trek in a

particular year climbed the 1222 steps, which are a total of 242 m high, in 7 min 58 s. Given

that her mass was 60 kg, at what rate was she using energy to overcome the gravitational

force alone? Use g = 9.8 m s−2.

SolutionThe work is against gravity so:

P = Ug

∆t =

mg∆h

∆t

m = 60 kg, g = 9.8 m s−2, ∆h = 242 m

∆t = (7 min × 60) + 58 s = 478 s

P = 60 × 9.8 × 242

478 = 3.0 × 102 W

In the special case of an applied force opposing friction or gravity and doingwork with no increase in the speed of the object, we can say:AsW = Fx then:

P = Fx

∆t

and as v = x

∆t then:

P = F avv

av

where P is power developed (W) F

av is average applied force (N)

v av

is average speed (m s−1)

i

Figure6.33 It is not the amount of energyrequired that stops the rest of us from winningthe 400 m sprint, but the rate at which we caneffectively convert it to useful work.



Motion216

• Whenever work is done, energy is converted from

one form into another.

• The efficiency of an energy transfer from one form to

the required form is:

efficiency (%) = energy outputenergy input

× 100

• Whenever energy is transformed, the total energy in

the system remains constant. This conservation of

energy is a fundamental natural principle.

• The total mechanical energy will remain constant in

an isolated system. That is:

Ek + U

s + U

g = constant

• Power, P (in watts, W), is the rate at which work is

done or energy transformed:

P = W

∆t =

∆E

∆t

• In the particular case of work being done to overcome

friction, with no resulting increase in speed:

P = Fav

vav

Energy transformation and power6.5 summary


11 Describe the energy transformations that take place

when:

a a car slows to rest

b a gymnast uses a springboard to propel themselves

into the air

c an archer draws back and then releases an arrow

vertically upward

d an athlete’s foot hits a track.

22 Draw an energy transformation flow chart for a

swimmer diving off a diving board and into a pool of

water.

33 A boy of mass 46 kg runs up a 12 m high flight ofstairs in 12 s.

a What is the gain in gravitational potential energy

for the boy?

b What is the average power he develops?

44 A coach is stacking shot-puts, from the shot-put

event, onto a shelf 1.0 m high following an athletics

meeting. Each shot-put has a mass of 500 g and all

are being lifted from the ground. The coach stacks

15 shot-puts, at the same level, in 2.0 minutes.

a How much useful work has been done in lifting

all the shot-puts?b What is the total gravitational potential energy of

all the shot-puts on the shelf?

c What was the coach’s average power output in

performing this task?

d The actual power output would be considerably

greater than the answer to part c. Suggest two

possible reasons for this difference.

55 One of the shot-puts in Question 4 rolls off the shelf

just after the coach has finished.a What is the gravitational potential energy of the

shot-put when it is halfway to the ground?

b What is the kinetic energy of the shot-put when it

is halfway to the ground?

c What happens to the kinetic energy of the shot-

put when it hits the ground?

66 Tarzan is running at his fastest speed (9.2 m s−1) and

grabs a vine hanging vertically from a tall tree in the

jungle.

a How high will he swing upwards while hanging

on to the end of the vine?

b What other factors that have not been considered

may affect your answer?

77 In high jumping, the kinetic energy of an athlete is

transformed into gravitational potential energy.

With what minimum speed must the athlete leave

the ground in order to lift his centre of gravity

1.80 m high with a remaining horizontal velocity of

0.50 m s−1?

88 A 100 g apple falls from a branch 5 m above the

ground.

a With what speed would it hit the ground if air

resistance could be ignored?

b If the apple actually hits the ground with a speed

of 3.0 m s−1, what was the average force of air

resistance exerted on it?

99 A 150 g ball is rolled onto the end of an ideal spring

whose spring constant is 1000 N m−1. The spring is

temporarily compressed.

Energy transformation and power6.5 questions




a The ball compresses the spring by a maximum

distance of 10 cm. How much elastic potential

energy is stored in the spring at this compression?

b How fast must the ball have been travelling just

before it began to compress the spring? Ignore anyfrictional effects.

c If in another trial the ball reached a speed of

5.0 m s−1 before compressing the spring, how far

would the spring be compressed?

110 As a 30 kg child compressed the spring of a pogo

stick, it stored 150 J of elastic potential energy.

Assuming the spring is 50% efficient:

a how much kinetic energy will the child be given

as the spring rebounds?b with what speed will the child rebound?

c ignoring air resistance, what gain in height will

the child achieve?


The following information relates to questions 1–4. A ball of mass

50 g strikes a brick wall. It compresses a maximum distance of

2.0 cm. The force extension properties of the ball are shown below.

800

700

600

500

400

300

200

100

F o r c e ( N )

0.01 0.02Compression (m)

1 What work does the wall do on the ball in bringing it to a stop?

2 How much …p is stored in the ball at its point of maximum

compression?

3 If the ball–wall system is 50% efficient, what is the rebound

speed of the ball?

4 At the instant that the ball had only been compressed by 1.0 cm,

had the wall done half of the work required to stop the ball?

Explain.

The following information relates to questions 5 and 6. An arrow with

a mass of 80 g is travelling at 80 m s−1

when it reaches its target.It penetrates the target board a distance of 24 cm before stopping.

5 Calculate the arrow’s kinetic energy just before impact.

6 Calculate the average net force between arrow and target.

7 A 70 kg bungee jumper jumps from a platform that is 35 m above

the ground. Assume that the person, the rope and the Earth form

an isolated system.

a Calculate the initial total mechanical energy of this system.

b Write a flow chart displaying the energy transformations

that are occurring during the first fall.

c The person can fall a distance of 10 m before the ropeattached to her feet begins to extend. How much kinetic

energy will the person have at this moment?

d After a fall of a further 15 m, the person momentarily stops

and the rope reaches its maximum extension. How much

elastic potential energy is stored in the rope at this moment?

e Since the bungee jumper bounces and eventually comes to

rest, this is not a truly isolated system. Explain.

8 A stone of mass 3 kg is dropped from a height of 5 m. Neglecting

air resistance, what will the kinetic energy of the stone be in

joules just before the stone hits the ground?

A 3B 5

C 15

D 147

E 150

The following information relates to questions 9 and 10. An object of

mass 2 kg is fired vertically upwards with an initial kinetic energy of

100 J. Assume no air resistance.

9 What is the speed of the object in m s−1 when it first leaves the

ground?

A 5

B 10

C 20

D 100

E 200

10 Which of A–E in Question 9 is the maximum height in metres that

the object will reach?

Chapter reviewMomentum, energy, work and power


Worked Solutions



Motion218

The following information relates to questions 11–14.

Casey and Mandeep go bobsledding in their holidays. Figure A shows

the bobsled stationary, with the handbrake engaged. Figure B shows

the bobsled moving down the slope at a constant velocity. The total

mass of the bobsled and occupants is 350 kg.

Casey

Mandeep Casey

Mandeep

30° 30°

A B

11 For the situation shown in Figure A, what is the net force acting

on the sled?

12 For the situation shown in Figure A, calculate the magnitude of

the support force of the slope on the bobsled.

13 For the situation shown in Figure A, calculate the magnitude of

the resistive force that is preventing the bobsled from sliding

down the slope.

14 The two students attempt to explain the situation shown in

Figure B.

• Casey says that as the sled is moving down the slope there

must be more force in that direction. She says that friction

can’t be as big as the ‘pull down the slope’.

• Mandeep says that friction is the same size as the ‘pull down

the slope’ but the sled can still keep moving.

Which student is correct? Explain why you think that student is

correct, making reference to any of Newton’s laws of motion.

The following information relates to questions 15 and 16.

Using one smooth lifting action, a weightlifter lifts a 240 kg barbell

from the ground to above his head. The lowest point of the barbell is

lifted to 1.70 m above the floor and held stationary for 3.00 seconds

before being dropped back to the ground.

15 How much gravitational potential energy did the barbell possess

while being held above the weightlifter’s head?

16 The complete lifting action was carried out in 0.40 seconds. What

power was developed during the lift?

The following information relates to questions 17–20. A roller-coaster

is shown in the following diagram. Assume no friction.

30 m

A

B

C

D25 m

12 m

17 Calculate the speed at points B, C and D, assuming an initial

speed of 4.0 m s−1 at point A.

18 Draw a graph of potential energy and kinetic energy against

vertical displacement for this motion. Use separate lines for each

form of energy and draw in a third line to represent the total

mechanical energy, assuming no frictional losses.

It is found that the roller-coaster actually just reaches point C

with no remaining speed.

19 What are the energy losses due to friction and air resistance

between A and C?

20 With what efficiency is the roller-coaster operating over this

section of track?

21 Two players collide during a game of netball. Just before impact

one player of mass 55 kg was running at 5.0 m s−1 while the

other player, of mass 70 kg, was stationary. After the collisionthey fall over together. What is the velocity as they fall, assuming

that momentum is conserved?

22 A 300 kg marshalling boat for a rowing event is floating at

2.0 m s−1 north. A starting cannon is fired from its bow, launching

a 500 g ball, travelling at 100 m s−1 south as it leaves the gun.

What is the final velocity of the marshalling boat?

23 A 150 g ice puck collides head on with a smaller 100 g ice puck,

initially stationary, on a smooth, frictionless surface. The initial

speed of the 150 g puck is 3 m s−1. After the collision the 100 g

ice puck moves with a speed of 1.2 m s −1 in the same direction.

What is the final velocity of the 150 g ice puck?

24 ‘When I jump, the Earth moves’. Is this true? Using reasonable

estimates and appropriate physics relationships explain your

answer.

The following information relates to questions 25–27. In a horrific

car crash, a car skids 85 m before striking a parked car in the rear

with a velocity of 15 m s−1. The cars become locked together and

skid a further 5.2 m before finally coming to rest. The mass of the

first car, including its occupants, is 1350 kg. The parked car has a

mass of 1520 kg.

25 What is the velocity of the two cars just after impact?

26 What is the impulse on each car during the collision?

27 What is the average size of the frictional force between road and

car that finally brings them to rest?

omentum, energy, work and power (continued)

Chapter QuizWorked Solutions



21Area of study review

The following information applies to questions 1–3. The acceleration

due to gravity may be taken as g = 9.8 m s−2 and the effects of air

resistance can be ignored. An Olympic archery competitor tests a

bow by firing an arrow of mass 25 g vertically into the air. The arrow

leaves the bow with an initial vertical velocity of 100 m s −1.

11 At what time will the arrow reach its maximum height?

22 What is the maximum vertical distance that this arrow reaches?

33 What is the acceleration of the arrow when it reaches its

maximum height?

44 Two students drop a lead weight from a tower and time its fall

at 2.0 s. How far does the weight travel during the 2nd second,

compared with the first second?

The following information applies to questions 5–7. A car with good

brakes, but smooth tyres, has a maximum retardation of 4.0 m s −2

on a wet road. The driver has a reaction time of 0.50 s. The driveris travelling at 72 km h−1 when she sees a danger and reacts by

braking.

55 How far does the car travel during the reaction time?

66 Assuming maximum retardation, calculate the braking time.

77 Determine the total distance travelled by the car from the time

the driver realises the danger to the time the car finally stops.

The following information applies to questions 8–11. Two physics

students, Helen and Emily, conduct the following experiment from a

skyscraper. Helen drops a platinum sphere from a vertical height of

122 m while at exactly the same time Emily throws a lead sphere with

an initial downward vertical velocity of 10.0 m s−1

from a verticalheight of 140 m. Assume g = 9.80 m s−2 and ignore friction.

88 Determine the time taken by the platinum sphere to strike the

ground.

99 Calculate the time taken by the lead sphere to strike the ground.

110 Determine the average velocity of each sphere over their

respective distances.

1111 In reality, will the diameters of the respective spheres affect the

outcome of the experiment?

The following information applies to questions 12–14. During a

physics experiment a student sets a multi-flash timer at a frequency

of 10 Hz. A nickel marble is rolled across a horizontal table. Thediagram shows the position of the marble for the first four flashes:

A, B, C and D.

Assume that when flash A occurred t = 0, at which time the marble

was at rest.

1.0 cm 3.0 cm

A B C D

5.0 cm

1212 Determine the average speed of the marble for the followingdistance intervals:

a A to B

bb B to C

cc C to D

1313 Determine the instantaneous speeds of the marble for the

following times:

a t = 0.05 s

b t = 0.15 s

cc t = 0.25 s

1414 Describe the motion of the marble.

The following information applies to questions 15–18. A tow-truck,pulling a car of mass 1000 kg along a straight road, causes its

velocity to increase from 5.00 m s−1 west to 10.0 m s−1 west in a

distance of 100 m. A constant frictional force of 200 N acts on the car.

1515 Calculate the acceleration of the car.

1616 What is the resultant force acting on the car during this 100 m?

1717 Calculate the force exerted on the car by the tow-truck.

1818 What force does the car exert on the tow-truck?

1919 A car that is initially at rest begins to roll down a steep road that

makes an angle of 11.3° with the horizontal. Ignoring friction,

determine the speed of the car in km h−1 after it has travelled a

distance of 100 m ( g = 9.8 m s−2).

The following information applies to questions 20–23. A 100 kg

trolley is being pushed up a rough 30° incline by a constant force

F . The frictional force F f between the incline and the trolley is 110 N.

F

g = 9.8 m s–2

30˚

220 Determine the value ofF that will move the trolley up the incline

at a constant velocity of 5.0 m s−1.

2121 Determine the value of F that will accelerate the trolley up the

incline at a value of 2.0 m s−2.

2222 Calculate the acceleration of the trolley if F = 1000 N.

2323 What is the value of F if the trolley accelerates up the incline at

10 m s−2?

AREA OF STUDY REVIEW Motion




Motion220

AREA OF STUDY REVIEW (continued)

2424 Two masses, 10 kg and 20 kg, are attached via a steel cable to a

frictionless pulley, as shown in the following diagram.

10 kg

20 kg

g = 9.8 m s–2 a

a

F T

F T

a Determine the acceleration of each mass.

b What is the magnitude of the tension in the cable?

2525 An 800 N force is applied as shown to a 20.0 kg mass, initially

at rest on a horizontal surface. During its subsequent motion

the mass encounters a constant frictional force of 100 N while

moving through a horizontal distance of 10 m.

20.0 kg

60°

F = 800 N

F f = 100 N

a Determine the resultant horizontal force acting on the

20.0 kg mass.

b Calculate the work done by the horizontal component of the

800 N force.

c Calculate the work done by the frictional force.

d Calculate the work done by the resultant horizontal force.

e Determine the change in kinetic energy of the mass.f What is the final speed of the mass?

g Describe the effect of the frictional force on the moving

mass.

2626 The figure shows the velocity–time graph for a car of mass

2000 kg. The engine of the car is providing a constant driving

force. During the 5.0 s interval the car encounters a constant

frictional force of 400 N. At t = 5.0 s, v = 40.0 m s−1.

5040302010

0

v ( m s

– 1 )

t (s)0 2 4 6

a How much kinetic energy (in MJ) does the car have at

t = 5.0 s?

b What is the resultant force acting on the car?

c What force is provided by the car’s engine during the 5.0 s

interval?

d How much work is done on the car during the 5.0 s interval?

e Determine the power output of the car’s engine during the

5.0 s interval.

f How much heat energy is produced due to friction during the

5.0 s interval?The following information applies to questions 27–30. The following

diagram shows the trajectory of a 2.0 kg shotput recorded by a

physics student during a practical investigation. The sphere is

projected at a vertical height of 2.0 m above the ground with initial

speed v = 10 m s−1. The maximum vertical height of the shotput is

5.0 m. (Ignore friction and assume g = 9.8 N kg−1.)

2.0 m

3.0 m

5.0 m A

B

C

2727 What is the total energy of the shotput just after it is released at

point A?

2828 What is the kinetic energy of the shotput at point B?

2929 What is the minimum speed of the shotput during its flight?

330 What is the total energy of the shotput at point C?

3131 A 5.0 kg trolley approaches a spring that is fixed to a wall. During

the collision, the spring undergoes a compression, ∆ x, and the

trolley is momentarily brought to rest, before bouncing back

at 10 m s−1. Following is the force–compression graph for the

spring. (Ignore friction.)

v

spring

5.0 kg

12.0

10.0

8.0

6.0

4.0

2.0

0.0

F o r c e

( k N )

Compression (cm)

0.0 1.0 2.0 3.0 4.0 5.0 6.0

a Calculate the elastic potential energy stored in the spring

when its compression is equal to 2.0 cm.

b What is the elastic potential energy stored in the spring when

the trolley momentarily comes to rest?

c At what compression will the trolley come to rest?

d Explain why the trolley starts moving again.



22Area of study review

e What property of a spring accounts for the situation

described above?

f Describe a situation in which the property of the spring in

this example could be used in a practical situation.

3232 A nickel cube of mass 200 g is sliding across a horizontal surface.One section of the surface is frictionless while the other is rough.

The graph shows the kinetic energy, …k, of the cube versus

distance, x, along the surface.

5.0

4.0

3.0

2.0

1.0

0.00.0 1.0 2.0 3.0 4.0 5.0 6.0 K

i n e t i c e n e r g y ( J )

x (cm)

a Which section of the surface is rough? Justify your answer.

b Determine the speed of the cube during the first 2.0 cm.c How much kinetic energy is lost by the cube between

x = 2.0 cm and x = 5.0 cm?

d What has happened to the kinetic energy that has been lost

by the cube?

e Calculate the value of the average frictional force acting on

the cube as it is travelling over the rough surface.

The following information applies to questions 33 and 34. The diagram

is an idealised velocity–time graph for the motion of an Olympic

sprinter.

Time (s)1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Velocity

(m s–1 )

10

9

8

7

6

5

4

3

2

1

0

3333 What distance was this race?

3434 Determine the average speed of the sprinter:a while she is racing to the finish line

b for the total time that she is moving.

The following information applies to questions 35–37. The diagram

gives the position–time graph of the motion of a boy on a bicycle.

The boy initially travels in a northerly direction.

A B C D E F G

Position

(m)

5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80

Time (s)

80

70

60

50

40

30

20

10

0

3535 During which of the section(s) (A–G) is the boy:

a travelling towards the north?

b stationary?

c travelling towards the south?

d speeding up?

e slowing down?

3636 For the boy’s 80 s ride, calculate:

a the total distance covered

b the average speed.

3737 Determine the velocity of the boy:

a when t = 10 s

b during section B

c when t = 60 s.

The following information applies to questions 38–40. A mass of

0.40 kg hangs from a string 1.5 m long. The string is kept taut and

the mass is drawn aside a vertical distance of 0.30 m, as shown

in the diagram below. A pencil is fixed in a clamp so that when the

mass is released it will swing down and break the pencil. The mass

swings on but now only moves through a vertical distance of 0.14 m.

(Assume g = 9.8 m s−2.)

1 . 5 0 m

0.30 m0.14 m




AREA OF STUDY REVIEW (continued)

3838 Calculate the velocity of the mass the instant before it strikes the

pencil.

3939 Calculate the work required to break the pencil.

440 Can you account for the loss in energy?The following information applies to questions 41–44. A small car

is found to slow down from 90 km h−1 to 60 km h−1 in 12 seconds

when the engine is switched off and the car is allowed to coast on

level ground. The car has a mass of 830 kg.

4141 What is the car’s deceleration (in m s−2) during the 12 s interval?

4242 What was the average braking force acting on the car during the

time interval?

4343 Determine the distance that the car travels during the 12 second

interval.

4444 Explain what happens to all the initial kinetic energy of the car.

4545 A spaceship with a mass of 20 tonnes (2.0 × 104 kg) is launched

from the surface of Earth, where g has a value of 9.8 N kg−1

downwards, to land on the Moon, where g is 1.6 N kg−1

downwards. What is the weight of the spaceship when it is on

Earth and when it is on the Moon?

4646 a Describe the motion of your chair when you stand up and

push it back from your desk.

b How would the chair behave if it were on castors?

c Explain how your answers to parts a and b are illustrations

of Newton’s laws.

4747 An engine pulls a line of train cars along a flat track with a steady

force, but instead of accelerating, the whole train travels at aconstant velocity. How can this be consistent with Newton’s first

and second laws of motion?

4848 Students were conducting an experiment to investigate the

behaviour of springs. Increasing masses (m) were hung from a

vertically suspended spring and the resulting force–extension

graph was plotted as shown.

10.0

8.0

6.0

4.0

2.0

0.0

F o r c e ( N )

Extension (mm)0 20 40 60 80 100 120

a Estimate the value of the spring constant.

b Use your answer to part a to calculate the elastic potential

energy stored in the spring when the extension is 100 cm.

c What other method could you have used to estimate the

energy stored in the spring when the extension is 100 cm?

Jordy is playing softball and hits a ball with her softball bat. The forceversus time graph for this interaction is shown below. The ball has a

mass of 170 g. Assume that the bat and ball form an isolated system

during the interaction.

4949 Determine the magnitude of the change in momentum of the

ball.

550 Determine the magnitude of the change in momentum of the bat.

5151 Determine the magnitude of the change in velocity of the ball.

500

Force (N)

0.015 0.040

Time (s)

newtons laws and energy

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