# MTH5P2B: Energy, work and power

##### This unit explains Energy, work and power This is a picture of a perpetual motion machine. What does this term mean and will this one work?

Energy and momentum
When describing the motion of objects in everyday language the words energy and momentum are often used quite loosely and sometimes no distinction is made between them. In mechanics they must be defined precisely.
For an object of mass m moving with velocity v:
Kinetic energy (this is the energy it has due to its motion)

Momentum = m v

Notice that kinetic energy is a scalar quantity with magnitude only, but momentum is a vector in the same direction as the velocity.
Both the kinetic energy and the momentum are liable to change when a force acts on a body and you will learn more about how the energy is changed in this chapter.

Work and energy

In everyday life you encounter many forms of energy such as heat, light, electricity and sound. You are familiar with the conversion of one form of energy to another: from chemical energy stored in wood to heat energy when you burn it; from electrical energy to the energy of a train’s motion, and so on. The S.I. unit for energy is the joule, J.
Mechanical energy and work
In mechanics two forms of energy are particularly important.
Kinetic energy is the energy which a body possesses because of its motion.
The kinetic energy of a moving object Potential energy is the energy which a body possesses because of its position. It may be thought of as stored energy which can be converted into kinetic or other forms of energy. You will meet this again on page 163.
The energy of an object is usually changed when it is acted on by a force. When a force is applied to an object which moves in the direction of its line of action, the force is said to do work. For a constant force this is defined as follows.
The work done by a constant force = force × distance moved in the direction of the force.
The following examples illustrate how to use these ideas.

A brick, initially at rest, is raised by a force averaging 40 N to a height 5 m above the ground where it is left stationary. How much work is done by the force?

SOLUTION
The work done by the force raising the brick is
40 × 5 = 200 J. Examples 9.2 and 9.3 show how the work done by a force can be related to the change in kinetic energy of an object.
Example 9.2 A train travelling on level ground is subject to a resisting force (from the brakes and air resistance) of 250 kN for a distance of 5 km. How much kinetic energy does the train lose?

SOLUTION
The forward force is −250 000 N. The work done by it is –250 000 × 5000 = −1250 000 000 J.

Hence −1250 000 000 J of kinetic energy are gained by the train, in other words + 1250 000 000 J of kinetic energy are lost and the train slows down. This energy is converted to other forms such as heat and perhaps a little sound. SOLUTION
Treating the car as a particle and applying Newton’s second law: Since F is assumed constant, the acceleration is constant also, so using Thus
work done by force = final kinetic energy − initial kinetic energy of car.
The work–energy principle
Examples 9.4 and 9.5 illustrate the work–energy principle which states that:
The total work done by the forces acting on a body is equal to the increase in the kinetic energy of the body.

A sledge of total mass 30 kg, initially moving at 2 m s−1, is pulled 14 m across smooth horizontal ice by a horizontal rope in which there is a constant tension of 45 N. Find its final velocity. SOLUTION
Since the ice is smooth, the work done by the force is all converted into kinetic energy and the final velocity can be found using work done by the force = final kinetic energy − initial kinetic energy.  The combined mass of a cyclist and her bicycle is 65 kg. She accelerated from rest
to 8 in 80 m along a horizontal road.
Calculate the work done by the net force in accelerating the cyclist and her bicycle.
Hence calculate the net forward force (assuming the force to be constant).   Work
It is important to realise that:
●work is done by a force
●work is only done when there is movement
●a force only does work on an object when it has a component in the direction of motion of the object.
It is quite common to speak of the work done by a person, say in pushing a lawn mower. In fact this is the work done by the force of the person on the lawn mower.
Notice that if you stand holding a brick stationary above your head, painful though it may be, the force you are exerting on it is doing no work. Nor is this vertical force doing any work if you walk round the room keeping the brick at the same height. However, once you start climbing the stairs, a component of the brick’s movement is in the direction of the upward force that you are exerting on it, so the force is now doing some work.

When applying the work–energy principle, you have to be careful to include all the forces acting on the body. In the example of a brick of weight 40 N being raised 5 m vertically, starting and ending at rest, the change in kinetic energy is clearly 0.

This seems paradoxical when it is clear that the force which raised the brick has done 40 × 5 = 200 J of work. However, the brick was subject to another force, namely its weight, which did −40 × 5 = 200 J of work on it, giving a total of 200 + (−200) = 0 J.

Conservation of mechanical energy
The net forward force on the cyclist in Example 9.5 is the girl’s driving force minus resistive forces such as air resistance and friction in the bearings. In the absence of such resistive forces, she would gain more kinetic energy; also the work she does against them is lost, it is dissipated as heat and sound. Contrast this with the work a cyclist does against gravity when going uphill.

This work can be recovered as kinetic energy on a downhill run. The work done against the force of gravity is conserved and gives the cyclist potential energy (see page 163). Forces such as friction which result in the dissipation of mechanical energy are called dissipative forces. Forces which conserve mechanical energy are called
conservative forces. The force of gravity is a conservative force and so is the tension in an elastic string; you can test this using an elastic band.

bullet of mass 25 g is fired at a wooden barrier 3 cm thick. When it hits the barrier it is travelling at 200 m s−1. The barrier exerts a constant resistive force of 5000 N on the bullet.
(i) Does the bullet pass through the barrier and if so with what speed does it emerge?
(ii) Is energy conserved in this situation?

SOLUTION The work done by the force is defined as the product of the force and the distance moved in the direction of the force. Since the bullet is moving in the direction opposite to the net resistive force, the work done by this force
is negative.

Work done = −5000 × 0.03
= 150 J
The initial kinetic energy of the bullet is A loss in energy of 150 J will not reduce kinetic energy to zero, so the bullet will still be moving on exit.
Since the work done is equal to the change in kinetic energy, So the bullet emerges from the barrier with a speed of Total energy is conserved but there is a loss of mechanical energy of This energy is converted into non-mechanical forms such as heat and sound.

An aircraft of mass m kg is flying at a constant velocity horizontally. Its engines are providing a horizontal driving force F N.
(i) Draw a diagram showing the driving force, the lift force L N, the air resistance (drag force) R N and the weight of the aircraft.

(ii) State which of these forces are equal in magnitude.

(iii) State which of the forces are doing no work.

(iv) In the case when m = 100 000, v = 270 and F = 350 000, find the work done in a 10-second period by those forces which are doing work, and show that the work–energy principle holds in this case.

At a later time the pilot increases the thrust of the aircraft’s engines to 400 000 N.

When the aircraft has travelled a distance of 30 km, its speed has increased to 300 m s−1.

(v) Find the work done against air resistance during this period, and the average resistance force.

SOLUTION Since the aircraft is travelling at constant velocity it is in equilibrium. Horizontal forces: F = R Vertical forces:

L = mg

(iii) Since the aircraft’s velocity has no vertical component, the vertical forces, L and mg, are doing no work.

(iv) In 10 s at 270 m s−1 the aircraft travels 2700 m.

Work done by force F = 350 000 × 2700 = 9 450 000 J
Work done by force R = 350 000 × −2700 = −9 450 000 J

The work–energy principle states that in this situation
work done by F + work done by R = change in kinetic energy.
Now work done by F + work done by R = (9 450 000 − 9 450 000) = 0 J, and change in kinetic energy = 0 (since velocity is constant), so the work–energy principle does indeed hold in this case.  Note
When an aircraft is in flight, most of the work done by the resistance force results in air currents and the generation of heat. A typical large jet cruising at 35 000 feet has a body temperature about 30°C above the surrounding air temperature. For supersonic flight the temperature difference is much greater. Concorde used to fly with a skin temperature more than 200°C above that of the surrounding air.

As you have seen, kinetic energy (K.E.) is the energy that an object has because of its motion. Potential energy (P.E.) is the energy an object has because of its position. The units of potential energy are the same as those of kinetic energy or any other form of energy, namely joules.

One form of potential energy is gravitational potential energy. The gravitational potential energy of the object in figure 9.6 of mass m kg at height h m above a fixed reference level, 0, is mgh J. If it falls to the reference level, the force of gravity does mgh J of work and the body loses mgh J of potential energy. A loss in gravitational potential energy is an alternative way of accounting for the work done by the force of gravity.

If a mass m kg is raised through a distance h m, the gravitational potential energy increases by mgh J. If a mass m kg is lowered through a distance h m the gravitational potential energy decreases by mgh J.

Calculate the gravitational potential energy, relative to the ground, of a ball of mass 0.15 kg at a height of 2 m above the ground.

SOLUTION
Mass m = 0.15, height h = 2.
Gravitational potential energy = mgh
= 0.15 × 10 × 2
= 3 J.
Note
If the ball falls:
loss in P.E. = work done by gravity
= gain in K.E.
There is no change in the total energy (P.E. + K.E.) of the ball.

Using conservation of mechanical energy
When gravity is the only force which does work on a body, mechanical energy is conserved. When this is the case, many problems are easily solved using energy.

This is possible even when the acceleration is not constant.

A skier slides down a smooth ski slope 400 m long which is at an angle of 30°to the horizontal. Find the speed of the skier when he reaches the bottom of the slope.

At the foot of the slope the ground becomes horizontal and is made rough in order to help him to stop. The coefficient of friction between his skis and the ground is 1/4 .
(i) Find how far the skier travels before coming to rest.
(ii) In what way is your model unrealistic?
SOLUTION
The skier is modelled as a particle. Since in this case the slope is smooth, the frictional force is zero. The skier is subject to two external forces: his weight mg and the normal reaction from the slope.

The normal reaction between the skier and the slope does no work because the skier does not move in the direction of this force. The only force which does work is gravity, so mechanical energy is conserved. The skier’s speed at the bottom of the slope is Notice that the mass of the skier cancels out. Using this model, all skiers should arrive at the bottom of the slope with the same speed. Also the slope could be curved so long as the total height lost is the same.

For the horizontal part there is some friction. Suppose that the skier travels a further distance s m before stopping.   So the distance the skier travels before stopping is 800 m.
(ii) The assumptions made in solving this problem are that friction on the slope and air resistance are negligible, and that the slope ends in a smooth curve atA. Clearly the speed of is very high, so the assumption that friction and air resistance are negligible must be suspect.

Ama, whose mass is 40 kg, is taking part in an assault course. The obstacle shown in figure 9.9 is a river at the bottom of a ravine 8 m wide which she has to cross by swinging on a rope 5 m long secured to a point on the branch of a tree, immediately above the centre of the ravine. Find how fast Ama is travelling at the lowest point of her crossing
(a) if she starts from rest
(b) if she launches herself off at a speed of (ii) Will her speed be faster throughout her crossing?

SOLUTION
(i) (a) The vertical height Ama loses is HB in the diagram.  Ama’s speed at the lowest point is only 0.08 in part (i)(b)
compared with that in part (i)(a), so she clearly will not travel faster throughout in part (i)(b).

Historical note
James Joule was born in Salford in Lancashire on Christmas Eve 1818. He studied at Manchester University at the same time as the famous chemist, Dalton.
Joule spent much of his life conducting experiments to measure the equivalence of heat and mechanical forms of energy to ever-increasing degrees of accuracy. Working with William Thomson, he also discovered that a gas cools when it expands without doing work against external forces. It was this discovery that paved the way
for the development of refrigerators.Joule died in 1889 but his contribution to science is remembered with the S.I. unit for energy named after him.

Work and kinetic energy for two-dimensional motion

Imagine that you are cycling along a level winding road in a strong wind.
Suppose that the strength and direction of the wind are constant, but because the road is winding sometimes the wind is directly against you but at other times it is from your side.
How does the work you do in travelling a certain distance – say 1 m – change with your direction?

Work done by a force at an angle to the direction of motion You have probably deduced that as a cyclist you would do work against the component of the wind force that is directly against you. The sideways component does not resist your forward progress.

Suppose that you are sailing and the angle between the force, F, of the wind on your sail and the direction of your motion is θ. In a certain time you travel a distance d in the direction of F, see figure 9.11, but during that time you actually travel a distance s along the line OP. Work done by F = Fd
Since d = s cos θ, the work done by the force F is Fs cos θ. This can also be written as the product of the component of F along OP, F cos θ, and the distance moved along OP, s.

F × s cos θ = F cos θ × s

(Notice that the direction of F is not necessarily the same as the direction of the wind, it depends on how you have set your sails.)

As a car of mass m kg drives up a slope at an angle α to the horizontal it experiences a constant resistive force F N and a driving force D N. What can be deduced about the work done by D as the car moves a distance d m uphill if:
(i) the car moves at constant speed?
(ii) the car slows down?
(iii) the car gains speed?

The initial and final speeds of the car are denoted by respectively.

(iv) Write in terms of the other variables.

SOLUTION
The diagram shows the forces acting on the car. The table shows the work done by each force. The normal reaction, R, does no work as the car moves no distance in the direction of R.  (i) If the car moves at a constant speed there is no change in kinetic energy so the total work done is zero, giving
Work done by D is
Dd = Fd + mgd sin α.
(ii) If the car slows down the total work done by the forces is negative, hence
Work done by D is
Dd < Fd + mgd sin α.

(iii) If the car gains speed the total work done by the forces is positive so
Work done by D is
Dd > Fd + mgd sin α.
(iv) Total work done = final K.E. − initial K.E. Power It is claimed that a motorcycle engine can develop a maximum power of 26.5 kW at a top speed of This suggests that power is related to speed and this is indeed the case.
Power is the rate at which work is being done. A powerful car does work at a greater rate than a less powerful one.
You might find it helpful to think in terms of a force, F, acting for a very short time t over a small distance s. Assume F to be constant over this short time.
Power is the rate of working so The power of a vehicle moving at speed v under a driving force F is given by Fv.
For a motor vehicle the power is produced by the engine, whereas for a bicycle it is produced by the cyclist. They both make the wheels turn, and the friction between the rotating wheels and the ground produces a forward force on the machine.

The unit of power is the watt (W), named after James Watt. The power produced by a force of 1 N acting on an object that is moving at Because the watt is such a small unit you will probably use kilowatts more often (1 kW = 1000 W).

A car of mass 1000 kg can produce a maximum power of 45 kW. Its driver wishes to overtake another vehicle. Ignoring air resistance, find the maximum acceleration of the car when it is travelling at This example shows why it is easier to overtake a slow moving vehicle.
A car of mass 900 kg produces power 45 kW when moving at a constant speed. It experiences a resistance of 1700 N.

(i) What is its speed?
(ii) The car comes to a downhill stretch inclined at 2° to the horizontal. What is its maximum speed downhill if the power and resistance remain unchanged?

SOLUTION

(i) As the car is travelling at a constant speed, there is no resultant force on the car. In this case the forward force of the engine must have the same magnitude as the resistance forces, i.e. 1700 N. (ii) The diagram shows the forces acting. At maximum speed there is no acceleration so the resultant force down the slope is zero. Historical note
James Watt was born in 1736 in Greenock in Scotland, the son of a house- and ship-builder. As a boy James was frail and he was taught by his mother rather than going to school. This allowed him to spend time in his father’s workshop where he developed practical and inventive skills.

As a young man he manufactured mathematical instruments: quadrants, scales, compasses and so on. One day he was repairing a model steam engine for a friend and noticed that its design was very wasteful of steam. He proposed an alternative arrangement, which was to become standard on later steam engines. This was the first of many engineering inventions which made possible the subsequent industrial revolution. James Watt died in 1819, a well known and highly respected man. His name lives on as the S.I. unit for power.

## STEM Elearning

We at FAWE have built this platform to aid learners, trainers and mentors get practical help with content, an interactive platform and tools to power their teaching and learning of STEM subjects, more

#### How to find your voice as a woman in Africa

© FAWE, Powered by: Yaaka DN.