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Transcript
CHAPTER
5
Newton’s laws
A
lthough 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 objects
were 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.
by the end of this 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.
5.1
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 of
situations 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.
(a)
(b)
Figure 5.1 (a) At the moment of impact, both the tennis ball and racquet strings are
distorted by the forces acting at this instant. (b) The rock climber is relying on the frictional
force between his hands and feet and the rock face. (c) A continual force causes the clay to
deform into the required shape. (d) The gravitational force between the Earth and the Moon is
responsible for two high tides each day. (e) The globe is suspended in mid-air because of the
magnetic forces of repulsion and attraction.
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 are
examples 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!
(c)
(d)
(e)
Chapter 5 Newton’s laws
143
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 a
comparison of the magnitude of some forces.
Table 5.1 A comparison of the magnitude of various forces
Force
Force on the electron in a hydrogen atom
Holding a small apple against gravity
Opening a door
Pedalling a bicycle
Magnitude (N)
10−7
1
10
300
Thrust of a Boeing 747 at take-off
106
Gravitational force between the Earth and the Sun
1022
Force: a vector quantity
Figure 5.2 The netball will only go through
the hoop if a force of the right magnitude and
direction is applied. Force is a vector; it can only
be completely specified if both the direction and
magnitude are given.
W
E
95 N
Figure 5.3 When drawing force diagrams, it is
important that the force is shown to be acting at
the correct location. In this example, the force on
the ball acts at the point of contact between the
ball and the foot.
144
Motion
In 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.
i
FORC… is measured in newtons (N) and is a vector quantity. It requires a
magnitude and a direction to describe it fully.
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
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, ΣF (shown as a doubleheaded arrow).
E
W
95 N
i
The N…T FORC… acting on a body experiencing a number of forces acting
simultaneously is given by the vector sum of all the individual forces acting:
ΣF = F1 + F2 + ... + Fn
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.
Figure 5.4 The two forces being considered
are acting at different locations and have
different strengths. The larger force is shown
as a longer vector.
95 N
95 N
=
+
20 N
ΣF = 75 N
Physics file
20 N
Figure 5.5 When the forces (95 N acting towards the east and 20 N acting to the west) are
added, the resultant or net force is 75 N towards the east. The ball will move as though this
resultant 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
(a)
Person 2
60 N
Person 1
20 N
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 twodimensional vector addition is included
in Appendix A.
North
80 N
View from above
(b)
80 N
80 N
+
60 N
=
ΣF = 100 N
60 N
Figure 5.6 (a) Two perpendicular forces are acting on the trolley. (b) The vector addition
of these two forces gives the resultant force (ΣF) that is acting on the trolley to be 100 N at
127°T. The trolley is treated as a point mass located at its centre of mass.
Chapter 5 Newton’s laws
145
To find the direction of the resultant force, trigonometry must be used:
60
= 0.75
tan θ =
80
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
Figure 5.7 The golf ball moves in the direction of
the applied force and is in the direction of the line
joining the centre of the club-head with the centre
of the ball. The force will be very large.
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
60 N
E
W
Physics file
There are two methods for describing
the direction of a vector in a twodimensional 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
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’.
N
180 N
b
560 N
For the vector addition, treat the boat as a point mass located at its centre of mass.
560 N
ΣF =
=
180 N
560 N
60 N
180 N
+
+
Σ F = 320 N
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 = FSally + FKen
W
E
F
220°T
or S40°W
S
100 N
+
150 N
=
ΣF
100 N
Figure 5.8 The direction 220°T lies 220°
clockwise from north. This direction can also
be written as S40°W meaning 40° west of
south.
146
Motion
60 N
150 N
ΣF = √1002 + 1502 = 180 N
150
= 1.5
tan θ =
100
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.
F = 45 N
20°
Figure 5.9 The pulling force acting on the cart
has a component in the horizontal direction
and a component in the vertical direction.
Fv = 45 sin 20°
= 15 N
45 N
20°
Fh = 45 cos 20°
= 42 N
Figure 5.10 The magnitudes of the vector
components Fh and Fv can be calculated using
trigonometry.
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
westerly component of
the force = 50 N
N
30n
W
force from
the hockey
stick = 100 N
northerly
component of
the force = 87 N
30n
E
S
Chapter 5 Newton’s laws
147
FW = 100sin30° = 50 N
FN = 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 down
with a force of 600 N. What is the actual force applied by Kate’s foot?
Solution
Q
F
Fv = 600 N down
Q
Fh = 400 N
5.1 summary
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 = Fhorizontal + Fvertical and a
vector diagram is needed.
Using Pythagoras’ theorem:
F = √Fh2 + Fv2 = √4002 + 6002 = √520 000 = 721 N
600
θ = tan−1
= tan−11.5 = 56°
400
So Kate supplies a force of 721 N backwards at 56° down from the horizontal.
Force as a vector
• 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:
ΣF = F1 + F2 + … + Fn
148
Motion
• 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 F cos θ parallel to the reference
direction, and F sin θ perpendicular to that reference
direction.
5.1 questions
Force as a vector
1 Which one or more of the following quantities are
vectors?
A mass
B velocity
C temperature
D force
7 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
2 Calculate the resultant force in each of the following
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
8 A small car is pulled by two people using ropes. Each
person 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?
400 N
40°
40°
3 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
d doing a chin-up exercise.
400 N
9 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
4 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
5 Convert the following into full circle bearings (i.e. °T).
a N60°E
b N40°W
c S60°W
d SE
e NNE
10 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?
6 Use the vectors below to determine the forces represented in the following situations.
Scale: 1 cm represents 20 N
a
b
c
F
F
F
Worked Solutions
Chapter 5 Newton’s laws
149
5.2
Newton’s first law of motion
Aristotle and Galileo
Interactive
(a)
Ideally you would expect the
ball to reach this height.
(b)
Ball ideally would roll further
and reach this position.
(c)
The ball ideally will keep
moving with a constant velocity.
Figure 5.11 Galileo used a thought experiment
to derive his law of inertia. This became Newton’s
first law and stated that the natural state of
bodies was to maintain their original motion.
This contradicted Aristotle’s idea that the natural
state of bodies was at rest.
150
Motion
The 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 place
where 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 section
was 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, nor
would 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
understanding that Newton was able to make. Any body will continue
with constant velocity if zero net force (ΣF = 0) acts upon it.
i
N…WTON’S FIRST LAW OF MOTION states that a body will either remain at rest or
continue with constant speed in a straight line (i.e. constant velocity) unless it
is acted on by an unbalanced force.
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.
Figure 5.12 An air-track glider moves with a constant velocity because there is zero net force
Physics file
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.
Figure 5.13 Ball bearings reduce friction and
enable wheels to work very efficiently.
acting on it. This is an example of inertia.
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.
Physics file
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.
Figure 5.14 Two Voyager spacecraft were launched from Cape Canaveral in 1977 with
the mission to investigate the outer planets of the solar system at close hand. Both craft
completed the mission successfully, passing Saturn in 1981, Uranus in 1986 and Neptune
in 1989. Voyager 1 and 2 have now left the solar system and since they have effectively
zero net force acting on them, they continue to travel away from the Earth with a constant
velocity.
Chapter 5 Newton’s laws
151
Σ F= 0
velocity doesn’t
change
FT
Fg
Figure 5.15 The forces acting on this book are
balanced—i.e. ΣF = 0—so the velocity of the
book will not change; the book will continue to
stay at rest. That is, the book’s velocity will be
constant at 0 m s−1.
Physics file
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.
Forces in equilibrium
Newton’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. This
applies 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 driving
forces must balance the retarding forces, i.e. ΣF = 0.
(a) deep space
(b)
Fground
v = 50 km h–1
Fdrag
Fdriving
Fgravity
no forces acting
YF = 0
velocity doesn’t change
forces are balanced
YF = 0
velocity doesn’t change
Figure 5.17 (a) There are no forces acting on the spacecraft. The net force is zero and so it
continues 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.
Worked example 5.2A
Figure 5.16 Steven Bradbury won the gold
medal as his opponents learned first-hand
about inertia and friction (or rather a lack of
friction!).
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. If
the 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.)
152
Motion
lift
F1
F2
applied
force
upwards
force from
ground on
roller
drag
Fthrust
weight, Fg
friction
weight of roller
Fdrag
Fg
F2
applied push
Flift
Fg
F1
Fg
Fthrust
Fg + F1 + F2 = 0
upward force
from ground
friction
weight of roller
Fthrust + Fg + Fdrag + Flift = 0
Figure 5.18 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 the
picture hanging on the wall, or continue with constant velocity, like the aircraft. A more complex situation involves the groundsman pushing the heavy
roller with constant velocity. The horizontal component of the force he applies along the handle exactly balances the frictional force that opposes the
motion of the roller in the horizontal direction. The vertical component of the applied force acts downwards, and adds to the weight of the roller, but
these two downward forces are balanced by an upward force provided by the ground.
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
Fdrag = 60 N
Fapplied = 60 N
ΣF = Fapplied + Fdrag = 0
Prac 21
SPARKlab
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.)
Chapter 5 Newton’s laws
153
Physics in action
Galileo Galilei—revolutionary
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, government
was 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 2nd
century 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
5.2 summary
Motion
red hair. Galileo made significant contributions to our understanding of
the forces that act on moving bodies. In his book the Principia, Newton
was quick to acknowledge his debt to Galileo’s genius.
This portrait was drawn 8 years before Galileo’s trial.
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. He
had 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.
Newton’s first law of motion
• 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
154
Figure 5.19 Galileo Galilei (1564–1642) was a short, active man with
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. ΣF = 0,
the forces are said to be balanced and are in
equilibrium.
5.2 questions
Newton’s first law of motion
1 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.
2 A billiard ball is rolling freely across a smooth
horizontal surface. Ignore drag and frictional forces
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?
A
B
F
F
C
D
F
F
3 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? Use
Newton’s first law of motion to explain what is
happening.
4 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?
5 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?
6 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 the
kart 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?
7 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?
8 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.
9 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.
10 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.
Worked Solutions
Chapter 5 Newton’s laws
155
5.3
Newton’s second law of motion
Figure 5.20 This sprinter is about to leave
the starting blocks. The starting blocks stop
his foot from slipping backwards, increasing
the size of the forward force acting on him and
increasing his forward acceleration.
Physics file
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!
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. ΣF ≠ 0; in other words, when
the forces are unbalanced.
When there is a non-zero net force acting on a body, the body will
accelerate 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:
a ∝ ΣF
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
a∝
m
If the two relationships are combined, we get:
1
a ∝ ΣF ×
m
or
ΣF
a∝
m
The relationship can be converted into an equality by including a constant
of proportionality, so:
ΣF
a=k×
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 ΣF = ma, a mathematical statement of
Newton’s second law of motion.
i
N…WTON’S S…COND LAW OF MOTION states that the acceleration of a body, a, is
directly proportional to the net force acting on it, ΣF, and inversely proportional
to its mass, m:
ΣF = ma
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).
ΣF = ma 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.
156
Motion
Solution
First, determine the acceleration of the athlete:
v = u + at
v−u
a=
t
12 − 0
=
5.0
a = 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.
Prac 22
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 = Fapplied + Fdrag
A vector addition gives ΣF = 30 N forwards.
So,
ΣF 30
=
a=
m 75
= 0.40 m s−2 in the direction of the applied force
Fapplied
+
Fdrag
50 N
20 N
ΣF
30 N
(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
ΣF = F1 + F2
ΣF = √F12 + F22
= √152 + 202 = √225 + 400 = √625 = 25 N
ΣF 25
a=
=
m 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:
F 20
tan θ = 2 = = 1.33
F1 15
so θ = tan−1 1.33 = 53°
The ball will travel in the direction N53°E or (053ºT).
F1 = 15 N north
F2 = 20 N east
F2
N
W
E
F1
ΣF
S
θ
Chapter 5 Newton’s laws
157
Physics file
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.
Figure 5.21 Scientists working at the
Australian synchrotron in Clayton have to take
account of these relativistic effects on the
electrons 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 or
3 × 107 m s−1), 0.8c and 0.99c.
Table 5.2 The mass of a 1 kg block
at 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.
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 these
expressions 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 of
mass 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
ΣF
horizontal direction, an acceleration is produced that is given by a =
.
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.
i
The (inertial) MASS of a body is its ability to resist acceleration when the body
is acted on by a net force. Mass is a constant property of the body.
on Earth
in deep space
F=1N
a = 1 m s–2
1 kg
F=1N
1 kg
a = 1 m s–2
Figure 5.22 Regardless of the external conditions, the inertial qualities of a mass remain the
same. A net force of 1 N will always produce an acceleration of 1 m s−2 for a 1 kg mass. In this
way, 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.
158
Motion
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,
Fg 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.
a = 9.8 m s–2
Prac 23
SPARKlab
a = 9.8 m s–2
Figure 5.23 If air resistance is ignored, the pumpkin and banana will fall side by side with an
acceleration of 9.8 m s−2 towards 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.
a = 1.6 m s–2
a = 1.6 m s–2
Figure 5.24 The pumpkin and banana will fall side by side with a uniform acceleration of
1.6 m s−2 towards the Moon.
i
The acceleration of a freely falling object due to gravity is known as g.
Chapter 5 Newton’s laws
159
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, Fg. Hence:
ΣF = Fg
ma = Fg
The acceleration of the pumpkin is 9.8 m s−2 or g, so Fg = mg.
a = 9.8 m s–2
Fg
i
Figure 5.25 The pumpkin is in free-fall. The
only force acting on it is gravity, Fg, and it
accelerates at 9.8 m s−2 towards the ground.
The W…IGHT of a body, W or Fg, is defined as the force of attraction on a body
due to gravity:
W = Fg = mg
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 the
gravitational field strength is newton/kg or N kg−1.
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 at
the 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.
10 kg
0.10 kg
Fg = 98 N
Fg = 0.98 N
Figure 5.26 The force dragging the pumpkin to the ground is much greater than the force
that is acting on the banana. However, the mass and inertia of the pumpkin is much greater
than the mass and inertia of the banana. The acceleration that results in both cases is
identical: 9.8 m s–2 down.
160
Motion
The force of gravity acting on the 10 kg pumpkin is:
W = Fg = mg = 10 × 9.8 = 98 N down
This is the only force acting on the pumpkin so the net force, ΣF, is also
98 N down.
The acceleration of the pumpkin can be calculated:
ΣF 98
a=
=
= 9.8 m s−2 down
m
10
The force of gravity acting on the 0.10 kg banana is:
W = Fg = mg = 0.10 × 9.8 = 0.98 N down
This is the only force acting on the banana so the net force, ΣF, is also 0.98
N down.
The acceleration of the banana can be calculated:
ΣF 0.98
a=
=
= 9.8 m s−2 down
m 0.10
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?
Figure 5.27 The weight of this boulder is the
force it experiences due to gravity given by
Fg = mg. This is approximately 2.5 × 105 N directed
to the centre of the Earth. The mass of the boulder
is approximately 25 000 kg. If the boulder were
taken to outer space where the gravitational field
strength is zero, the boulder would still have the
same mass but no weight.
Solution
a The cart and mass experience a net force equal to the weight of the falling mass. So
ΣF = Fg = mg = 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.
1.5 kg
2.5 kg
ΣF
m
24.5
=
= 6.1 m s−2 to the right
4.0
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,
ΣF 16.0
=
and a =
= 4.0 m s−2 to the right.
m 4.0
a=
b
24.5 N
8.5 N
Chapter 5 Newton’s laws
161
Physics in action
Terminal velocity
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 dropped
objects. 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),
, and gravity, g. Since his speed is low, the drag force is
a
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 ∝ v2. This results in a smaller net
force downwards (Figure 5.28b). His downwards acceleration
is therefore reduced. It is important to remember that he is
still 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
(a)
(c)
Fa
ΣF
Fa
v
Fg
Fg
(d)
(b)
Fa
v
Fa
ΣF
ΣF = 0
v
Fg
Fg
Figure 5.28 As the jumper falls, the force of gravity does not change,
but the drag force increases as he travels faster. Eventually these two
forces will be in equilibrium and the jumper will fall with a constant or
terminal velocity.
force will also remain constant and the motion of the jumper
will 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.
Figure 5.29 Skydivers can change their speed by changing their body profile. If they assume a tuck position they will fall faster and if they
spreadeagle they will fall slower. This enables them to meet and form spectacular patterns as they fall.
162
Motion
ΣF = 0
v
5.3 summary
Newton’s second law of motion
• Newton’s second law of motion states that the
acceleration a body experiences is directly proportional to the net force acting on it, and inversely
proportional to its mass:
ΣF = ma
where m is measured in kilograms (kg), a is
measured in metres per second squared (m s−2), and
ΣF in newtons (N).
5.3 questions
• 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 Fg is defined as the force
of attraction on a body due to gravity. This will be
given by W = Fg = mg where m is the mass of the body
and g is the strength of the gravitational field.
Newton’s second law of motion
Use g = 9.8 m s−2 when answering these questions.
1 State whether the forces are balanced (B) or
unbalanced (U) for each of these situations.
a a netball falling towards the ground
b 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
2 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?
3 Use Newton’s laws to explain why a 1.0 kg shot-put
can be thrown further than a 1.5 kg shotput.
4 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.
5 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.
a 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.
c Find the net horizontal force acting on the canoe.
d Calculate the magnitude of the acceleration of the
canoe.
7 On the surface of the Earth, a geological hammer has
a mass of 1.5 kg. Determine its mass and weight on
Mars where g = 3.6 m s−2.
8 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?
9 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.
10 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?
a What is Mary’s mass?
b Calculate Mary’s weight.
Worked Solutions
Chapter 5 Newton’s laws
163
5.4
Newton’s third law of motion
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 the
pair 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.
(a)
(b)
force that
nail exerts
on hammer
Figure 5.30 A hammer hitting a nail is a
good example of an action–reaction pair and
Newton’s third law.
force that
hammer exerts
on nail
force that
nail exerts
on hammer
force that
hammer exerts
on nail
Figure 5.31 (a) As the hammer gently taps the nail, both the hammer and nail experience
small forces. (b) When the hammer smashes into the nail, both the hammer and nail
experience large forces.
Now consider what will happen if the hammer is smashed into the nail.
The nail now experiences a large downwards force as the hammer smashes
into 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.
i
N…WTON’S THIRD LAW OF MOTION states that for every action force (object A on
B), 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.
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
164
Motion
(a)
(b)
force that foot
exerts on ball
force that floor
exerts on girl
force that girl
exerts on floor
force that ball
exerts on foot
(c)
(d)
gravitational force
that Earth exerts
on brick (Fg)
force that bus
exerts on car
gravitational force
that brick exerts
on Earth
force that car
exerts on bus
Figure 5.32 Some action–reaction force pairs. Notice that these can be contact or noncontact forces. (a) The girl exerts a downwards force on the floor and the floor exerts an equal
upwards force on the girl. (b) The foot exerts a forwards force on the ball and the ball exerts
an equal size backwards force on the foot. (c) The bus exerts a backwards force on the car
and the car exerts an equal size forwards force on the bus. (d) The Earth exerts a downwards
gravitational force on the brick and the brick exerts an equal size upwards force on the Earth.
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 on
the 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.
Figure 5.33 The soccer ball and the player’s
head exert equal forces on each other during this
collision, but only the player will experience pain!
Worked example 5.4A
In each of the following diagrams, one of the forces is given.
(a)
i
ii
iii
(b)
(c)
Describe each given force using the phrase ‘force that
exerts on
’.
Describe the reaction pair to the given force using the phrase ‘force that
exerts on
’.
Draw each reaction force on the diagram, carefully showing its size and location.
Chapter 5 Newton’s laws
165
Solution
a i force that bat exerts on ball
ii force that ball exerts on bat
iii see diagram at right
b
i
ii
iii
c iii
ii
iii
force that ball exerts on floor
force that floor exerts on ball
see diagram at right
gravitational force that Earth exerts on astronaut
gravitational force that astronaut exerts on Earth
see diagram at right
Motion explained
F (reaction)
Fforwards
θ
Fup
F (action)
θ
θ
F (reaction)
Figure 5.34 Walking relies on an action and
reaction force pair in which the foot will push down
and backwards with an action force. In response, the
ground will push upwards and forwards. The forward
component of the reaction force is actually friction.
This is responsible for the body moving forward as a
whole, while the back foot remains at rest.
166
Motion
Newton’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 = ma, 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 supply
the 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.
Table 5.3 All motion can be explained in terms of action and reaction
force 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.
1200 kg
Solution
a This is the forwards force that the road exerts on the front tyre and could be called a
frictional force.
b
c
force A
d
force B
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 force
When 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 (Fg = W = mg) is
still acting. Since the bin is at rest, there must be another force (equal in
magnitude 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 FN or N, (often abbreviated to the normal force).
i
The NORMAL FORC… is a reaction force supplied by a surface at 90° to its plane.
Chapter 5 Newton’s laws
167
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.
Figure 5.35 (a) When the bin is in mid-air,
there is an unbalanced force acting on it so it
accelerates towards the ground. (b) When the
empty bin is resting on the table, there is a small
upwards force from the table acting on it. The
bin remains at rest, so the forces are balanced.
(c) The forces acting on the full bin are also
balanced. The weight of the bin is greater, so the
normal force exerted by the table is larger than
for the empty bin.
(a)
(b)
(b)
Fg
FN
FN
Fg
Fg
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 and
the other force acts on the other object. Let us identify the reaction pair to
each of the forces shown in Figure 5.36.
(a)
Figure 5.36 (a) The reaction force pair to
the weight of the bin is the gravitational force
of attraction that the bin exerts on the Earth.
(b) The reaction force pair to the normal force
on the bin is the force that the bin exerts on the
table. The force pairs are equal and opposite
but they do not cancel out because they are
not acting on the same object.
ΣF = Fg + FN = 0
FN
Fg
(b)
Fg = gravitational force
Earth exerts on bin
FG = gravitational force bin
exerts on Earth
FN = force table
exerts on bin
FT = force bin exerts
on table
In Figure 5.36a the action force shown is the force of gravity Fg 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 FG. In Figure 5.36b, the action force shown is the normal force on
the bin, FN. 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 FT.
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 the
table.
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 be Fg = mg = 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 = Fg + FN = 0.
168
Motion
b
c
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.
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 plane
The 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 Fg and the normal force FN supplied
by the surface. The weight force always acts downwards and is given by
Fg = mg. The normal force is supplied by the surface and will vary depending
on 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: Fg = 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.
(a)
(b)
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.
Prac 24
SPARKlab
ΣF = Fg + FN
ΣF = Fg + FN
=0
FN
Figure 5.37 (a) Where the surface is
FN
FN
Fg
FN
body remains
at rest
Fg
Physics file
Fg
θ
ΣF
θ
Fg
perpendicular to the weight force, the normal force
will act directly upwards and cancel the weight
force. (b) On an inclined plane, FN is at an angle to
Fg and as long as no friction acts, there will be a net
force down the incline. The body will accelerate.
From the vector diagram of the forces:
ΣF = Fg + FN = Fg sin θ = mg
g sin θ
From Newton’s second law, the net force is:
ΣF = ma
so,
ma = mgg sin θ
or
a = g sin 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 a = g sin θ
= 9.8 × sin 20° = 3.4 m s−2—quite a rate!
Chapter 5 Newton’s laws
169
Physics file
In February 2003, a train driver pulled
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
train simply rolled off down the hill,
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
Jacana
Glenroy
Oak Park
Pascoe Vale
Glenbervie
Strathmore
Moonee Ponds
Essendon
Ascot Vale
Newmarket
Kensington
North Melbourne
Spencer Street Station
0
5 km
Figure 5.38 An empty three-carriage
train took 16 minutes to roll downhill from
Southern Cross) station. The track was like a
long inclined plane and the train accelerated
along it after the driver forgot to put the
handbrake on!
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.
YF = Fg + FN
FN
Fg
FN 30\$
Fg
30\$
Solution
ΣF = Fg + FN
From the vector diagram:
ΣF = Fg sin θ
So, ma = mg 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 = Fg + FN. This is a vector addition. From the vector
diagram:
ΣF = Fg sin θ
So, a = g sin 35°
= 9.8 sin 35°
= 5.6 m s−2
ΣF
FN
Fg
35°
ice patch
80 m
Fg
170
Motion
FN
35°
b
If this acceleration continues over 80 m, Kristie’s final speed would be:
v2 = u2 + 2ax
so v2 = 0 + 2 × 5.6 × 80 = 900
v = 30 m s−1 (110 km h−1)
Kristie is travelling with a constant velocity, so ΣF = 0, i.e. the force of friction Ff would
balance the component of the weight force parallel to the incline.
So Ff = Fg sin 35 = 340 N up the incline.
Tension
Another 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.
(a)
(b)
tension -T
tension -T
-T = 490 N
Σ- = 0
T = 50 kg
-g = 490 N
Figure 5.39 (a) The stretched cable exerts an upwards force on the person and an equal size
downwards force on the beam. (b) The forces acting on the person are balanced, so the tension
force is equal in magnitude to the weight force (490 N).
At the top end of the cable, the tension force FT pulls down on the beam.
At the other end, the cable exerts an upwards force FT 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.
Chapter 5 Newton’s laws
171
The system of the rope and the monkey is now subjected to a net force of:
ΣF = 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:
ΣF 49
=
= 2.0 m s−2
a=
Σm 25
(a)
(b)
(c)
FT = 117 N
FT = 118 N
15 kg
a
147 N
ΣF = 15 × 2.0
= 30 N
ΣF = 10 × 2.0
= 20 N
10 kg
98 N
Fg =147 N
Fg = 98 N
ΣF = 49 N
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 the
rope is found by considering the forces that act on each weight.
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 ΣF = Fg + FT.
So,
FT = ΣF − Fg
= 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,
ΣF = Fg + FT
So,
FT = ΣF − Fg = 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.
172
Motion
Physics in action
Frictional forces
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 than s is needed to
keep the book moving. This force is called the kinetic friction
force, and is represented by k.
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 is
required 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
Applied force
maximum
static friction
Fs
kinetic friction
Fk
Time
motion
Fk
Fs
PHYSICS
11
overcoming friction
PHYSICS
11
constant velocity
Figure 5.41 To get things moving, the static friction between an
object and the surface must be overcome. This requires a larger
force than that needed to maintain constant velocity.
Figure 5.42 This magnetic levitation train in China rides 1 cm above
the track, so the frictional forces are negligible. The train is propelled by
a magnetic force to a cruising speed of about 430 km h−1.
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 brakepads 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.
Chapter 5 Newton’s laws
173
5.4 summary
Newton’s third law of motion
• 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 objects
and so should never be added together.
• Whenever a force acts against a fixed surface, the
surface provides a normal force, FN 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.
5.4 questions
Newton’s third law of motion
1 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.
2 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.
3 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 the
astronaut 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?
174
Motion
• 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
a = 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.
4 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?
5 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.
6 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), determine 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?
7 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?
8 Discuss and compare the size of the normal force
that acts on the skater as he travels down the halfpipe from A to B to C as shown.
x
x
x
A
9 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 but
opposite 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?
10 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
grabs onto the other end of the
rope to lower himself.
a When the student steps off
the ladder, what is the net
30 kg
external force on the rope?
b With what acceleration will
50 kg
the system move?
c What is the tension in the
rope?
B
C
Worked Solutions
Chapter review
Newton’s laws
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
b
c
Calculate the size of the drag force acting on the parachutist
at this speed.
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?
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?
Continued on next page
Chapter 5 Newton’s laws
175
Newton’s laws (continued)
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 and
the 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?
Applied
force
(N)
0.5
0.2
0
1
t (s)
2
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?
A
FN
Fg
B
F
FN
Fg
C
FN
Fdrag
F
Fg
D
176
Motion
Fg
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.
120 N
30 kg
50 kg
a
b
With what acceleration will the trolleys move?
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:
horizontal force on the girl
horizontal force on the skateboard
acceleration of the girl
b
acceleration of the skateboard
final velocity of the girl
c
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?
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
1
car accelerates at 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
14 The force diagram below shows the forces acting on a full
water tank.
b
FN
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 they
collide. Who is correct? Explain.
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.
Fg
a
b
Are these forces an action–reaction pair as described by
Newton’s third law?
Worked Solutions
Chapter Quiz
Chapter 5 Newton’s laws
177
CHAPTER
6
Momentum, energy,
work and power
I
s skydiving on your list of things to do in your future? Basejumping? 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 thread
of 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!
by the end of this chapter
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.
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:
Σ net = m
and using the relationship for acceleration:
−
m( − )
Σ =
=
Δt
Δt
where a is the acceleration during the collision (m s−2)
Δt 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 Δt = m( − )
or:
Σ Δt = mΔ
This relationship introduces two important ideas.
• The product of net force and the contact time is referred to as impulse,
I. The idea of impulse is commonly applied to objects during collisions
when 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:
=
−1
where is momentum (kg m s )
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 moving
with 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.
Interactive
Figure 6.1 When two footballers collide, they
exert an equal and opposite force on each
other. The effect this force will have on the
velocity of each footballer can be investigated
using the concept of momentum.
Figure 6.2 The enormous mass of a large ship
endows it with very large momentum despite
its relatively slow speed. After turning off its
engines the ship can continue against the
resistance of the water for more than 4 km if
no other braking is applied.
Chapter 6 Momentum, energy, work and power
179
That is:
Δp
ΣF = Δt
where ΣF is the average net force applied to the object during the collision,
in newtons (N)
Δp is the change in momentum during contact for a time Δt.
i
The change in momentum of a body is proportional to the net force applied to it:
Δp ∝ ΣF
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 (Δp) of the object will be given by:
Change in momentum = final momentum − initial momentum
Δp = pf − pi
i
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:
b
180
Motion
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:
Δp = 0 − 656 = −656 kg m s−1 towards the pole
or:
Δp = 6.6 × 102 kg m s−1 away from the pole
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 (100) 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:
Δp
F=
Δt
656
F = −1 = 6560 ≈ 7 × 103 N away from the pole, i.e. a retarding force.
10
Impulse
Think 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 Δt),
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 then
decrease 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. I = FavΔt. The total impulse during the time the ball is in
contact will be I = Fav × t, where Fav 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 IMPULS… affecting an object during a collision is the product of the net
average applied force and the time of contact and is equivalent to the area
under a force–time graph.
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
momentum
From the derivation earlier in the chapter, the impulse is also equal to the
change of momentum for an object. Previously we had:
mΔ
F=
Δt
Impulse = Fav × ¬t
= area under graph
(b)
F (N)
i
(a)
0
0.03 0.06 0.09 0.12 0.15
t (s)
Figure 6.3 (a) When a tennis player hits a ball, an
unbalanced force is applied to the ball, producing
a change in its momentum; hence an impulse is
applied to the ball. The magnitude of the force will
change over time. (b) The impulse can be found
from the area under the force–time graph since
area = x-axis × y-axis = Fav × Δt = impulse.
Chapter 6 Momentum, energy, work and power
181
Multiplying by the time interval Δt:
FavΔt = mΔ
or:
I = FavΔt = Δp
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
b
c
Final momentum: pf = mv = 0.058 × 0 = 0 kg m s−1
Change in momentum: Δp = 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.
Impulse = change in momentum:
I = 3.19 ≈ 3.2 N s
Using I = FavΔt
I
then Fav =
Δt
3.19
so Fav =
= 21.27 N ≈ 21 N in the opposite direction to the ball’s travel.
0.15
Physics in action
Forces during collisions
(a)
(b)
Figure 6.4 A simple example of the effect on the applied force of the
stopping time during a collision can be achieved with nothing more
complicated than a hammer, a can of fruit and your finger. (a) A rigid
object, such as the hammer, will stop quickly. The applied force will be
large. (b) A can will experience the same change in momentum, but,
having a simple crumple zone, will stop more slowly, thus reducing
the applied force to a tolerable amount.
182
Motion
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.
I = Δp = mΔv
= 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:
I
10
= 100 N
F=
=
Δt 0.1
—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:
I
10
= 20 N
F=
=
Δt 0.5
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 prove
this 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 reduced
the 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.
Table 6.1 Relative size of forces associated with
some common movements in sport
Movement
Footwear
Ratio of normal to
weight 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
They offer good cushioning but are more responsive, allowing
faster take-offs than grass.
Vehicle safety design
Designing 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 airbag 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.
Figure 6.5 The forces developed between track and foot can do
considerable damage to a runner’s body unless they are reduced by
increasing the stopping time through cushioning of the foot. Modern
track shoes incorporate sophisticated design principles to increase
stopping time and thus decrease the forces generated.
Continued on next page
Chapter 6 Momentum, energy, work and power
183
Forces during collisions (continued)
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
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 compression
or 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:
F
P=
A
where P is the pressure (N m−2)
F 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 also
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.
(a)
Applied force (N)
(b)
no air bag
with air bag
0
184
Motion
20
40
60
80
Time (ms)
100
120
140
Figure 6.6 (a) The air bag is one of a number of passive safety
features incorporated into the design of modern cars. It extends
the stopping time, significantly reducing the forces on the head
and neck during a collision. It also distributes the force required to
decelerate the mass of the driver or passenger over a larger area
than a seat belt. (b) The deflation rate of the bag is governed by the
size of holes in the rear of the air bag, and is designed to provide
the optimum deceleration of the head for a large range of impacts.
The relationship between momentum and force
• 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 kg
v is in m s−1
Momentum is a vector quantity.
• The change in momentum (Δp) = final momentum
(pf) − initial momentum (pi).
• Impulse is the product of the net force during a
collision and the time interval Δt during which the
6.1 questions
force acts: I = FavΔt. 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, Δp, caused by the action of the net
applied force:
FavΔt = Δp
• Extending the time over which a collision occurs will
1
decrease the average net force applied since Fav ∝ .
Δt
This is the principle behind many safety designs.
The relationship between momentum and force
Use g = 9.8 m s−2 where required.
1 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?
2 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?
3 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?
4 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 a
mass of 3.4 kg
d that experiences a net force of magnitude 45 N, if
the net force is applied for 3.5 s.
5 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 the
racquet 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?
6 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 ball
during the contact time?
c What is the net average force acting on the bat
during the contact time?
7 The following graph shows the net vertical force
generated as an athlete’s foot strikes an asphalt
running track.
Force (N)
6.1 summary
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.
Continued on next page
Chapter 6 Momentum, energy, work and power
185
The relationship between momentum and force (continued)
b Estimate the total impulse during the contact
time.
8 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?
9 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
b Would a rigid ‘shell’ be as successful? Explain.
10 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.
Worked Solutions
186
Motion
6.2
Conservation of momentum
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:
F1 = −F2
From Newton’s second law, ΣF = ma, so the forces could be expressed as:
m1a1 = −m2a2
v−u
we get:
and using a =
Δt
v − u1
v − u2
m1 1
= −m2 2
Δt
Δt
where a1 and a2 are the respective accelerations of the two objects during
the collision in m s−2
Δt is the time of contact (s)
u1 and u2 are the velocities of the objects prior to collision (m s−1)
v1 and v2 are the velocities after collision (m s−1)
Since the time that each object is in contact with the other will be the same,
Δt will cancel out:
m1(v1 − u1) = −m2(v2 − u2)
or:
m1u1 + m2u2 = m1v1 + m2v2
In other words, the total momentum before colliding is the same as the total
momentum after the collision.
i
TH… LAW OF CONS…RVATION OF MOM…NTUM 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; that is, the total initial
momentum will equal the total final momentum:
Σpi = Σpf
or
m1u1 + m2u2 = m1v1 + m2v2
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 nearfrictionless 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
Physics file
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 be
investigated 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 equal
momentum in the opposite direction.
Rockets destined for space carry their
own oxygen supply, while jet engines use
the surrounding air supply.
Chapter 6 Momentum, energy, work and power
187
6.0 m s–1
6.0 m s–1
Figure 6.7 When two skaters collide on a nearfrictionless skating rink they exert equal and
opposite forces on each other. The total momentum
of the two skaters before the collision will equal
the total momentum of the two skaters after the
collision. Because no other large horizontal forces
are involved other than those in the collision, this
can be considered as an isolated system.
Prac 25
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 with
a 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:
b
Figure 6.8 Newton’s cradle, or Newton’s balls
to some, is an instructive ‘executive toy’ based
on the principle of conservation of momentum
extended over a number of objects.
188
Motion
Σpi = Σpf
or m1u1 + m2u2 = m1v1 + m2v2
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, u2 = −6.0 m s−1, v1 = 0 m s−1, v2 = ?
Substituting into equation:
80 × 6.0 + 70 × (−6.0) = 80 × 0 + 70 × v2
and v2 = +0.86 m s−1.
The 70 kg skater bounces back in the opposite direction with a speed of 0.86 m s−1.
Treating the two skaters as one mass after the collision:
80 × 6.0 + 70 × (−6.0) = (80 + 70) × v2
and now v2 = +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.
Physics in action
Collisions and pedestrians
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 is
travelling 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:
=m
car
1000
)
= 1400 × (60 ×
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)
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
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.
2. 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’s
head 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 little
impact-absorbing ability. It is, therefore, logical to ban bull
bars in residential areas.
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 reduced
speed 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
Figure 6.9 This Nissan car has a pop-up bonnet that has been
designed to result in less damage to a pedestrian in a collision,
by making space between the bonnet and the engine.
Chapter 6 Momentum, energy, work and power
189
6.2 summary
Conservation of momentum
• 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 = Σpf
6.2 questions
• Conservation of momentum can be extended to
any number of colliding objects within an isolated
system.
Conservation of momentum
1 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?
2 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?
3 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?
4 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?
5 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
6 A superhero stops a truck simply by blocking it with
his outreached arm.
a 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.
7 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?
8 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
9 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.
10 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.)
Worked Solutions
190
Motion
6.3
Work
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 the
observations 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 greater
effort 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.
Figure 6.10 In each situation involving work,
a load can clearly be specified.
Chapter 6 Momentum, energy, work and power
191
Physics file
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!
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 determining 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.
(a) I’m doing no
work on the
crate since
x=0
I’m doing no work
on the crate since
x=0
W = Fx = 0
(b)
I’m doing
work!
work is done
W = Fx
crate speeding up
Figure 6.11 (a) No work is done on the crate since its energy is not altered. (b) The
energy of the crate is changing, so work is being done on the crate.
Work done by a constant force
If 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:
Figure 6.12 During the fall the force due to
gravity does work on the person and produces
a displacement.
i
The net WORK done on an object is defined as the product of the net force on
the object and its displacement in the direction of the net force.
When the force and displacement are in the same direction, the work done by
the 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 of
1 newton moves an object through a distance of 1 metre in the direction of the
net force.
Physics file
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
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).
192
Motion
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.
W = Fx cos θ
where θ is the angle between the applied force and the direction of motion.
i
(a)
(b)
(b)
F
F
F
θ
F cos θ
Direction of motion
Direction of motion
Direction of motion
Figure 6.13 (a) If a force is applied in the direction of motion of the cart, then the force is at its most effective in moving
the cart. (b) When the force is applied at an angle θ to the direction of motion of the cart, the force is less effective. The
component of the force in the direction of the displacement, F cos θ, is used to calculate the work. (c) When the angle at
which the force is acting is increased to a right angle (θ = 90°), then the component of the force in the direction of the
intended 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.
Work and friction
If 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 approach
would 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 the
applied 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.
ΣF = 200 N
Net work
done on crate
= 1000 J
Fapplied
= 300 N
frictional
force Ff
= 100 N
Figure 6.14 The object slides across a
displacement of 5 m. Due to friction the net work
done on the object is less than the work done by the
person on the object.
object’s energy is unchanged
Fapplied
= 100 N
frictional
force Ff
= 100 N
Figure 6.15 Due to friction the net work done on
the object is zero since the object has no increase
in kinetic energy.
Chapter 6 Momentum, energy, work and power
193
Figure 6.16 The Empire State Building Run-up
is one of a number of races to the tops of tall
buildings around the world. The current record
for the 320 m (vertical) race is 9 minutes and
33 seconds.
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 = Δh = 320 m
Force applied = weight = mg = 60 × 9.8 = 588 N
W = 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
Fapplied = 50 N, θ = 30°, Ff = 10 N, x = 10 m
ΣF = Fapplied × cos 30° − Ff
= 50 × 0.866 − 10
= 33.3 N
Now W = ΣFx
= 33.3 × 10 = 330 J
194
Motion
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.
i
Whenever the net force is perpendicular to the direction of motion no (net) work
is done on the object.
Force–displacement graphs
A 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.
Physics file
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.
Units for area = N r m
=Nm
=J
i.e. area = work done
Force (N)
Figure 6.17 The area under a force–displacement
Displacement (m)
graph is equivalent to the work done by a force
acting in the direction of the displacement. Where
the net applied force is changing, the area can be
found by counting squares or by dividing the area
into segments. The area of each segment then
equals the work done by a constant force during
that small displacement and the total area will
represent the total work.
Chapter 6 Momentum, energy, work and power
195
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.
100
90
80
Force (N)
70
Solution
60
50
40
30
20
Displacement (m)
0.008
0.007
0.006
0.005
0.004
0.003
0.002
0.001
10
This 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
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, ΔE.
Summarising:
• Impulse is equal to F × Δt, is equivalent to Δp, 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 ΔE, has the units joules (J), and can
be determined from the area under a force–displacement graph.
6.3 summary
Work
• 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 the
displacement is given by Fx cos θ where θ is the
angle between the force and the direction of the
196
Motion
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.
Work
Where appropriate use g = 9.8 m s−2.
1 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?
2 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.
3 Is the quantity you calculated in Question 2 the only
work that the bushwalker has done? Explain.
4 The work done by a force is:
i calculated by multiplying the force by the
distance moved
ii measured in joules
iii 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
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 lifting
the 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?
6 If a lift of mass 500 kg is raised through a height of
15 m by an electric motor:
i the weight of the lift is 4900 N
ii the useful work done on the lift is 73 500 J
iii 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
7 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
spring
balance
sandbag
8 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?
9 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.
10 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.
10
Force (N)
6.3 questions
5
10
20
Extension (cm)
Worked Solutions
Chapter 6 Momentum, energy, work and power
197
6.4
Mechanical energy
Interactive
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 to
motion. 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.
i
…N…RGY is the ability to do work.
Some comparative energy transformations are included in Table 6.2.
Table 6.2 Comparison of various energy transformations
Energy use
Household in 1 day
Fan heater in 1 hour
Adult food intake in 1 day
Amount of energy
150 MJ
8.6 MJ
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 energy
An 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’).
Figure 6.18 Mechanical energy
exists in many forms.
198
Motion
Figure 6.19 The kinetic energy of any object
depends on its mass and the square of its speed.
Doubling the velocity will increase the kinetic
energy by a factor of four.
Physics file
If a moving object of mass, m, and initial velocity, u, experiences a
constant net force, F, 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 F = ma we get:
W = max
. . . . . . (i)
Using one of the earlier equations of motion: v2 = u2 + 2ax
v2 − u2
and rearranging: x =
2a
v2 − u2
Substitute this for x in equation (i): W = ma ×
2a
1
1
Rearranging gives: W = 2 mv2 − 2 mu2
but W = ΔE
If it is accepted that the work done results in a change in kinetic energy,
1
then an object of mass m with a speed v has kinetic energy equal to 2 mv2.
i
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.
The KIN…TIC …N…RGY, …k, of a body of mass m and speed v is:
…k = 12 mv2
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
Earth in orbit
Orbiting satellite
Large car
Mass (kg)
Average speed
(m s–1)
6 × 1024
3 × 104
2.7 × 1033
100
8 × 103
3 × 109
…k (J)
5.5 × 105
1400
28
Netball player
60
8
1900
Footballer
90
8
2900
Chapter 6 Momentum, energy, work and power
199
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.
i
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 = Δ…k
= …k(final) − …k(initial)
= 12 mv2 − 12 mu2
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 = Δ…k = 12 m(v 2 – u 2)
Interactive
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 ΔEk = W = Fav x.
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 = 2 mv2:
…k = 12 × 60 × 8.02
≈ 1900 J.
1
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
b
200
Motion
into the aorta.
m = 0.10 kg, v = 0.15 m s−1, u = 0 m s−1
1
Using W = Δ…k = 2 m(v2 − u2)
1
W = 2 × 0.10 × (0.152 − 02)
W = 1.125 × 10−3 J = 1.1 mJ
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
Potential energy
An 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, gravitational, 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, Δh. Replacing F and x with these equivalent
terms gives:
W = mgΔh
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:
ΔUg = mgΔh
Figure 6.20 The energy gained or lost due to a
change in height within a gravitational field is called
gravitational potential energy. An increase in height
will require the transformation of energy from
other sources. A decrease will usually increase the
kinetic energy of the body.
Figure 6.21 This photograph of a pole-vaulter
illustrates that elastic potential energy is stored in
the pole. This energy is largely converted to kinetic
energy and then the gravitational potential energy
of the athlete.
Chapter 6 Momentum, energy, work and power
201
Physics file
The 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
1
g∝
r2
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.
i
The change in gravitational potential energy is due to the work done against a
gravitational field and is given by:
ΔUg = mgΔh
where ΔUg is the change in gravitational potential energy measured in joules (J)
m is the mass of the body (kg)
Δh is the change in height (m)
g is the acceleration due to gravity (m s–2)
The ΔUg 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, Δh.
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. For
example, 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.
A (260 m)
B (389 m)
C (0 m)
a
b
c
What is her gravitational potential energy at points B and C relative to A?
What is the change in the ranger’s potential energy as she walks from B to 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, hB = 389 m, hC = 0 m
Potential energy change from A to B:
ΔUg = mg(hB − hA)
= 60 × 9.8 × (389 − 260)
= 7.6 × 104 J
202
Motion
b
c
Potential energy change from A to C:
ΔUg = mg(hC − hA)
= 60 × 9.8 × (0 − 260)
= −1.5 × 105 J
There is no need to calculate ΔUg from B to C separately as the difference between the
two previous results can be used.
Potential energy change from B to C:
ΔUg = ΔUg (A to C) − ΔUg (A to B)
= −1.5 × 105 − 7.6 × 104
= −2.3 × 105 J
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:
ΔUg = mgh − mgh0
= 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 energy
The 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 law
Springs 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, Δx, occurs.
Chapter 6 Momentum, energy, work and power
203
Applied force (N)
Extension (m)
Figure 6.22 Ideal materials obey Hooke’s law:
F ∝ kΔx.
Prac 26
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, Δx, can be graphed against the applied force, F. You can imagine
to it so that it stretches. Many items, such as well-designed springs, will (at
least 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 Δx 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 ∝ Δx, we can say F = kΔx. 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 extended
or 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.
i
HOOK…’S LAW states that the force applied by a spring is directly proportional,
but opposite in direction, to the spring’s extension or compression. That is:
F = −kΔx
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)
Δx is the amount of extension or compression of the ideal spring (m)
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.
204
Motion
i
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:
Us = 12 kΔx2
where Us is the elastic potential energy stored during compression or
extension (J)
k is the stiffness constant of the material (N m−1)
Δx is the extension or compression (m)
area under graph
= work done
Force applied (N)
Work done = area under F–x graph
= area of a triangle
1
= 2 × base × height
= 12 × Δx × F
But since F = kΔx:
1
Work done = 2 × Δx × kΔx
W = 12 kΔx2
= 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–Δx graph
must be used to determine the elastic potential energy stored.
Extension (m)
Figure 6.23 Elastic potential energy is a form
of mechanical energy. Work is done as elastic
potential energy is stored, as indicated by the
area under the F–Δx graph.
Physics file
Take care! The two forms of potential
energy, elastic (or spring) potential
energy, Us, and gravitational potential
energy, Ug, 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!
Worked example 6.4D
120
A
B
100
F applied (N)
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–Δx 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 Δx = 40 mm, which spring will have stored the most
elastic potential energy? Justify your choice.
80
60
40
C
20
10 20 30 40 50 60
x (mm)
Solution
a Ideal springs produce an extension that is consistently proportional to the applied
force. Since spring B has an F–Δx graph which is a straight line emerging from the
origin, it is behaving ideally and obeying Hooke’s law until an extension of ∼30 mm
is reached.
Spring C does not obey Hooke’s law, that is, F is not directly proportional to Δx, 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.
Chapter 6 Momentum, energy, work and power
205
b
c
d
e
k = gradient of F–Δx graph = rise/run = 90/0.030 = 3.0 × 103 N m−1
1
Since spring B obeys Hooke’s law, the equation Us = 2 kΔx2 can be applied.
W = ΔUs = Us[final] − Us[initial]
As there is initially zero energy stored:
1
W = 2 kΔx2
1
= 2 × 3.0 × 103 × 0.0252
= 0.94 J
Spring C does not obey Hooke’s law, so the work done must be calculated using the area
under the F–Δx 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
Elastic potential energy is given by the area under the F–Δx 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.
Physics in action
James Joule
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.
206
Motion
Figure 6.24 James Prescott Joule.
water
pulley
falling masses
Figure 6.25 Joule’s original apparatus for
investigating the mechanical work equivalent
of heat energy. The falling weights caused
the paddle to turn. The friction between the
wheel and the water created heat energy in
the water. For the first time, heat energy could
be measured and related to other forms of
energy.
6.4 summary
Figure 6.26 As a result of the considerable amount of mechanical work being done
on the water of a waterfall, the temperature of the water at the bottom of the falls is
usually 1°C or 2°C higher than at the top.
Mechanical energy
• Energy is the ability to do work. Whenever work
is done, energy is transformed from one form to
another.
• Kinetic energy is the energy a body has because of
1
its motion. Ek = 2 mv2 where Ek 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: ΔUg = mgΔh where Ug is the gravitational
potential energy in joules (J), m is the mass in
kilograms (kg) and Δh 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 = −kΔx
• The elastic potential energy, Us, stored in an item is
given by the area below the force–extension graph
1
for that item, or Us = 2 kx2 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 = ΔUg or W = ΔEk or W = ΔUs
or W = area under F–Δx graph
Chapter 6 Momentum, energy, work and power
207
Mechanical energy
Where appropriate use g = 9.8 m s−2.
1 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.
2 Calculate the gravitational potential energy relative
to 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.
3 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?
4 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
5 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?
6 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.
300
F applied (N)
6.4 questions
A
B
200
100
C
0.05 0.10 0.15 0.25
x (m)
7 Calculate the spring constant for each spring and
determine which is the stiffest spring.
8 Each spring has a force of 100 N applied to it.
Calculate the elastic potential energy stored by each
spring.
9 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?
10 The gymnasium equipment described in Question
9 is adjusted so that its spring constant is now
3000 N m−1. If 12.5 J of energy is now stored in
the spring, by what distance has the spring been
compressed?
Worked Solutions
208
Motion
6.5
Energy transformation and power
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.
Interactive
Transformation of energy
Energy 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 at
the 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.
i
Work is done whenever energy is transformed from one form to another.
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.
(a)
(b)
gravitational
potential energy
work
done
work
done
heat
h
kinetic energy
work
done
sound
work
done
heat
Figure 6.27 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 take
place. (b) An energy-conversion flow diagram can be useful in visualising the transformations
that 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.
Chapter 6 Momentum, energy, work and power
209
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 transformations
The 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 useful output × 100
=
total energy supplied
total input
Table 6.4 Efficiencies of some common energy transfers
Figure 6.28 In each of these situations below
Device
Desired energy transfer
an energy transformation is taking place. Can you
identify the forms involved in each transformation?
Large electric motor
Electric to kinetic
Efficiency (%)
90
Gas heater or boiler
Chemical to heat in water
75
Steam turbine
Heat to kinetic
45
High-efficiency solar cell
25
Coal-fired electric generator
Chemical to electric
30
Compact low-energy
fluorescent light
Electric 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.
210
Motion
Physics in action
Air resistance in sport
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. Fa ∝ v2). 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 crosssection 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 the
pedals. This allows the rider to race bent forward, reducing the
area presented to the air.
Drag (N)
20
10
0
5
9.0
13.5
Velocity (m s−1)
Figure 6.30 Technological advances in bike design, pioneered by
RMIT University, have led to efficient designs such as that used by
Australian riders in recent international competition.
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 helmets
and 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.
Figure 6.29 The efficiency of a cyclist is affected by the velocity
of the bike. As the velocity increases, the drag or air resistance
can use as much as 90% of the energy the cyclist’s input.
Conservation of energy
No matter what energy transformation occurs overall, no energy is gained
or 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
Chapter 6 Momentum, energy, work and power
211
All E p
equal to the potential energy at the top. The total energy in the system
remains the same. Energy has been conserved.
i
1
—
2 Ep
1
—
2 Ek
All E k
Figure 6.31 A diver loses gravitational potential
energy but gains kinetic energy during the fall.
Prac 27
212
Motion
SPARKlab
The TOTAL …N…RGY in an isolated system is neither increased nor decreased by
any transformation. Energy can be transformed from one kind to another, but
the total amount stays the same.
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.
i
Kinetic energy + potential energy = 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:
1
total mechanical energy = 2 mv2 + 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 + Ep 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 its
speed 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
this total could be maintained or even increased until the gymnast finally
ran out of available energy himself.
81
102
185
64
43
20
122
125
194
159
174
207
218
147
flight on bed
Energy (J)
4000
flight
on bed
flight
gravitational
potential energy
plus kinetic energy
3000
gravitational
potential
energy
2000
kinetic
energy
1000
20
40
60
80 100
120 140
Number of frames
160
180
200
220
Figure 6.32 During each airborne stage of a gymnast’s trampoline routine (indicated on
the graph by shading), mechanical energy is conserved. The graph shows the relationship
between total energy and gravitational potential energy and kinetic energy. Each time the
gymnast lands, energy is transferred to the trampoline. The energy returning from the
springs after each landing allows the routine to continue.
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 bar
with 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
mu2 = …k + Ug (at peak height)
2
= 12 mv2 + mgΔh
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 to
make sense remember that all objects fall at the same rate regardless of their mass.)
1 2
u = 12 v2 + gΔh
2
Substituting the values from the question:
1 2
u = 12 × 1.22 + 9.8 × 1.5
2
1 2
u = 15.42
2
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.
Chapter 6 Momentum, energy, work and power
213
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.
Solution
m = 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 = 12 mv2 = 12 × 75 × 3.22 = 384 J
Total energy transformed to forms other than gravitational potential and kinetic energy:
Δ… = …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 = Δ… = Ff × x
and:
Δ… 32 691
Ff =
= 726 N ≈ 730 N
=
45
x
Power
Physics file
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.
Why 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.
i
POW…R =
work done energy transformed
=
time taken
time taken
or
W Δ…
=
Δt Δt
where P is the power developed in watts (W) resulting from an energy
transformation Δ… occurring in time Δt. Δ… is measured in joules (J), time is
measured in seconds (s).
P=
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.
214
Motion
i
In the special case of an applied force opposing friction or gravity and doing
work with no increase in the speed of the object, we can say:
As W = Fx then:
Fx
P=
Δt
x
and as v = then:
Δt
P = Favvav
where
P
is
power
developed
(W)
This is useful when finding the power required to produce a constant
average applied
force (N)
Fav ais frictional
speed against
or gravitational
force.
vav is average speed (m s−1)
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 length
of time. Table 6.5 includes comparative figures for the power developed in
various activities and devices.
Figure 6.33 It is not the amount of energy
required that stops the rest of us from winning
the 400 m sprint, but the rate at which we can
effectively convert it to useful work.
Table 6.5 Average power ratings for various human activities and
machines
Activity or machine
Power rating (W)
100
300
Cycling (not racing)
500
Standard light globe
Television
Fast-boil kettle
Family car
60
200
2400
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.
Solution
The work is against gravity so:
U
mgΔh
P= g =
Δt
Δt
m = 60 kg, g = 9.8 m s−2, Δh = 242 m
Δt = (7 min × 60) + 58 s = 478 s
60 × 9.8 × 242
P=
= 3.0 × 102 W
478
Chapter 6 Momentum, energy, work and power
215
6.5 summary
Energy transformation and power
• 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:
energy output
efficiency (%) =
× 100
energy input
• Whenever energy is transformed, the total energy in
the system remains constant. This conservation of
energy is a fundamental natural principle.
6.5 questions
Energy transformation and power
Use g = 9.8 m s−2 where required.
1 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.
2 Draw an energy transformation flow chart for a
swimmer diving off a diving board and into a pool of
water.
3 A boy of mass 46 kg runs up a 12 m high flight of
stairs in 12 s.
a What is the gain in gravitational potential energy
for the boy?
b What is the average power he develops?
4 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
d The actual power output would be considerably
greater than the answer to part c. Suggest two
possible reasons for this difference.
216
Motion
• The total mechanical energy will remain constant in
an isolated system. That is:
Ek + Us + Ug = constant
• Power, P (in watts, W), is the rate at which work is
done or energy transformed:
W ΔE
P=
=
Δt Δt
• In the particular case of work being done to overcome
friction, with no resulting increase in speed:
P = Favvav
5 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 shotput when it hits the ground?
6 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
7 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?
8 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?
9 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.
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 any
frictional 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?
10 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?
Worked Solutions
Chapter review
Momentum, energy, work and power
Use g = 9.8 m s−2 where required.
Force (N)
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
0.01
0.02
Compression (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
b
c
The person can fall a distance of 10 m before the rope
attached 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 3
B 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?
Calculate the initial total mechanical energy of this system.
Write a flow chart displaying the energy transformations
that are occurring during the first fall.
Continued on next page
Chapter 6 Momentum, energy, work and power
217
Momentum, energy, work and power (continued)
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.
A
Casey
Mandeep
B
Casey
Mandeep
30°
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?
30°
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.
A
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 collision
they 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
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?
C
30 m
25 m
B
D
12 m
Worked Solutions
218
Motion
Chapter Quiz
AREA OF STUDY REVIEW Motion
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.
1 At what time will the arrow reach its maximum height?
2 What is the maximum vertical distance that this arrow reaches?
3 What is the acceleration of the arrow when it reaches its
maximum height?
4 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 driver
is travelling at 72 km h−1 when she sees a danger and reacts by
braking.
A
B
1.0 cm
C
3.0 cm
D
5.0 cm
12 Determine the average speed of the marble for the following
distance intervals:
a A to B
b B to C
c C to D
13 Determine the instantaneous speeds of the marble for the
following times:
a t = 0.05 s
b t = 0.15 s
c t = 0.25 s
14 Describe the motion of the marble.
5 How far does the car travel during the reaction time?
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.
6 Assuming maximum retardation, calculate the braking time.
15 Calculate the acceleration of the car.
7 Determine the total distance travelled by the car from the time
the driver realises the danger to the time the car finally stops.
16 What is the resultant force acting on the car during this 100 m?
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 vertical
height of 140 m. Assume g = 9.80 m s−2 and ignore friction.
8 Determine the time taken by the platinum sphere to strike the
ground.
17 Calculate the force exerted on the car by the tow-truck.
18 What force does the car exert on the tow-truck?
19 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 Ff between the incline and the trolley is 110 N.
9 Calculate the time taken by the lead sphere to strike the ground.
g = 9.8 m s–2
10 Determine the average velocity of each sphere over their
respective distances.
11 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. The
diagram 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.
F
30˚
20 Determine the value of F that will move the trolley up the incline
at a constant velocity of 5.0 m s−1.
21 Determine the value of F that will accelerate the trolley up the
incline at a value of 2.0 m s−2.
22 Calculate the acceleration of the trolley if F = 1000 N.
23 What is the value of F if the trolley accelerates up the incline at
10 m s−2?
Continued on next page
Area of study review
219
AREA OF STUDY REVIEW (continued)
24 Two masses, 10 kg and 20 kg, are attached via a steel cable to a
frictionless pulley, as shown in the following diagram.
a
g = 9.8 m s–2
FT
FT
10 kg
20 kg
a
d
e
How much work is done on the car during the 5.0 s interval?
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.)
a Determine the acceleration of each mass.
b What is the magnitude of the tension in the cable?
25 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.
F = 800 N
20.0 kg
Ff = 100 N
2
4
6
v (m s–1)
t (s)
a
b
c
220
Motion
2.0 m
28 What is the kinetic energy of the shotput at point B?
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.
26 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.
0
5.0 m
3.0 m
27 What is the total energy of the shotput just after it is released at
point A?
a
50
40
30
20
10
0
B
A
How much kinetic energy (in MJ) does the car have at
t = 5.0 s?
What is the resultant force acting on the car?
What force is provided by the car’s engine during the 5.0 s
interval?
29 What is the minimum speed of the shotput during its flight?
30 What is the total energy of the shotput at point C?
31 A 5.0 kg trolley approaches a spring that is fixed to a wall. During
the collision, the spring undergoes a compression, Δx, 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
Force (kN)
60°
C
a
b
c
d
12.0
10.0
8.0
6.0
4.0
2.0
0.0
0.0 1.0
2.0 3.0 4.0 5.0
Compression (cm)
6.0
Calculate the elastic potential energy stored in the spring
when its compression is equal to 2.0 cm.
What is the elastic potential energy stored in the spring when
the trolley momentarily comes to rest?
At what compression will the trolley come to rest?
Explain why the trolley starts moving again.
e
Kinetic energy (J)
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.
32 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
0.0
0.0 1.0
Position
(m)
A
B C
D
E
F
G
80
70
60
50
40
30
20
2.0
3.0 4.0
x (cm)
5.0 6.0
10
a
b
c
Determine the speed of the cube during the first 2.0 cm.
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.
Velocity
(m s–1)
10
9
8
7
6
5
4
3
2
1
0
Time (s)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Time (s)
0
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80
35 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?
36 For the boy’s 80 s ride, calculate:
a the total distance covered
b the average speed.
37 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.)
34 Determine the average speed of the sprinter:
1.5
0m
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.
0.30 m
0.14 m
Continued on next page
Area of study review
221
AREA OF STUDY REVIEW (continued)
38 Calculate the velocity of the mass the instant before it strikes the
pencil.
39 Calculate the work required to break the pencil.
40 Can you account for the loss in energy?
48 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
Force (N)
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.
0
41 What is the car’s deceleration (in m s−2) during the 12 s interval?
42 What was the average braking force acting on the car during the
time interval?
43 Determine the distance that the car travels during the 12 second
interval.
a
b
c
44 Explain what happens to all the initial kinetic energy of the car.
45 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?
46 a
Describe the motion of your chair when you stand up and
push it back from your desk.
20
40 60 80 100 120
Extension (mm)
Estimate the value of the spring constant.
energy stored in the spring when the extension is 100 cm.
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 force
versus 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.
Force (N)
500
b
c
How would the chair behave if it were on castors?
of Newton’s laws.
47 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 a
constant velocity. How can this be consistent with Newton’s first
and second laws of motion?
Time (s)
0.015
0.040
49 Determine the magnitude of the change in momentum of the
ball.
50 Determine the magnitude of the change in momentum of the bat.
51 Determine the magnitude of the change in velocity of the ball.
Worked Solutions
222
Motion