Test yourself1 Name three different types of energy.2 What unit is used for measuring work?3 What quantity is defi ned as work done per unit time?4 Give a single word that means the ‘forces in a stretched

spring’.

IntroductionThe 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.

Module 3 Work and energy

UNIT1

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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

Module contents

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1.3 1 Work and the jouleBy the end of this spread, you should be able to . . .

✱ Defi ne work and the joule.✱ Calculate the work done by a force.

WorkWork, as the word is used in physics, is defi ned by the equation:

work = force × distance moved in the direction of the force

As the above defi nition 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 defi nition 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 workThe SI unit of work is the joule.

Like all units, it is defi ned in terms of other units. 1 joule = 1 newton metre. This is not newtons per metre: you have to multiply, not divide.

Typical work valuesIn 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

Force at an angle to the direction of movementSo 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.

Figure 1 Barrel on a ramp

25° 200 N

4.26 m

1.8 m

Key defi nition

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.

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A barrel of weight 200 N is raised by a vertical distance of 1.8 m by being moved along the ramp.

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:

distance moved along the ramp = 1.8 m _______ sin 25º

= 4.26

force required = 360 J _______ 4.26 m

= 84.5 N

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.

Questions1 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.

Man

Canal

DragBarge 30°

400N

340N

Horse

200N

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.

Module 3Work and energy

Work and the joule

Figure 2 The Earth exerts a large gravitational force on the Moon, but does no work on it

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1.3 2 The conservation of energyBy the end of this spread, you should be able to . . .

✱ Describe examples of energy in diff erent 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.

EnergyEnergy and its uses are fundamental to any physics course. Indeed, physics can be described as the study of energy. So fi rst 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 fl y 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 specifi c 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% effi cient. Most modern cars have effi ciencies below 50%. This problem is due to the very nature of energy-to-work conversions. There is a theoretical limit on the effi ciency 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 fi elds.

However different energy names are frequently used for specifi c 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 fi elds.

• Gravitational potential energy: where an object is at a high level in the Earth’s gravitational fi eld.

• 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.

Key defi nition

Energy is the stored ability to do work.

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The conservation of energyIn 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.

Module 3Work and energy

The conservation of energy

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.

Questions1 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.

Figure 1 Sankey diagram

4.8 MJ chemicalenergy from fuel

0.9 MJ Work done on road –road gets hot

Energy of exhaust gases.They are hot too.

0.06 MJ light energy

2.6 MJ

0.04 MJ sound energy

Kinetic energy ofturbulence insurrounding air

1.2 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.

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.

Where did the energy go?

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1.3 3 Potential and kinetic energiesBy 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 energyEverything on this planet exists in the Earth’s gravitational fi eld. Near the Earth’s surface, the strength of this fi eld 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:

force × distance = Wh = mgh

This gives the general equation, when g is a constant, as:

gravitational potential energy = mgh

Gravitational potential energy is gained when rising and lost when falling.

Kinetic energyThe kinetic energy of a moving body equals the work it can do as a result of its motion.

If you work through this worked example again, but this time using a different stopping force, you will fi nd 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 fi rst place.

Key defi nition

Kinetic energy is the work an object can do by virtue of its speed.

Key defi nition

Gravitational potential energy is the energy stored in an object (the work an object can do) by virtue of its position in a gravitational fi eld.

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.

The acceleration of the lorry = F __ m = – 84 000 N ________ 30 000

kg

= –2.8 m s–2

Applying v2 = u2 + 2as gives: 02 = 202 + 2 × (– 2.8) × sand so: 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

Worked example

Figure 1 Skydivers converting gravitational potential energy into kinetic energy

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Falling objectsTime-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 = v2 or v = √____

2gh

Module 3Work and energy

Potential and kinetic energies

Questions1 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.

Zero

0.016 s

0.289 s

s

40.7 cm

DistancesTimes

Light gate

Light gate

Figure 2 Dropping a ball bearing through light gates

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

g = 0.814 _________ 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.

Worked example

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Work1 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°?

2 A box of books weighing 200 N is pushed steadily along a smooth fl oor 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)?

Conservation of energy3 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.

Work1 A person in a car travelling at 30 m s

Work

1.3 Further questions B

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Module 3Work and energy

Further questions B

Potential and kinetic energy5 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

40 m

X

Y

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

(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.

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.

Falling objects9 A boy of mass 55 kg, swinging from the branch of a tree,

drops a distance of 1.5 m to the ground.

(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.

10 Two balls of different materials are dropped from 1.0 m 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?

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.

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1.3 4 Power and the wattBy the end of this spread, you should be able to . . .

✱Defi ne and use power and the watt.✱Calculate the cost of using an electrical appliance.

PowerPower is defi ned as the rate of doing work.

In equation form this is expressed as:

power = work done __________ time taken

Power is measured in joules per second, also known as watts.

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 fl owing through the fi lament 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 Time Total Example of total

speed 80 km h–1 4 h distance 320 km

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.

Figure 1 An athlete performing press-ups

Key defi nition

Power is the rate of doing work.

Key defi nition

One watt (W) is equal to one joule per second.

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 effi ciency 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.

Worked example

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Questions1 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.

Module 3Work and energy

Power and the watt

Figure 2

15

1

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

(b) Power required = 63 600 J ________ 360 s

= 176 W

(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.

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!

Human power and horse power

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1.3 5 Effi ciencyBy the end of this spread, you should be able to . . .

✱ Calculate the effi ciency of a device and explain why it is always less than 100%.✱ Draw and interpret Sankey diagrams.

Effi ciencyEnergy 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-fi red 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 effi ciency of a device.

Effi ciency is expressed as:

effi ciency = useful output energy

__________________ total input energy

× 100%

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:

effi ciency = useful output power

__________________ total input power

× 100%

Table 1 contains some examples of common devices for which effi ciency is important. It shows their various input and output energy types, along with the typical effi ciencies for these devices.

Device Energy input Energy output Typical effi ciency (%)

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 effi ciency is important

It is worth noting that using electrical energy to produce some other form of energy can be highly effi cient – the only reason why the effi ciency 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, ineffi cient. 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 effi cient 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 diffi cult. The science of thermodynamics deals with such problems.

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Module 3Work and energy

Efficiency

Figure 4 Power transmission lines from a power station

Figure 2 Didcot power station with its cooling towers

Figure 3 Sizewell power station with the sea to provide cooling water

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).

Questions1 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.

Power fromburning fuel3000 MW

Electricalpower output

50 MWPower to runstation equipment

5 MWFriction lossesin generator1 MW

Transformerlosses

1944 MW

Heat losses incooling towers

1000 MW

Figure 1 Sankey diagram for a power station

Sankey diagramsWe 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).

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1.3 6 Deformation of materialsBy 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.✱ Defi ne the terms elastic deformation and plastic deformation.

Elastic and plastic deformationSolids 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.

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 defi nition, is plastic.

Tensile and compressive forcesIn 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 fi xed, 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.

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.

An experiment to stretch a wireFigure 2 shows a long, thin copper wire held fi rmly 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 fi xed 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.

Figure 1a Stretching a wire

Two equal and opposite tensileforces stretching a wire

Figure 1b Squeezing a spring

Two equal and opposite compressiveforces squeezing a spring

Figure 2 Experiment to stretch a wire

MarkerLength of wire under testClamp

Ruler

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.

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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.

The graph shows two distinct regions. The initial straight-line portion of the graph is followed by a curved section, showing much greater extension. The 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.

Questions1 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?

Module 3Work and energy

Deformation of materials

Figure 4 Presses forming body panels for cars

Figure 3 Graph plotted using data from stretch experiment

Tens

ion/

N

8

6

4

2

040 60 80 100 120200

Extension/mm

Elasticlimit

Figure 5

Tens

ion/

N

Tens

ion/

N

Tens

ion/

N

Tens

ion/

N

200

00 2

Extension/mm4

400

1000

00 0.3

Extension/mm0.6

2000(i) (ii)

20

00 200

Extension/mm400

40

5

00 100

Extension/mm200

10(iii) (iv)

B A

Figure 6

Tens

ion/

N

Tens

ion/

N

10

00 100

Extension/mm200

20

50

00 150

Extension/mm300

100

Mass on hanger /kg

Tension in wire /N

Reading on ruler /mm

Extension /mm

0 0 26.0 0

0.050 0.49 29.0 3.0

0.100 0.98 32.0 6.0

0.150 1.47 35.5 9.5

0.200 1.96 38.5 12.5

0.250 2.45 42.0 16.0

0.300 2.94 45.5 19.5

0.350 3.43 48.5 22.5

0.400 3.92 53.0 27.0

0.500 4.90 66.5 40.5

0.600 5.88 87? 61?

0.700 6.86 122? 96?Table 1

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1.3 7 Hooke’s lawBy 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 constantIn 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.

The work done to stretch a wireWe 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:

E = ½Fx = ½kx2

Figure 1 Straight-line graph of tension against extension

Tens

ion/

N

Extension/mmx0

F

Figure 2 Child’s toy showing a spring within which elastic potential energy may be stored

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Energy stored in a plastic deformationThe 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.

Questions1 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.

(b) Plot a graph of how the maximum height risen by the spring, h, varies with the distance that the spring is compressed, y, up to y = 40 mm.

Module 3Work and energy

Hooke’s law

Forc

e/N

Extension/mmx0

AB

Figure 3 Stretching a wire beyond its elastic limit

Fixed support

4.0 N

Figure 4

Figure 5

y

h

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?

Answer(a) Since E = ½kx2

250 = ½ × k × 0.102

k = 2 × 250 _______ 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

Worked example

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1.3 8 The Young modulusBy the end of this spread, you should be able to . . .

✱ Defi ne and use the terms stress, strain and Young modulus.✱ Describe an experiment to determine the Young modulus of a metal wire.

Stress and strainIn physics, these two terms have completely different meanings, and are expressed in different units.

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. 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.

The Young modulusStress 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:

Young modulus (Y) = stress ______ strain

= force _____ area _________

extension _________ length

=

force × length _______________

area × extension

The experiment described in spread 1.3.6, in which a wire was stretched, can be modifi ed 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.

The calculation for the experiment, using Figure 1 on this page and Figure 3 from spread 1.3.6, is:

force _________ extension

= F/x

= 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 mdiameter of wire (d) = 0.081 mm = 8.1 × 10–5 m

so, area of cross section of wire = πd2

____ 4 = 5.15 × 10–9 m2

Y = Fl ___ Ax

= F __ x × l __ A

= 150 × 4.05 __________ 5.15 × 10–9

= 1.18 × 1011 Pa

Examiner tip

Take care when using the Young modulus equation. It is usually safer to calculate the stress and strain separately and use the ratio to fi nd 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.

Key defi nition

Stress is force per unit cross-sectional area.

Strain is extension per unit length.

Figure 1 Experiment to stretch a wire

MarkerLength of wire under testClamp

Ruler

diamond 12 × 1011

iron 2.1 × 1011

copper 1.2 × 1011

aluminium 0.71 × 1011

lead 0.18 × 1011

rubber 0.0002 × 1011

Table 1 Young modulus for some common materials/Pa.

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Questions1 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?

Module 3Work and energy

The Young modulus

Stre

ss (1

08 Pa

)

Strain (103)0 1 2 3 4

4

3

2

1

Figure 2

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.

Uncertainty

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1.3 9 Categories of materialsBy the end of this spread, you should be able to . . .

✱ Defi ne and use the term ultimate tensile strength.✱ Describe the shapes of stress–strain graphs for ductile, brittle and polymeric materials.

Material varietyGone 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, fi rst manufactured during the 1930s, are now used widely, while new materials are being designed for highly specifi c 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 fl exible 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 fi breglass, in which glass fi bres 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 fi re 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 materialsDuctile 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 fl ow of the copper. As you pull, the copper wire increases in length, straightens, and its area of cross-section decreases before it eventually breaks.

The reduction in the area of cross-section 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 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.

Figure 1 Using pencils to stretch a copper wire

Pencil

Pencil

Thincopperwire

Figure 2 Stress–strain graph for a ductile material

Strain

Stre

ss

0 0.1 0.2 0.30

Breaks

Elastic

Plastic

Maximum tensile stress

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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 materialsFigure 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.

Polymeric materialsIn 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.

Module 3Work and energy

Categories of materials

Figure 3 Stress–strain graph for a brittle material

Strain

Stre

ss

0 0.1 0.2 0.30

Breaks

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.

Questions1 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.

Note the high values of strainStrain

Stre

ss

0 1 2 30

Breaks

Figure 4 Stress–strain graph for a rubber band

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Work andenergy

P = Wt

W = Fx cos U

Joule

Elastic limit

Efficiency =Useful output energy

Total input energy�100

mgh = mv 212

Gravitational potential energy = mgh

Kinetic energy = mv 212

Chemical energyElectrical potential energyElectromagnetic wave energyGravitational potential energyInternal energyNuclear energySound energyKinetic energy

Forms ofenergy

Transfer andtransform

Create and destroy

Sankey diagrams

Conservation ofenergy

Ductile Polymeric Brittle

Stre

ss

Strain

Stre

ss

Strain

Stre

ss

Strain

Strain = ExtensionLength

Stress = FA

Young modulus = StressStrain

W = F � Distance moved in the direction of the force

1 W = 1 J s – 1

Watt

Power

Forc

e

Extension

Gradient = k

Area under line =Work done

W = Fx12

Work done

F = kx

Hooke’s law

ElasticPlastic

Materials

Efficiency

1.3 Work and energy summary

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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 runs on a narrow flexible track. It has been formed into a vertical circular loop as shown in Figure 1.

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

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.

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.

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.

(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.

Practice questions

Module 3Work and energy

Practice questions

C

v

B

u

A

u

r

Figure 1

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78

1 Figure 1 illustrates a conveyor belt for transporting young children up a snow-covered bank so that they can ski back down.

Figure 1

Conveyer belt

24m

4.0m

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 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.

Figure 2

EscapelaneBed of loose gravel

Road

Downh

ill

Downhill

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]

(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)

3 Figure 3 shows a spring that is fi xed at one end and is hanging vertically.

Figure 3

Fixed endof spring

Mass M

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 4

1000 200 300 400x /mm

F /N

4.0

3.0

2.0

1.0

0

1 Figure 1 illustrates a conveyor belt for transporting young children up a snow-covered bank so that they can ski back

1 Figure 1 illustrates a conveyor belt for transporting young

1.3 Examination questions

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Module 3Work 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.

Figure 5

s/m

m

200

100

0

–100

–200

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4t (s)

1.6 1.8

(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)

5 Figure 6 shows the path of a ball that has been thrown by a girl towards a vertical wall.

4.9 m

53°

10 m s–1

Path of ball

3.3 m

Wall

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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