UNIT
1 Module 3
Work and energy
Introduction
The photograph of a pole vaulter illustrates how a surge of energy is possible from a
person. During the jump, large forces are exerted by the vaulter in order to convert
the kinetic energy he has acquired during his run up into elastic potential energy in
the bending of the pole, and then into as much gravitational potential energy as
possible. All this has to be done while he has control of his movement, so avoiding
hitting the bar. We use energy all the time to do work. Even when we are asleep, we
use energy to breathe and to keep warm.
You no doubt know that the joule is the SI unit of work and energy. You probably do
not realise how cheap it is to buy energy. Gas companies, in 2007, supply gas on
some domestic tariffs that provide 3 600 000 J for only 2.1p. Put another way,
1 joule of energy costs 0.000 000 58 pence.
Admittedly, a joule is a small unit of energy, but even if you work hard you would
struggle to maintain a work rate of 100 joules per second for any length of time. It is
not therefore surprising that, in general, people waste a huge amount of energy.
Global warming would be much less worrying if 1 joule cost £0.000 000 58. Gas bills
of £100 would then be £10 000!
In this module you will learn:
• how to measure energy
• about the different types of energy
• to use the terms energy and power correctly
• to construct Sankey diagrams showing power usage
• about the law of conservation of energy and
• how to calculate the energy in a spring or deformed material.
Test yourself
1
2
3
4
Name three different types of energy.
What unit is used for measuring work?
What quantity is defined as work done per unit time?
Give a single word that means the ‘forces in a stretched
spring’.
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Module contents
1 Work and the joule
2 The conservation of energy
3 Potential and kinetic energies
4 Power and the watt
5 Efficiency
6 Deformation of materials
7 Hooke’s law
8 The Young modulus
9 Categories of materials
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1.3
1
Work and the joule
By the end of this spread, you should be able to . . .
✱ Define work and the joule.
✱ Calculate the work done by a force.
Work
Work, as the word is used in physics, is defined by the equation:
work = force × distance moved in the direction of the force
As the above definition involves a direction of movement and since force itself has
direction, it would appear logical that work itself must be a vector. In fact, it is a scalar.
In spread 1.1.4, where we discussed adding vectors together, it was explained that the
result is also a vector. When a vector is multiplied by a scalar, such as velocity × time, the
result is always a vector: in this case it would be displacement. In the definition of work, a
vector is multiplied by another vector but in this case it produces a scalar quantity. Work
itself does not have direction. For example, when you run fast your body may be moving
in a particular direction, but the work being done by your body involves making you hot
and sweaty, i.e. not a direction. The work that you have done has been transformed into
heat energy, which does not have a direction.
The joule, the SI unit of work
The SI unit of work is the joule.
Key definition
1 joule is the work done when a force
of 1 newton moves its point of
application 1 metre in the direction of
the force.
Typical work values
In the following examples of work done, you can see that large numerical values are
often involved. In all these examples, it is assumed that the activity described is being
done at a constant speed. What happens where work is done to increase the speed of
an object will be dealt with in spread 1.3.3.
Work done:
• picking up a pen = 0.2 N × 0.1 m = 0.02 J
• lifting an apple to your mouth = 1 N × 0.5 m = 0.5 J
• walking upstairs = 400 N × 3 m = 1200 J
• climbing a 1000 m mountain = 400 N × 1000 m = 4 × 105 J
• by a crane lifting a 40 tonne container 30 m out of the hold of a ship
= weight of container × 30 m
= 40 000 kg × 9.81 × 30 m
= 1.18 × 107 J
• launching a 200 kg satellite into space = 1.24 × 1010 J
• getting a 300 tonne plane up to an altitude of 10 000 m
= weight of plane × 10 000 m
= 300 000 kg × 9.81 × 10 000 m
= 2.94 × 1010 J
1.8 m
6m
4.2
25°
Like all units, it is defined in terms of other units. 1 joule = 1 newton metre. This is not
newtons per metre: you have to multiply, not divide.
200 N
Figure 1 Barrel on a ramp
Force at an angle to the direction of movement
So far we have mainly looked at lifting an object at a constant speed. However, in practice
this can be somewhat more complex. Objects do not always move at constant speed in
the direction they are pushed. The plane in the above example does not rise vertically to
10 000 m; a sailing boat seldom moves in the direction of the wind. As illustrated in Figure
1, the weight of a barrel on a ramp acts vertically downwards. However, if you want to lift
the barrel, it is easier to push it parallel to the ramp.
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Module 3
Work and energy
A barrel of weight 200 N is raised by a vertical distance of 1.8 m by being moved along
the ramp.
Work and the joule
The work done against gravity will be 200 N × 1.8 m = 360  J.
If the ramp is at an angle of 25º to the horizontal, then the force required will be less but
the total work done must, if friction is negligible, be the same. So:
1.8 m
distance moved along the ramp = _______
​
​
= 4.26
sin 25º
360 J
​
= 84.5 N
force required = ​ _______
4.26 m
A simpler way is to use the vertical component of the distance moved along the slope.
This directly gives:
work done = 200 N × 4.26 m × cos 65°
= 360 J
65° is the angle between the force and the distance moved.
In other words:
work done = force × distance moved in the direction of the force
= F d cos �
where d is the distance travelled and � is the angle between the force and the direction
of travel.
Note that if the force and the direction of travel are at right angles to one another, then no
work is done as cos 90° is zero. This may seem rather irrelevant, since at first sight a force
at right angles to a direction of travel seems impossible. However, the force of gravity on
the Moon, as it orbits the Earth, is at right angles to the Moon’s direction of travel. Despite
the large gravitational force the Earth is exerting on the Moon, the Earth is not doing any
work on the Moon, and so the Moon moves at a constant speed for a very long time.
Questions
Figure 2 The Earth exerts a large
gravitational force on the Moon, but does
no work on it
1 A caravan of mass 400 kg is pulled at constant speed along a horizontal road against
a drag force of 240 N.
(a) Calculate the work done to pull the caravan 500 m.
(b) Explain why the weight of the caravan does not enter into the calculation.
2 A barge is pulled by a rope tied to a horse on the towpath of a canal. There is a drag
force on the boat in the water. A man on the other bank applies a sideways force via a
rope to maintain the course of the barge along the canal. Figure 3 shows the
magnitudes and directions of these forces.
Horse
400 N
340N
30°
Barge
Drag
Canal
200 N
Man
Figure 3
(a) Write down the components of each of these forces along the canal.
(b)How much work is done by each of these forces when the barge moves 10 m
along the canal?
(c)How much of the work done on the barge by the horse is not work done against
the drag force?
(d)Describe how the motion of the barge changes.
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1.3
2
The conservation of energy
By the end of this spread, you should be able to . . .
✱ Describe examples of energy in different forms, its conversion and conservation.
✱ State and apply the principle of conservation of energy.
✱ Use the idea that work done is equal to the transfer of energy to solve problems.
Energy
Key definition
Energy is the stored ability to do work.
Energy and its uses are fundamental to any physics course. Indeed, physics can be
described as the study of energy. So first of all, let’s examine the term energy.
Whenever people such as politicians, businessmen or journalists talk about energy, it is
usually in terms of supplying it to industrial and domestic markets, for example as
quantities of oil, gas, coal or uranium. These are fuels used throughout the world to
supply energy to industrial and domestic markets. There is a huge demand for this
energy, because it is used to do work. This includes work to fly planes, to drive cars, to
run machinery in factories and homes, to operate televisions, computers, telephones,
telecommunication systems. The list goes on!
Let’s look at one specific use of the chemical energy stored in oil. A car’s engine is able
to release this energy from the oil and some of the energy does work to move the car
along the road. Unfortunately, the conversion of chemical energy into work is never
100% efficient. Most modern cars have efficiencies below 50%. This problem is due to
the very nature of energy-to-work conversions. There is a theoretical limit on the
efficiency of a car engine that depends on the high temperature of the burning fuel and
the low temperature of the surroundings.
At a basic level energy is either kinetic energy or potential energy.
• Kinetic energy: where movement is taking place.
• Potential energy: regions where electric, magnetic, gravitational and nuclear forces
exist. Regions such as this are called fields.
However different energy names are frequently used for specific situations. We have just
looked at the form that is called chemical energy.
The following is a list of different forms of energy together with some details of how the
energy is stored.
• Chemical energy: energy can be released when the arrangement of atoms is altered.
• Electrical potential energy: for example, a positive charge is pushed close to
another positive charge. This will often be called electrical energy.
• Electromagnetic energy: includes all the waves that travel at the speed of light in a
vacuum – gamma rays, X-rays, ultraviolet, light, infrared, microwaves and radio waves.
These waves hold their energy in electric and magnetic fields.
• Gravitational potential energy: where an object is at a high level in the Earth’s
gravitational field.
• Internal energy: the molecules in all objects have random movement and have some
potential energy when they are close to one another. This type of energy is often
called heat energy.
• Kinetic energy: when an object has speed.
• Nuclear energy: energy can be released by reorganising the protons and neutrons in
an atom’s nucleus. This form of energy is also known as atomic energy.
• Sound energy: in the movement of atoms.
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Module 3
Work and energy
The conservation of energy
The conservation of energy
In physics, energy conservation does not refer to the need to use less fuel to prevent
global warming. The principle of conservation of energy is one of the fundamental laws
and is stated as:
In any closed system energy may be converted from one form into another, but cannot
be created or destroyed.
The Sankey diagram is used to illustrate this principle. A simplified Sankey diagram for a
car travelling at night at a constant speed is shown in Figure 1. Here, the ‘closed system’
referred to in the above statement is the car, the ground and the air around the car.
0.9 MJ
Key definition
Conservation of energy describes the
situation in any closed system, where
energy may be converted from one form
into another, but cannot be created or
destroyed.
Work done on road –
road gets hot
4.8 MJ chemical
energy from fuel
2.6 MJ
Energy of exhaust gases.
They are hot too.
0.06 MJ light energy
1.2 MJ
0.04 MJ sound energy
Kinetic energy of
turbulence in
surrounding air
Figure 1 Sankey diagram
It is assumed that the energy supplied in one minute by the burning fuel is 4.8 MJ. Some
of this energy is used to drive the car. However, since the car is not accelerating, none of
this energy is converted into kinetic energy for the car. Much of the energy is converted
into kinetic energy (1.2 MJ) of the turbulent air caused by the car’s movement, 2.6 MJ
becomes internal energy of the exhaust gases and 0.9 MJ becomes internal energy of
the road itself, as the forces moving on the road cause an increase in its temperature.
The remainder is converted into light energy for headlights, etc. (0.06 MJ) and sound
energy from the noise the car makes (0.04 MJ).
The importance of such a diagram is not the figures themselves, but the fact that you
can account for the total energy produced and it is exactly equal to the energy supplied.
No energy has been lost.
Questions
1 Suggest a Sankey diagram for:
(a) an electric motor that raises people on an escalator;
(b) a television.
2 Describe the energy changes involved when a pole-vaulter does a jump using a
flexible pole, from the run-up to the landing.
3 Sketch how you think that the magnetic potential energy changes when you pull two
touching bar magnets apart. Does the potential energy increase or decrease? To
answer this, consider the change in kinetic energy when you release them, allowing
them to move back together.
Where did the energy go?
In the nineteenth century, several
eminent scientists could not
understand how it was possible not
to lose energy. It was the English
physicist James Joule who finally
proved this principle with a series
of meticulous experiments. He
measured the rise in water
temperature produced when the
water was stirred by a simple
paddle wheel. He designed and
made his own thermometers which
were accurate to 0.01 °C. He
repeatedly found that 100% of the
work done by the paddles could be
accounted for in the corresponding
rise in temperature.
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1.3
3
Potential and kinetic energies
By the end of this spread, you should be able to . . .
✱
✱
✱
✱
Derive and use the equation for gravitational potential energy close to the Earth’s surface.
Use the equation for kinetic energy.
Solve problems involving exchange between kinetic and gravitational potential energy.
Apply the principle of conservation of energy to determine the speed of an object falling.
Gravitational potential energy
Everything on this planet exists in the Earth’s gravitational field. Near the Earth’s surface,
the strength of this field is almost constant. Look at the skydivers in Figure 1. For each of
them, their own weight is pulling them down towards the ground. For a single skydiver of
mass m, using F = ma we get:
W = mg
So, the gravitational potential energy lost in falling a distance h is:
Figure 1 Skydivers converting
gravitational potential energy into kinetic
energy
force × distance = Wh
= mgh
This gives the general equation, when g is a constant, as:
gravitational potential energy = mgh
Key definition
Gravitational potential energy is gained when rising and lost when falling.
Gravitational potential energy is the
energy stored in an object (the work an
object can do) by virtue of its position in
a gravitational field.
Kinetic energy
The kinetic energy of a moving body equals the work it can do as a result of its motion.
Worked example
As a loaded lorry with a mass of 30 000 kg starts to exit the motorway, it has a
speed of 20 m s–1. The upward slope of the exit road along with the friction provide
a constant stopping force of 84 000 N. The lorry stops at the top of the exit road.
84 000 N
F
__
= –________ kg
The acceleration of the lorry =
m
30 000
= –2.8 m s–2
Applying v2 = u2 + 2as gives:
and so:
02
= 202 + 2 × (– 2.8) × s
400 = 5.6s
s
= 71.4 m
The work done by the lorry against the stopping force is therefore:
F × s = 84 000 N × 71.4 m
= 6 000 000 J = 6.0 MJ
Key definition
Kinetic energy is the work an object
can do by virtue of its speed.
If you work through this worked example again, but this time using a different stopping
force, you will find that the work done by the lorry is still 6.0 MJ. This is because in the
calculation the stopping force cancels out (see question 1). Using algebra, the expression
given for the work done against the stopping force is always ½mv2, and is only
dependent on the mass and the speed of the lorry. This quantity is known as the kinetic
energy of the lorry.
The kinetic energy of a body of mass m, travelling with speed v, is given by
kinetic energy (k.e.) = ½mv2
Using the law of conservation of energy, kinetic energy is the work that needs to be
done on an object in order to get it up to that speed in the first place.
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Module 3
Work and energy
Falling objects
Potential and kinetic energies
Time-lapse photography (see spread 1.1.8) makes it easier to appreciate the acceleration
occurring when something falls under gravity. Before Galileo, it was assumed that heavy
objects fell faster than light objects. This is still a common perception since air resistance
has a greater effect on light objects than it has on heavier ones. For example, if a flat
sheet of paper and a pen are dropped at exactly the same time, the pen will hit the floor
first. If, however, you screw the paper into a tight ball and repeat the experiment, then
the two objects will hit the floor together. In this spread we will assume that air resistance
is negligible.
An object of mass m, falling from rest, loses gravitational potential energy. From the
principle of conservation of energy, it gains an exactly equivalent amount of kinetic
energy as a result of the work being done on it by gravity. So:
mgh = ½mv2
where v is its speed and h the distance fallen. m cancels to give:
____
​ 2gh
​
2gh = v2 or v = √
Worked example
Figure 2 shows a ball bearing of mass 5.38 gram being dropped from an
electromagnet at time zero. It passes through two light gates, separated by a
distance of 40.7 cm, at times 0.016 s and 0.289 s.
Calculate:
(a) g, the acceleration due to gravity;
(b) the loss of potential energy of the ball between the two gates;
(c) the kinetic energy of the ball as it passes the lower gate.
Answer
(a) Using:s = ut + ½at2, for the fall to the first light gate,
s = 0 + ½g(0.016)2
and for the fall from start to the second light gate:
s + 0.407 = 0 + ½g(0.289)2
(Note: use 0.407 m not 40.7 cm)
Then eliminate s by subtracting the first equation from the second, to get:
0.407= ½g(0.2892 – 0.0162)
0.814= g(0.083 521 – 0.000 256)
= g × 0.083 265
0.814
g
= _________
​
​
0.083 265
= 9.78 m s–2 to 3 sig. figs.
(b) Loss of potential energy = mgh = 0.005 38 × 9.78 × 0.407 = 0.0214 J.
(c) We now have to find the speed of the ball at the second light gate.
speed= acceleration × time
= 9.78 m s–2 × 0.289 s = 2.83 m s–1
Therefore the kinetic energy when passing the lower gate = ½mv2
= ½ × 0.005 38 × 2.832
= 0.0215 J
Note that the value for (c) must be larger than that for (b), but only by a small
amount. It is always worth checking your answer is reasonable. Since the ball
hasn’t taken long to reach the first light gate, it will have been travelling slowly
and so will not have much kinetic energy at that gate. Most of the kinetic
energy at the second light gate will have been acquired between the two
gates.
Times
Zero
Distances
s
0.016 s
Light gate
40.7 cm
0.289 s
Light gate
Figure 2 Dropping a ball bearing through
light gates
Questions
1 An object of mass m, travelling
with speed v, is stopped by a
constant force F. Calculate:
(a)the acceleration of the
object;
(b)the distance the object
travels before stopping;
(c)the work done by the body
against the stopping force;
(d)the kinetic energy of the
body at the start.
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1.3
Further questions B
Work
1 A person in a car travelling at 30 m s–1 has a mass of
80 kg.
(a) How much work does the person do against the
braking force on him before stopping?
(b) What provides this braking force on the person?
(c) What is the value of the braking force, assumed
constant, in order to stop the person in a distance
of 50 m?
(d) What is the new value of the braking force if the
50 m is travelled up a slope with a gradient of 9.0°?
Conservation of energy
3 Describe the energy changes that occur when a rubber
ball is dropped to the ground and bounces many times
before coming to a standstill.
4 Draw a Sankey diagram for:
(a) the energy from batteries supplying the electric
motors that move a submerged submarine;
(b) the energy from coal burned in a power station to
supply the electricity to boil a kettle.
2 A box of books weighing 200 N is pushed steadily along
a smooth floor for 10 m.
(a) How much work is done against gravity?
The box is then lifted into a car boot 0.7 m off the
ground.
(b) How much work is done against gravity?
(c) How much work would be done if the box were
pushed up a ramp at an angle of 30° instead?
(Ignore work done against friction.)
(d) How much force parallel to the ramp is needed in
(c)?
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Module 3
Work and energy
Further questions B
Potential and kinetic energy
Falling objects
5 Figure 1 shows a Ferris wheel at a fairground. The wheel is
rotating in a vertical plane with each car of mass 120 kg
moving at a speed of 0.6 m s–1. Calculate
9 A boy of mass 55 kg, swinging from the branch of a tree,
drops a distance of 1.5 m to the ground.
Y
(a)By considering the energy changes as he falls,
calculate his kinetic energy as his feet touch the
ground.
(b)How fast is he travelling at the point of impact?
(c)Use an energy argument to explain why the force of
the impact on his legs will be much less if he bends his
knees as he lands.
40 m
10 Two balls of different materials are dropped from 1.0 m
X
Figure 1
(a)the kinetic energy of each car
(b)the change in potential energy of a car as it moves
from X to Y
above a hard surface. At each bounce ball A loses one
quarter of its kinetic energy, ball B loses one quarter of its
speed. How high does each ball rise after its second
bounce? Will they bounce in time with each other?
(c)the total energy change of a car as it moves from
X to Y
(d)the total energy change of the wheel as it makes half a
revolution.
11 A coin of mass 20 g dropped down a well shaft soon
reaches a terminal velocity of 12 m s–1. Calculate
(a) its kinetic energy at the bottom 8.0 m down,
(b) its change in gravitational potential energy, and
(c) the energy lost to the air during the fall.
6 Sketch a graph of how the kinetic energy of a cyclist varies
with her speed from 0 to 10 m s–1.
7 The speed at which an object must be fired upwards to
escape from the Earth’s surface into space is 11.2 km s–1,
assuming that there is no air resistance.
(a)Calculate how much kinetic energy a satellite of mass
20 kg needs to escape.
(b)Imagine the satellite is fired from a gun 100 m long.
Assume that a constant force is applied to the satellite
whilst inside the barrel. Calculate the magnitude of this
force needed to launch the satellite.
(c)Estimate how fast the satellite would be travelling at a
height of 1200 km above the Earth’s surface.
8 A buffer consists of a spring that requires a force of
200 000 newtons per metre for compression. A train of
mass 300 000 kg hits the buffers when travelling at
3.0 m s–1.
Calculate
(a) the kinetic energy of the train,
(b)the work done on the spring by the train before it
stops,
(c) the distance the spring is compressed by the train.
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1.3
4
Power and the watt
By the end of this spread, you should be able to . . .
✱ Define and use power and the watt.
✱ Calculate the cost of using an electrical appliance.
Power
Key definition
Power is the rate of doing work.
Power is defined as the rate of doing work.
In equation form this is expressed as:
work done
power = __________
time taken
Key definition
Power is measured in joules per second, also known as watts.
One watt (W) is equal to one joule per
second.
The power generated by engines and for electrical appliances is usually expressed in
kilowatts (kW).
1 kW = 1000 W
Large-scale power generation is expressed in megawatts (MW).
1 MW = 1000 kW = 1 000 000 W
To power a 100 watt light bulb, an electric current must be flowing through the filament
of the bulb. It supplies energy at a rate of 100 joules per second. This implies that in one
hour, the energy supplied to the bulb must equal 100 J s–1 × 3600 s = 360 000 J.
Electrical energy is sold to domestic users in units called kilowatt-hours (kWh) –
equivalent to the use of 1000 W of power for an hour. For example, 1 kWh would be the
energy supplied to a 100 W lamp over 10 hours, or to a 4 kW cooker in use for ¼ hour.
1 kWh = 1000 J s–1 × 3600 s = 3 600 000 J
Today the cost of 1 kWh of energy to domestic users is about 15p.
Be careful to distinguish between rates and totals. For example, you cannot buy a kW of
power; you pay for energy. You can pay for 1 kW used for 6 hours – 6 kWh. Table 1
illustrates the relationship between rates and totals for several units.
Rate
Example of rate
–1
Time
Total
Example of total
4h
distance
320 km
speed
80 km h
power
3 kW
200 s
energy
600 kJ
current
25 mA
1000 s
charge
25 C
Table 1 Relationship between rates and totals
Power calculations do not just refer to electrical power. They can also be applied to the
power used by a human body as the following worked example shows.
Worked example
Figure 1 An athlete performing press-ups
An athlete of mass 90 kg performs press-ups at the rate of 50 per minute for a
time of 6 minutes. For each press-up he raises his centre of gravity a distance of
24 cm. Calculate:
(a) the total work done;
(b) the mean power required;
(c) the energy which he has used, given that the conversion efficiency from
energy used to work done by muscles is 20%.
(d) Use your answers to explain why the athlete gets hot while doing press-ups.
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Module 3
Work and energy
Power and the watt
Answer
(a) Gain in potential energy for each press up = mgh
= 90 kg × 9.8 m s–2 × 0.24 m = 212 J
Number of press ups in 6 min = 50 × 6 = 300
Total work done = 300 × 212 J = 63 600 J
63 600 J
(b) Power required = ________
​
​
= 176 W
360 s
(c) Let energy used be E
20% of energy used = 63 600 = 0.20E.
So, E = 63 600 × 5 = 318 000 J
(d) The energy not used to lift his body is 318 kJ – 63.6 kJ = 254 kJ. This energy
is expended as internal energy in his body. That is, molecules in his body
move faster; he feels hotter. Some of the energy he does use to raise his
body will also be wasted in his muscles when he lowers himself, making him
even hotter.
Note: none of these answers will be correct to 3 sig. figs. Even the second figure
will be open to doubt. Working is done to this number of figures to ensure that
too much rounding up does not take place.
Human power and horse power
176 W is a high rate of work that only a fit person could sustain for any length of time.
Most people would find it difficult to work continuously at a rate of 70 W.
Horse power is still used to express some power ratings. For example, puissance en
chevaux (horsepower) is used to calculate tolls for cars on some bridges in France.
1 horse power is equal to 746 W – though apparently this is a somewhat optimistic
measure of the output achieved by real horses!
Questions
1 A student counts 14 risers on a flight of stairs, each riser being 18 cm in height. On
the bathroom scales her mass is 52 kg. She ran up the stairs three times taking 3.1 s,
3.0 s and 3.2 s as measured by another student. Calculate her average power while
performing this exercise. Do you think that this is a fair way to measure the mechanical
output power of a person?
2 The engine of a car of mass 900 kg produces a driving force of 300 N while travelling
at a constant speed of 20 m s–1 along a straight level road.
(a)Suggest a reason why the car is not accelerating.
(b) Calculate the power required to drive the car at 20 m s–1.
(c)The car comes to a hill with a gradient of 1 in 15 as shown in Figure 2. Calculate
the additional power required to maintain the same speed of 20 m s–1.
3 To accelerate from rest to 20 m s–1 the car of the previous question takes 10 s.
(a)Calculate the average rate at which kinetic energy is supplied to the car during the
acceleration.
(b)How does your answer to (a) compare with the power in the previous question?
Explain why the values are different.
(c)To accelerate the car the engine provides power at a constant rate. Starting from
the fact that power = (force) × (velocity) explain why the acceleration of the car is
not constant.
15
1
Figure 2 65
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1.3
5
Efficiency
By the end of this spread, you should be able to . . .
✱ Calculate the efficiency of a device and explain why it is always less than 100%.
✱ Draw and interpret Sankey diagrams.
Efficiency
Energy of one sort is often required to obtain a different type of energy as an output. For
example, chemical energy is supplied to an oil-fired power station, while electrical energy
is the output. However, sometimes the same type of energy is both the input and the
output. A car’s alternator, for instance, supplies the battery with electrical energy (input),
and the battery supplies this electrical energy (output) to the radio. Whatever the type of
output energy required, it needs to be as high a percentage of the input energy as
possible. This is determined by the efficiency of a device.
Efficiency is expressed as:
useful output energy
efficiency = __________________ × 100%
total input energy
When the output energy is being provided at the same time as the input is being
supplied, then the time over which the output energy is provided is the same as that for
the input supply. Under these conditions we get the equation:
useful output power
efficiency = __________________× 100%
total input power
Table 1 contains some examples of common devices for which efficiency is important.
It shows their various input and output energy types, along with the typical efficiencies for
these devices.
Device
Energy input
Energy output
Typical efficiency (%)
electric motor
electrical
kinetic/potential
85
solar cell
light
electrical
10
rechargeable battery
electrical
electrical
30
electric radiator
electrical
internal
100
power station
nuclear
electrical
40
car (petrol)
chemical
kinetic/potential
45
car (diesel)
chemical
kinetic/potential
55
steam engine
chemical
kinetic/potential
8
Table 1 Examples of common devices for which efficiency is important
It is worth noting that using electrical energy to produce some other form of energy can
be highly efficient – the only reason why the efficiency of the rechargeable battery in the
table is only 30% is because the energy is stored as chemical energy.
It is easy to convert electrical energy into heat – you just need a resistance. Getting
electrical energy from internal energy is, in practical terms, inefficient. The problem can
be viewed in terms of order. Obtaining an output of disordered energy, i.e. the chaotic
movement of atoms in a substance, from an electrical input is an efficient process – as
with the electric radiator. However, reversing this process to obtain an output of ordered
energy (e.g. electrical) from a disordered energy input (e.g. internal energy) is difficult. The
science of thermodynamics deals with such problems.
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Module 3
Work and energy
Sankey diagrams
Efficiency
We discussed these types of diagrams in spread 1.3.2. They are usually used to
represent power rather than energy problems. With Sankey diagrams, it is usual to
collect the different losses together. This is shown in the Sankey diagram for a typical
power station supplying 1000 MW electrical output (Figure 1).
Electrical
power output
1000 MW
Power from
burning fuel
3000 MW
50 MW
Power to run
station equipment
1 MW
Transformer
losses
5 MW
Friction losses
in generator
Figure 2 Didcot power station with its
cooling towers
1944 MW
Heat losses in
cooling towers
Figure 1 Sankey diagram
for a power station
Before this 1000 MW power can be supplied to customers, further losses are incurred. The
transmission cables used to carry this power supply have resistance, and so heat up when
the electric current passes through them. One of the reasons why overhead, rather than
underground, cables are used is that the surrounding air acts both to electrically insulate
the cables and to keep them cool. When underground cables are required, they have to be
electrically well insulated and have cooling oil continuously pumped through them.
All power stations require cooling. Some use cooling towers (Figure 2), while others use
water from adjacent rivers or the sea (Figure 3).
Questions
1 Any mechanical system, such as a machine, has an efficiency rating. Consider a
mechanical car jack. To change a wheel it may need to lift a load of 2000 N. Suppose
that one rotation of the screw raises the car by 10 mm. You apply a force to the end
of a lever 0.5 m long through a full circle. The jack is only 40% efficient. Calculate the
force that you have to apply.
2 A combined heat and power station (CHP) makes use of the hot cooling water to heat
the houses and factories near the power station. A CHP station is built on a much
smaller scale than a 1000 MW power station which is designed purely to supply the
Grid with electricity.
For an input of 500 MW of fossil fuel, the CHP station gives 150 MW of electricity to
the Grid and 250 MW to water for local area heating and the rest goes up the chimney
or is lost in the generators, etc.
(a) Draw a Sankey diagram for a CHP station.
(b) (i)Calculate the efficiency of the CHP station by considering electricity as the
only useful output.
(ii)How does this compare with the 1000 MW station considered in Figure 1
above?
(c) (i)Calculate the efficiency of the CHP station by considering both electrical and
local heating as useful output.
(ii) How does this compare with the 1000 MW station?
(d) Explain why it is more efficient to burn a fossil fuel directly at the home than to heat
the house electrically from the 1000 MW station.
Figure 3 Sizewell power station with the
sea to provide cooling water
Figure 4 Power transmission lines from
a power station
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1.3
Deformation of materials
6
By the end of this spread, you should be able to . . .
✱ Describe how deformation is caused by a force.
✱ Describe the behaviour of springs and wires in terms of force, extension and elastic limit.
✱ Define the terms elastic deformation and plastic deformation.
Elastic and plastic deformation
Solids are made of huge numbers of tightly packed molecules, so when a solid is
squeezed or stretched the spacing of the molecules within it is altered. While the force
between individual molecules is small, because there are so many of them, a large force
is necessary for an appreciable change in the size or shape of the solid. Usually, when a
force changing the shape of an object is removed, it springs back to its original shape.
Of the many objects that you exert forces on (touch) during the course of a day, almost
none of them change shape permanently. You lift a pen, move a book, eat with a fork,
bite with your teeth, walk on a pavement – but none of these objects are permanently
deformed by the forces you exert on them. In fact all objects change shape slightly,
before returning to their original shape once an applied force is removed. It is easy to see
that when you sit on a chair the cushion changes shape. Less noticeable is the fact that
the chair legs get shorter too – although by a very small amount. But when you stand up
the cushion and the legs return to their original shape. We describe such objects as
‘elastic’. The word elastic has a different meaning in physics from its everyday meaning.
The same is true for the word plastic, which has the opposite meaning to elastic.
Examiner tip
The word elastic can be applied to a
collision. In an elastic collision no
kinetic energy is lost. This can only
happen when there is no permanent
distortion of the objects colliding,
because if there is permanent distortion
some energy must have been used to
create the distortion. Collisions which
are not elastic collisions are not usually
called plastic collisions but inelastic
collisions. More information about
collisions will be given in your A2
course.
When a distorting force is applied to an object and then removed, the object exhibits:
• elastic behaviour when it regains its original shape; or
• plastic behaviour when it is permanently distorted.
You may see the word ‘inelastic’ used on occasions. This means that an object is not
elastic – and therefore, by definition, is plastic.
Tensile and compressive forces
In this spread we will be dealing with forces that stretch objects such as wires, springs
and rubber bands. These are called tensile forces because they cause tension in the
object. Clearly, for there to be a tension in a fixed, stretched wire it must have equal and
opposite forces on it at either end. (The weight of the wire is usually negligible compared
with the size of the forces applied.) This is shown in Figure 1a.
Two equal and opposite tensile
forces stretching a wire
Figure 1a Stretching a wire
With a spring it is possible to reduce its length by squeezing it (see Figure 1b). The forces
being applied in this instance are called compressive forces. Again, two equal and
opposite forces are required, assuming that the spring is not accelerating.
Two equal and opposite compressive
forces squeezing a spring
Figure 1b Squeezing a spring
An experiment to stretch a wire
Clamp
Length of wire under test
Marker
Ruler
Figure 2 Experiment to stretch a wire
Figure 2 shows a long, thin copper wire held firmly in a clamp at
one end. The other end supports a hanger weight after passing
over a pulley. The hanger must be just heavy enough to keep
the wire taut. A marker is attached to the wire and a ruler is fixed
in position below the marker. Table 1 illustrates the position of
the marker as weights are added progressively to the hanger.
Note: There is a danger that the wire may snap in this experiment.
Safety goggles should be worn, and a box should be placed
beneath the hanger – to catch the falling weight, and also to
ensure that you do not stand directly underneath it.
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Module 3
Work and energy
Deformation of materials
8
6
Tension/N
The graph in Figure 3 plots tension (y-axis)
against extension (x-axis). After the two
final readings were taken it was noted that
the degree of extension was still
increasing – hence the question marks in
columns three and four. This process is
known as ‘creep’. The wire broke soon
after the last reading was taken.
4
Elastic
limit
Mass on
hanger
/kg
0
0.050
0.100
0.150
0.200
0.250
0.300
0.350
0.400
0.500
0.600
0.700
2
The graph shows two distinct regions.
0
0
20
40
60
80
100
The initial straight-line portion of the
Extension/mm
graph is followed by a curved section,
Figure 3 Graph plotted using data from stretch
showing much greater extension. The
experiment
end of the straight line is the limit of
proportionality. It is usually very close to
this point where the wire ceases to be elastic. In other words, if the load had been
removed for any value up to the elastic limit, the wire would have regained its original
length. Once the elastic limit has been passed, the stretch becomes permanent.
Figure 4 shows body panels of a car being
pressed from sheet metal. A template of the
required shape is attached to the press, which
exerts a huge force on a flat piece of steel. The
steel must retain its altered shape. The force
applied must therefore be sufficiently large to
ensure that the steel goes beyond its elastic limit
and becomes plastic. Plastic deformation is an
integral part of many manufacturing processes.
120
Tension
in wire
/N
0
0.49
0.98
1.47
1.96
2.45
2.94
3.43
3.92
4.90
5.88
6.86
Reading
on ruler
/mm
26.0
29.0
32.0
35.5
38.5
42.0
45.5
48.5
53.0
66.5
87?
122?
Extension
/mm
0
3.0
6.0
9.5
12.5
16.0
19.5
22.5
27.0
40.5
61?
96?
Table 1 Figure 4 Presses forming body panels
for cars
Questions
10
0
0
100
Extension/mm
200
(ii)
2000
Tension/N
Tension/N
400
200
0
0
2
Extension/mm
A
1000
0
4
(iii)
B
0
0.3
Extension/mm
0.6
0
100
Extension/mm
200
(iv)
10
Tension/N
40
20
0
0
200
Extension/mm
400
5
0
Figure 5
100
Tension/N
Tension/N
20
(i)
Tension/N
1 The four graphs in Figure 5 show how the lengths of four different
materials vary as they are stretched. The original length and area of
cross-section of each specimen is the same. Graph (i) is for a
copper wire, (ii) is for two steel wires having different concentrations
of carbon, (iii) is for polyethylene and (iv) is for rubber.
(a)How do the graphs show that the atomic structure of metals
is different to the molecular structure of polymeric
substances? Why is the behaviour of mild (A) and high
carbon (B) steel different?
(b)Which of the graphs show(s) no evidence of plastic
deformation?
(c)Copy each graph and add a return curve to show how the
extension decreases when the tension is removed.
2 Figure 6 shows two possible answers to 1(c) for graph (iv) in
Figure 5. Explain the significance of the difference. What does the
area between the curves signify?
50
0
0
150
Extension/mm
300
Figure 6
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1.3
7
Hooke’s law
By the end of this spread, you should be able to . . .
✱ Describe the behaviour of springs and wires in terms of Hooke’s law and the force constant.
✱ Explain that the area under a force against extension/compression graph is equal to work done
by the force.
✱ Use the equations for elastic potential energy.
Hooke’s law and the force constant
In the previous spread we saw how an elastic deformation usually produces a
straight-line graph when tension is plotted against extension. This is summarised by
the following statement, known as Hooke’s law.
• The extension of an elastic body is proportional to the force that causes it.
In equation form this becomes:
F = kx
where F is the force causing extension x, and k is known as the force constant (or
spring constant, particularly for springs).
The force constant is expressed in newtons per metre. k tells us how much force is
required per unit of extension. For example, a k of 6 N mm–1 means it takes 6 N to cause
an extension of 1 mm. Note that the force constant can only be used when the material
is undergoing elastic deformation. When deformation becomes plastic, the force per unit
extension is no longer constant.
Tension/N
F
The work done to stretch a wire
0
Extension/mm
x
Figure 1 Straight-line graph of tension
against extension
We usually plot the cause of a change on the x-axis of a graph, and the effect of that
cause on the y-axis. The graph of tension against extension plotted in spread 1.3.6
breaks that convention – but for good reason. If you plot extension on the x-axis, the
area beneath the line is equal to the work required to stretch the wire. Figure 1 shows a
straight-line graph of tension against extension for the elastic part of a deformation.
The extension produced by tension F is x. The work done to produce this extension is
not Fx, however, because F is not constant. The wire stretches with the application of a
small force initially, but F gradually increases as the degree of extension increases.
The work done to reach this extension is the area beneath the graph and is shaded in
Figure 1.
work done = area of triangle = ½Fx
Since F = kx, the work done can also be expressed in terms of k, giving:
work done = ½kx2
Work having been done on the wire to stretch it, the wire itself stores elastic potential
energy, which can be released when required. This applies equally to the extension or
compression of springs, in which the release of stored elastic potential energy is more
apparent. Many children’s toys incorporate springs which store energy upon
compression, and which drive the toy forward when released.
In the case of elastic deformations, the elastic potential energy E equals the work done,
giving:
Figure 2 Child’s toy showing a spring
within which elastic potential energy may
be stored
E = ½Fx = ½kx2
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Module 3
Work and energy
Hooke’s law
Worked example
A car’s shock absorbers make the ride more comfortable, by using a spring that
absorbs energy when the car goes over a bump. One of these springs, placed
next to a wheel, needs to store 250 J of energy when compressed by a distance
of 10 cm.
(a) What value of force constant is required for the spring?
(b) How much energy would be stored if it were compressed by 20 cm?
Force/N
Answer
(a) Since E =½kx2
250=½ × k × 0.102
2 × 250
k =​ _______
​
0.102
= 50 000 N m–1
(b) The extent of compression has been doubled. Provided the spring is still
elastic, the energy is proportional to the square of the extension. Twice the
extension will therefore equal four times the energy.
energy stored = 250 × 4 = 1000 J
Energy stored in a plastic deformation
The graph shown in Figure 3 could be produced by stretching a copper wire beyond its
elastic limit. The work done stretching the wire is given by area A + B. If the tension is
then reduced to zero, the wire behaves elastically, contracting to a permanent extension
x. As the tension is reduced, energy equivalent to area B is released from the wire. The
net result of the wire having work A + B done on it, but only releasing energy B, is that
the wire becomes hot to the touch.
B
0
Extension/mm
x
Figure 3 Stretching a wire beyond its
elastic limit
Questions
1 A spring which obeys Hooke’s law has a force constant of 160 N m–1.
(a) (i) Calculate the extension in mm of the spring produced by a force of 4.0 N.
(ii) Calculate the strain energy stored in the spring.
(b)A second identical spring is connected in parallel with the first as shown in
Figure 4.
(i) Calculate the extension in mm of each spring produced by a force of 4.0 N.
(ii) Calculate the total strain energy stored in the springs.
(c)The second spring is now connected to the end of the first spring to make a single
spring of twice the length.
(i)Calculate the extension in mm of each spring produced
by a force of 4.0 N.
(ii) Calculate the total strain energy stored in the springs.
2 In Figure 5 an open-wound spring is compressed a distance y of
40 mm. When released the spring rises to a maximum height h
of 800 mm. The mass of the spring is 2.0 g.
(a)By considering the total change of strain energy to
gravitational potential energy, calculate the force constant of
the spring.
h
(b)Plot a graph of how the maximum height risen by the spring,
h, varies with the distance that the spring is compressed, y,
y
up to y = 40 mm.
A
Fixed support
4.0 N
Figure 4 Figure 5 71
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1.3
The Young modulus
8
By the end of this spread, you should be able to . . .
✱ Define and use the terms stress, strain and Young modulus.
✱ Describe an experiment to determine the Young modulus of a metal wire.
Stress and strain
Key definition
In physics, these two terms have completely different meanings, and are expressed in
different units.
Stress is force per unit cross-sectional
area.
Stress is force per unit cross-section area. Stress is therefore expressed in the SI unit
newton per square metre, N m–2. This unit is called the pascal (Pa), which is also used to
quantify pressure. Pressure is used when applied to liquids and gases, whereas stress is
usually applied to solids undergoing deformation.
Strain is extension per unit length.
Examiner tip
Strain is extension per unit length. As a result, strain does not have a unit, since it is a
length divided by a length. It is sometimes quoted as a percentage. A strain of 2% is the
same as a strain of 0.02 and implies that a material has extended by 2 cm for each
metre of its original length.
Take care when using the Young
modulus equation. It is usually safer to
calculate the stress and strain
separately and use the ratio to find Y.
Young modulus values for some
common materials are given in Table 1.
The numerical values are usually large
because the two large numbers appear
on the top of the equation and the two
small numbers are on the bottom. The
very high value for diamond indicates
that only a small change in shape
occurs for the application of a very high
stress. For rubber a large change in
shape can be achieved much more
easily.
diamond
12 × 1011
iron
2.1 × 1011
copper
1.2 × 1011
aluminium
0.71 × 1011
lead
0.18 × 1011
rubber
0.0002 × 1011
The Young modulus
Stress on a material causes strain. How much strain is caused depends on how stiff it is.
A stiff material, such as cast iron, will not alter its shape much when a stress is applied to
it, but a relatively small stress will cause a substantial strain in a soft material, such as
clay. The ratio between stress and strain in a material is known as its Young modulus.
Like stress, it is expressed in pascals. The Young modulus of a material is calculated as
follows:
force
_____
force × length
area
stress _________
______
= _______________
=
Young modulus (Y) =
extension area × extension
strain _________
length
The experiment described in spread 1.3.6, in which a wire was stretched, can be
modified to measure the Young modulus of the material from which the wire is made.
The gradient of the straight-line section of the force/extension graph gives us some of
the information required to do this. We also need to measure the length of the wire
section for which the extension is being measured (i.e. from the clamp to the marker).
The area of cross section of the wire can be calculated by measuring its diameter using
a micrometer screw gauge.
Table 1 Young modulus for some
common materials/Pa.
Clamp
Length of wire under test
Marker
Ruler
Figure 1 Experiment to stretch a wire
The calculation for the experiment, using Figure 1 on this page
and Figure 3 from spread 1.3.6, is:
force
_________
= F/x
extension
= gradient of straight-line part of graph
3.0 N
= ________
0.020 m
= 150 N m–1
length (l) of wire between clamp and marker = 4.05 m
diameter of wire (d) = 0.081 mm = 8.1 × 10–5 m
πd 2
so, area of cross section of wire = ____ = 5.15 × 10–9 m2
4
4.05
Fl
F
l
Y = ___ = __ × __ = 150 × __________
= 1.18 × 1011 Pa
Ax x A
5.15 × 10–9
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Module 3
Work and energy
The Young modulus
Uncertainty
The main uncertainties in this experiment arise from the difficulties involved in
measuring accurately the length of stretch and the area of cross-section of the wire.
These can be overcome as follows:
• The extension can be measured more accurately using a travelling microscope. This
is capable of measuring the distance moved by the marker to within a hundredth of
a millimetre.
• The wire used in this experiment was labelled s.w.g. 44 (standard wire gauge 44).
Manufacturers of copper wire provide tabulated data on their products, which in this
case tell us that the area of cross-section for s.w.g 44 is 0.005 189 mm2. This
information means we do not have to measure the wire’s diameter and square that
figure (squaring a value automatically doubles the uncertainty).
Finally, remember to measure the wire only from the marker to the clamp. A frequent
error made by students performing this experiment is to measure the entire length of
the wire, instead of the section under test.
Questions
4
3
Stress (108 Pa)
1 The unit of pressure, stress and the Young modulus is the pascal. Express the pascal
in terms of the base units kg, m and s. Explain why the Young modulus has the same
unit as stress.
2 (a)The greatest stress that steel can support without breaking is 1.5 × 109 N m2.
What is the greatest load in N which can be applied to a wire of cross-section
1.0 × 10–8 m2 without breaking?
(b)A steel wire of length 1.5 m is given a strain of 0.005. How much longer is it?
3 Copy and complete the following sentences, choosing the most suitable word in each
case from: stress, strain, tension, extension, Young modulus and force constant.
For an experiment to measure the Young modulus of copper several reels of different
gauge (diameter) copper wires are available.
(a)Whatever the length of wire from a reel that is used, a given load attached to the
end of the wire will always produce the same ………… and ………………
(b)For a given length of wire from any reel, the …....…. will be the same if the strain is
the same.
(c)Whatever the dimensions of the wire that is used, the gradient of the stress–strain
graph will always be .........................
4 Figure 2 shows the stress–strain graph for a sample of copper wire. The stress in the
wire was increased slowly from zero to 2.8 × 108 Pa and then reduced slowly back to
zero.
(a)Use the stress–strain graph to find the Young modulus of copper.
(b)Why did the strain not return to zero?
(c)Explain how you could use the graph to estimate the amount of energy which was
dissipated in the experiment. Assume that you have data giving the length and
area of cross-section of the wire.
(d)Another length of copper wire, which has twice the diameter of the wire used in
Figure 2, is stretched until its strain is 2.0 × 10–3.
(i)How would the stress–strain graph for this wire compare with Figure 2?
(ii)How much greater is the tension in the
second wire than the first at the same strain?
2
1
0
1
2
3
Strain (103)
4
Figure 2
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1.3
Categories of materials
9
By the end of this spread, you should be able to . . .
✱ Define and use the term ultimate tensile strength.
✱ Describe the shapes of stress–strain graphs for ductile, brittle and polymeric materials.
Material variety
Gone are the days when engineers and designers faced a simple choice between wood
or metal when planning a construction project. Today, they can choose from a
bewildering variety of materials. Plastics, first manufactured during the 1930s, are now
used widely, while new materials are being designed for highly specific applications.
Alloys of aluminium were traditionally used in the construction of aeroplane wings, for
example, but wings are now being manufactured using so-called ‘composite’ materials
that are stronger, lighter and more flexible than any aluminium alloy. In a composite
material, several different materials, each with its own advantage, are bonded together.
One of the earliest composites was fibreglass, in which glass fibres are embedded in a
polyester plastic and used, for example, to make the hulls of yachts.
Apart from the strength, a range of other properties may affect the choice of material
used in a particular project. Ductility, brittleness, stiffness, density, elasticity, plasticity,
toughness, fatigue resistance, conductivity and fire resistance are some of the many
properties that may shape choices, while cost, ease of shaping and customer appeal
may also enter into the equation.
The properties of individual material types can be illustrated clearly by sketching graphs
of stress against strain.
Ductile materials
Ductile materials, such as copper, can be
drawn out into a wire. Only materials with a
large plastic region can have their shape
altered in this way. Try wrapping the ends of
a thin copper wire around two pencils as
shown in Figure 1. If you pull steadily on the
pencils to stretch the wire, you will feel the
plastic flow of the copper. As you pull, the
copper wire increases in length, straightens,
and its area of cross-section decreases
before it eventually breaks.
Maximum tensile stress
Stress
Plastic
Breaks
Elastic
0
0
0.1
0.2
0.3
Strain
Figure 2 Stress–strain graph for a ductile
material
Pencil
Thin
copper
wire
The reduction in the area of cross-section
Pencil
explains why we use stress–strain graphs
rather than force–extension graphs to analyse
material properties. Where force is constant,
a decrease in the area under the graph
Figure 1 Using pencils to stretch a copper wire
implies an increase in stress.
The shape of the stress–strain graph for a ductile material is shown in Figure 2. Note that
it has a large plastic region within which the material will continue to stretch, even if the
stress is reduced. The graph shows the maximum stress that can be applied to the
material before it will break. This is known as the ultimate tensile stress of a material.
Most metals are ductile, and can be pulled into wires or beaten into sheets. In addition to
copper, wires made from steel and aluminium are widely available, while a range of alloys
are used in electrical circuit wiring. Silver, gold and platinum are used in wire and sheet
form by manufacturers of jewellery.
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Module 3
Work and energy
Categories of materials
Tungsten, which is used in electric lamp filaments because of its high melting point, is
not ductile. This means that it cannot be stretched to form the fine wire used in filaments.
Instead, these are made by squashing finely powdered tungsten into a wire form.
Brittle materials
Polymeric materials
Breaks
Stress
Figure 3 shows a stress–strain graph typical of a brittle material. Note the contrast
between this and the graph for a ductile material (Figure 2). Brittle materials distort very
little, but will snap if subjected to a sufficiently large stress. The graph ceases abruptly,
and the area beneath the graph is small, indicating that very little elastic potential
energy has been stored by the material. Biscuits and concrete are both brittle, but
illustrate the difference in the amount of stress required to break different materials
exhibiting this property.
0
0
0.1
0.2
0.3
Strain
In most stress–strain graphs, the degree of strain is a few percent. For a ductile material
it may be as high as 50%, but for certain polymeric materials it may reach 300%. Rubber
is a common example. Figure 4 plots a stress–strain graph for a rubber band. A 6 cm
rubber band may be stretched to 24 cm quite easily, but is difficult to stretch any further,
even if it does not snap. This is due to the nature of rubber molecules, which are
arranged in a mass of squashed long chains. When a stress is applied to rubber, these
chains straighten, resulting in a large strain.
Figure 3 Stress–strain graph for a brittle
material
Stress
Breaks
0
0
1
2
Strain
Note the high values of strain
3
Figure 4 Stress–strain graph for a rubber band
The flexible properties of rubber have been harnessed in the manufacture of vehicle
tyres. Natural rubber becomes weak and sticky when warm, however, so it is subjected
to a process known as vulcanisation. This involves the addition of impurities, such as
sulfur, to bind the chains of rubber molecules together, making them harder and
stronger, i.e. less easy to stretch.
Questions
1 Here is a list of some mechanical properties of materials: brittle, ductile, elastic, plastic,
strong and tough. Choose as many of these properties as are suitable to describe
each of the following common materials: bone, cast iron, ceramic, concrete, copper,
glass, lead, polyethylene, rubber, wood.
2 When a rubber band is cooled down it becomes much stiffer but it maintains the same
ultimate tensile stress. Imagine that the temperature is such that the strain is half of
that shown in Figure 4 for the same stress. Sketch the shape of the new graph.
Suggest an explanation on a molecular scale for such a significant variation of strain
with temperature.
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Stress
Stress
Work and energy summary
Stress
1.3
Stress =
Strain
Strain
Strain
Ductile
Polymeric
Brittle
Useful output energy
Efficiency =
� 100
Total input energy
F
A
Strain =
Extension
Length
Elastic
Plastic
Sankey diagrams
Conservation of
energy
Materials
Hooke’s law
Efficiency
Elastic limit
Create and destroy
Transfer and
transform
Force
Work and
energy
Power
Area under line =
Work done
W = 12 Fx
Work done
Watt
P= W
t
W = Fx cos U
1 W = 1 J s –1
Gravitational potential energy = mgh
Kinetic energy = 12 mv 2
F = kx
Gradient = k
Extension
Forms of
energy
Chemical energy
Electrical potential energy
Electromagnetic wave energy
Gravitational potential energy
Internal energy
Nuclear energy
Sound energy
Kinetic energy
Stress
Strain
Young modulus =
Joule
W = F � Distance moved in the direction of the force
mgh = 12 mv 2
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Module 3
Work and energy
P r a c t ic e q u e s t io n s
Practice questions
1 The table shows how the braking distance (see spread
1.2.8) for a car of mass 800 kg varies with the initial speed
of the car when a constant braking force F is applied.
Speed (m s–1)
0
10
20
30
40
Distance (m)
0
6
24
x
96
(a)Calculate the kinetic energy in J of the car when it is
travelling at 20 m s–1.
(b)How much work in J is done by the braking force to
bring the car to rest from 20 m s–1?
(c)Calculate the value of the braking force in N.
(d)Calculate the braking distance x in m of the car from
an initial speed of 30 m s–1.
(e)Write down a general equation which will allow you to
calculate the braking distance of the car from any initial
speed.
(f)One simplistic method of measuring the severity of a
car crash is by the amount of kinetic energy which
must be dissipated. Using this method, determine
whether a car hitting a wall at 20 m s–1 is a worse crash
than two identical cars each travelling at 10 m s–1 in
opposite directions making a head-on collision.
2 A model car
Figure 1
runs on a
narrow flexible
track. It has
been formed
into a vertical
circular loop
as shown in
Figure 1.
C
v
3 A student takes a single measurement to measure the
Young modulus of steel. He stretches a fine wire of length
1.5 m and cross-sectional area 5.2 × 10–8 m2 using a force
of 20 N. He measures the extension to be 2.8 mm.
(a)Calculate (i) the stress in the wire, (ii) the strain caused
and (iii) the value of the Young modulus of steel from
these data.
(b)A very tall building requires a series of lifts to reach all
floors of the building. The reasons for this arrangement
include convenience and logistics as well as physics.
The following calculation suggests one of the physical
problems. The cables of length 70 m supporting a lift
consist of two steel ropes each of 100 strands giving a
total cross-sectional area of 1.0 × 10–4 m2. Consider a
full lift to carry 8 passengers of average mass 75 kg.
Calculate by how much an empty lift moves
downwards when it is entered by 8 passengers.
4 The gravitational force between the Earth and the Moon is
2.0 × 1020 N. To try to visualise the magnitude of this force
imagine that the force holding the Earth and Moon together
is provided instead by a steel cable.
(a)Calculate the minimum diameter of such a cable. Use
the ultimate tensile stress of steel, 1.5 × 109 N m–2.
(b)Your answer to (a) should be about 400 km. Show that
the mass of this cable would be greater than the mass
of the Moon which is 7 × 1022 kg. The distance
between the Earth and the Moon is 4 × 108 m and the
density of steel is 7.9 × 103 kg m–3.
r
u
u
A
B
5 The ultimate tensile stress is not the value looked at by
engineers when considering safety in their design of a
building or aircraft or whatever. It is the yield stress at
which permanent or plastic deformation starts. The basic
rule is that the maximum stresses in the design should be
no more than one quarter of the yield stress.
Assume that there are no frictional forces. The car is not
driven by any motive force. It just freewheels.
(a)The car approaches the loop at speed u. Explain why
the speed v at C must be less than u.
(b)Using the law of conservation of energy show that u
and v are related by the equation
(a)The yield stress of steel is 3 × 108 N m–2 and the
ultimate breaking stress is about 109 N m–2. How much
smaller is the maximum stress used in design than the
ultimate breaking stress for steel?
(b)A tug boat assists a tanker to dock at a quay using a
steel cable in which the maximum safe tension is
6 × 105 N. Calculate the diameter of this steel cable.
(c)Explain why it is better to use a long cable to tow a
boat or a car, rather than a short one. Hint: First
consider the extension and increase in strain when
there are sudden changes in motion, e.g. a jerk.
(d)Calculate the strain energy stored in 100 m of cable at
the maximum safe tension. Take the Young modulus of
steel to be 2.1 × 1011 N m–2.
u2 = v2 + 4gr
where g is the acceleration due to gravity.
(c)There is a minimum value of v necessary ____
for the car to
reach C. Unless u is greater or equal to ​√5gr ​ the car
will fall off the track before reaching C. Write down an
expression for the minimum value of v.
(d)For a loop of radius 20 cm, find the minimum height to
which the end of the track, beyond the left-hand side
of the diagram, must be raised for the car to complete
the circle, when released from rest at the end of the
track.
(e)The boy playing with this toy only has enough track to
raise the end 40 cm above the floor. Calculate the
minimum speed at which he must release the car if it is
to perform the stunt.
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1.3
1
Examination questions
(ii) The escape lane is horizontal. The layer of loose
gravel provides a resistive force against the wheels
of the lorry. The lorry is brought to rest 50 m along
the escape lane. Show that the average resistive
force of the gravel on the lorry is 32 kN.
[2]
(iii) In practice, the resistive force provided by the
gravel is unlikely to be constant. Suggest one factor
which may cause a change in the resistive force
as the lorry decelerates, and explain how this will
affect the stopping distance.
[2]
(b) A better design for an escape lane is to give the lane an
uphill gradient from its entry point.
(i) By considering the energy changes taking place,
explain why a lorry of the same mass and speed as
that in (a) can now be brought to rest in a shorter
distance. Assume the average resistive force from
the gravel remains the same.
[2]
(ii) When the lorry of mass 8000 kg enters this escape
lane with kinetic energy 1.6 MJ, it rises through a
vertical height of 4.5 m before stopping. Calculate
the distance the lorry travels through this gravel bed
before stopping. Assume the resistive force remains
at 32 kN. Take g = 9.8 N kg–1.
[5]
(OCR 2861 Jun05)
Figure 1 illustrates a conveyor belt for transporting young
children up a snow-covered bank so that they can ski back
down.
24 m
Conveyer belt
4.0m
Figure 1
A child of mass 20 kg travels up the conveyor belt at a
constant speed. The distance travelled up the slope is 24 m
and the time taken is 55 s. The vertical height climbed in
this time is 4.0 m.
(a) For the child on the conveyor belt, calculate (i) her
speed in m s–1, (ii) her kinetic energy in J and (iii) the
increase in her potential energy in J for the complete
journey up the slope.
[6]
(b) (i) The conveyor belt is designed to take a maximum
of 15 children at any one time. Calculate the power
in W needed to lift 15 children of average mass
20 kg through a height of 4.0 m in 55 s.
[2]
(ii) The belt is driven by an electric motor. State two
reasons why the motor needs a greater output
power than that calculated in (b)(i).
[2]
(OCR 2821 Jan06)
2
3
Figure 3 shows a spring that is fixed at one end and is
hanging vertically.
Fixed end
of spring
This question is about using a gravel bed to stop a runaway
lorry. Figure 2 shows an escape lane positioned on the bend
of a steep hill.
Mass M
Escape
lane
Bed of loose gravel
Downhill
Do
wn
hil
l
Figure 3
Road
A mass M has been placed on the free end of the spring and
this has produced an extension of 250 mm. The weight of
the mass M is 2.00 N.
Figure 4 shows how the force F applied to the spring varies
with extension x up to an extension of x = 250 mm.
Figure 2
4.0
3.0
F /N
The escape lane provides a safe exit from the hill for any
vehicle whose brakes may have failed while descending the
hill.
(a) A lorry of mass 8000 kg enters the escape lane at a
speed of 20 m s–1.
(i) Show that the kinetic energy of the lorry as it enters
the escape lane is 1.6 MJ.
[2]
2.0
1.0
0
0
100
200
x/mm
300
400
Figure 4
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Module 3
Work and energy
Examination questions
(a) (i) Calculate the spring constant of the spring.
[3]
(ii)Calculate the strain energy in J in the spring when
the extension is 250 mm.
[2]
(b) The mass M is pulled down a further 150 mm by a force
F additional to its weight.
(i)Determine the force F in N.
(ii) State any assumption made.
[2]
(c)The mass M is now released and it oscillates up and
down. Figure 5 shows the displacement s against time t
for these oscillations.
5 Figure 6 shows the path of a ball that has been thrown by a
girl towards a vertical wall.
Wall
Path of ball
3.3 m
10 m s–1
53°
200
s/mm
100
0
4.9 m
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
t (s)
1.8
–100
–200
Figure 5
(i) 1 By looking at Figure 5 state a time when the
mass M has maximum downward velocity.
2 Use the graph to determine this maximum
downward velocity in m s–1 of the mass.
[3]
(ii) 1 Use Figure 5 to state a time when the mass M
has maximum resultant force acting on it.
2 Explain your choice of time.
[2]
(OCR 2821 Jun05 Q6)
4 (a) Define the Young modulus.
[1]
(b)The wire used in a piano string is made from steel. The
original length of wire used was 0.75 m. Fixing one end
and applying a force to the other stretches the wire. The
extension produced is 4.2 mm.
(i) Calculate the strain produced in the wire.
[2]
(ii) The Young modulus of the steel is 2.0 × 1011 Pa and
the cross-sectional area of the wire is 4.5 × 10–7 m2.
Calculate the force in N required to produce the
strain in the wire calculated in (i).
[3]
(c)A different material is used for one of the other strings
in the piano. It has the same length, cross-sectional
area and force applied. Calculate the extension in mm
produced in this wire if the Young modulus of this
material is half that of steel.
[2]
(OCR 2821 Jun04)
Figure 6
The girl throws the ball, of mass 5.0 × 10–2 kg, with a
velocity of 10 m s–1 at 53° to the horizontal. In this question,
ignore air resistance.
(a) (i)Show that the horizontal component of the velocity
is 6.0 m s–1. [1]
(ii)In moving to the wall, the ball travels 4.9 m
horizontally and 3.3 m vertically. Calculate the
time in s taken for the ball to travel from the girl’s
hand to the wall.
[2]
(iii)Calculate the gain in potential energy in J of the
ball from leaving the girl’s hand to when it hits the
wall.
[3]
(b)The ball is moving horizontally at 6.0 m s–1 when it hits
the wall. The ball is in contact with the wall for 0.16 s
and rebounds horizontally at 4.0 m s–1. Calculate, for
the time that the ball is in contact with the wall
(i)the change in velocity in m s–1 of the ball
[1]
(ii)the horizontal acceleration of the ball (assumed to
be constant)
[2]
(iii)the magnitude in N and direction of the horizontal
force acting on the ball
[3]
(iv)the loss in kinetic energy in J of the ball when
rebounding from the wall.
[3]
(OCR 2821 Jun06)
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