M2 – CHAPTER 3 – WORK, ENERGY AND POWER
WORK
A force does work if it alters the motion or position of an object. The amount of work depends on
the magnitude of the force and the distance through which its point of application moves.
If the point of application of a force, F, undergoes a displacement, s, in the direction of the force,
then:
Work done by the force = F x s
The SI unit of work is the joule (J). 1 joule is the work done when a force of 1 newton is displaced
through 1 metre.
What matters is the displacement in the direction of the force.
Example 1
Find the work done by a horizontal force of magnitude 0.7N which pushes a particle a distance of
1.2m across a horizontal surface.
Example 2
A box is pulled at a constant speed across a horizontal surface by a horizontal rope. When the box
has moved a distance of 9m the work done is 54J. Find the constant resistance to the motion.
Example 3
Find the work done against gravity when a particle of mass 2kg is raised a vertical distance of 8m.
Example 4
A particle of mass 6kg is pulled at a constant speed a distance of 15m up a rough surface inclined
at 30 to the horizontal. The coefficient of friction between the particle and the surface is 0.2.
Assuming the particle moves up the line of greatest slope find the work done against friction and
the work done against gravity.
Example 5
A rough surface is inclined to the horizontal at 25°. A body of mass 20kg is pulled for 15m up the
line of greatest slope. The coefficient of friction between the body and the surface is 0.25. Find
the total amount of work done on the body if:
(a) The body travels all the way at constant speed
(b) The body starts from rest and takes 6s to travel the 15m with constant acceleration.
ENERGY
Suppose you raise an object of mass 20kg by 10m. The work done against gravity is (20g x 10) = 1969J.
If you let the object sink back to its original level, the force due to gravity will do 1960J of work.
By raising the object you store 1960J of work, which you can retrieve by lowering the object.
Stored work is called ENERGY. The unit of energy is the joule.
Energy of position is gravitational potential energy (GPE). It depends on the mass of the object and the
height of the object relative to a chosen zero level. In a given problem you set a zero level and measure
GPE relative to that.
The work done against gravity in raising an object of mass m kg from the zero level to a height of h metres
is mgh J.
Gravitational potential energy (GPE) = mgh (J)
You also store work when you accelerate an object. For example, suppose you apply a force of
100N to an object of mass 20kg, initially at rest, while it travels 10m.
The work done is (100 x 10) = 1000J
The object accelerates to 10ms-1.
You would need to apply another force to stop the object. Work would be done against that
force. For example, if you apply a frictional force of 200N, the object slows to rest in 5m.
The work done against friction is (200 x 5) = 1000J.
By giving the object a speed of 10ms-1, you stored 1000J of work, which was retrieved when the
object was brought to rest.
The work capacity of an object due to its motion is called kinetic energy (KE). It depends on the
mass of the object and its velocity.
Suppose you apply a force F N to a stationary object of mass m kg for a distance s m. It accelerates
at a ms-2 to a velocity v ms-1.
v2 = u2 + 2as (But u = 0)
as = ½v2 (1)
Work done = Fs = mas (2)
Substitute from (1) into (2)
Work done = ½ mv2 J
An object of mass m kg travelling at v ms-1 has:
Kinetic Energy = ½ mv2 J
Example 6
Find the change in potential energy of a man of mass 70kg ascending a vertical distance of 100m
Example 7
Find the kinetic energy of a bullet of mass 15g moving at 500ms-1.
Example 8
A child of mass 30kg slides 5m down a playground slide inclined at 40 to the horizontal. By
modelling the child as a particle calculate the potential energy lost by the child.
The conservation of energy
KE and GPE are forms of mechanical energy. There is a third form of mechanical energy – elastic
potential energy – but this isn’t covered in this module.
The mechanical energy of a system is changed if an external force does work on it. However in
some situations external forces are negligible and so no work is done on the system.
Suppose for example that you suspend a particle by a light string and set it swinging as a
pendulum. Assume there is no air resistance. Consider the stages of motion of the particle.
On the downswing, the height of the
particle and therefore its GPE decrease,
but its speed and therefore its KE
increase.
Zero KE
Max GPE
Zero KE
Max GPE
Max KE
Min GPE
When the particle reaches its lowest
position, its GPE is at a minimum
and its KE is at a maximum
Zero KE
Max GPE
Zero KE
Max GPE
Max KE
Min GPE
As the particle swings up from its
lowest point, work is done against
gravity. The particle gains height
(GPE increases) and slows down
(KE decreases).
Zero KE
Max GPE
Zero KE
Max GPE
Max KE
Min GPE
The particle stops, at which point it
has zero KE and maximum GPE.
Zero KE
Max GPE
Zero KE
Max GPE
Max KE
Min GPE
The particle then swings down again and the
Pattern starts again.
Zero KE
Max GPE
Zero KE
Max GPE
Max KE
Min GPE
In this idealised model, the sequence repeats forever. On the way up, KE is converted into GPE.
On the way down, GPE is converted into KE. The total energy of the particle remains exactly the
same at all points.
The total mechanical energy of a system only changes because of one of the following:
 An external force, other than gravity, acts on the system in such a way that work is done.
 There are sudden changes in the motion of the system. This happens if parts of the system
collide, or if strings connecting parts of the system are jerked taut.
This is the principle of conservation of mechanical energy.
If the external force does work on the system, its energy increases. If the system does work
against the external force (e.g. against friction) its energy decreases.
The total mechanical energy of a system remains constant provided no external work is done and
there are no sudden changes in the motion of the system.
If external forces act on a system so that work is done, mechanical energy is not conserved.
However, you can still make use of energy to solve problems provided you know the total amount
of work done on the system.
The work-energy principle states that the total work done on a system equals the change in
mechanical energy of the system.
Example 9
A stone of mass 1kg is dropped from the top of a vertical cliff. If it hits the ground with a speed of
20ms-1 find
(a)
the kinetic energy gained
(b)
the potential energy lost, stating an assumption you have made
(c)
the height of the cliff
Example 10
A particle of mass 4kg is projected up the line of greatest slope of a rough plane inclined at
arcsin 135 to the horizontal with a velocity of 8ms-1. Given that the coefficient of friction between
the single particle and the plane is
1
4
find, by energy considerations, the distance the particle
moves up the plane before coming to rest.
Example 11
A cyclist and his machine have a combined mass of 90kg. The cyclist freewheels down a hill
inclined at 30 to the horizontal. He increases his speed from 5ms-1 to 25ms-1. By modelling the
cyclist and his bicycle as a single particle and assuming that resistances can be neglected, calculate
the potential energy lost by the cyclist and the distance he travels.
Example 12
Particles P and Q, of mass 6kg and 2kg respectively, are connected by a light string of length 8m
passing over a smooth pulley at the top of a double 30° slope, as shown. The coefficient of friction
for both particles is 0.1. The system starts from rest with the string taut and P at the top of the
slope. Find the velocity of the particles when Q reaches the top of the top of the slope.
P
4m
Q
30°
30°
POWER
If a person of mass 90kg climbs a flight of stairs taking them to a height of 15m, they do
90g x 15 = 13230 J of work. This is the same whether they run up the stairs or walk up slowly.
However, the effects on their breathing and heart rate vary depending on how quickly they work.
This is the power they exert.



Power is the rate at which work is done
The SI unit of power is the watt (W)
1 watt is the rate of working of 1 joule per second.
Example 13
A crane lifts a load of 50kg to a height of 12m at a steady speed of 0.6ms-1. Find the power
required.
Example 14
A car of mass 900kg moves at a steady speed of 15ms-1 up a slope inclined to the horizontal at 10°
against a constant resistive force of 400N. Find the power output of the engine.
Example 15
A pump raises water from a tank through a height of 3m and expels it through a circular nozzle of
radius 3cm at 8ms-1. Find the rate at which the pump is working. Ignore any resistive forces.
For a vehicle, the power output by the engine depends on the driving force of the engine and the
velocity of the vehicle.
Suppose a car exerts a constant driving force F N so that it has a constant speed of v ms -1.
In 1 second, the car travels vm.
You have work done in 1 second = Fv J, so, power output by engine = Fv W
For a vehicle travelling at v ms-1 under a driving force F N, power = Fv.
Example 16
A car is being driven at a constant speed of 20ms-1 on a level road against a constant resistance
force of 260N. Find the power output of the engine.
Example 17
A car of mass 800kg is driven along a level road against a constant resistance of 450N. The output
of the engine is 7kW. Find:
(a) the acceleration when the speed is 10ms-1.
(b) The maximum speed of the car.
Example 18
A car of mass 900kg travels up a hill, inclined at 10° to the horizontal, against a constant resistance
force of 250N. Its maximum speed is 45 kmh-1. Find:
(a) The power output of the engine
(b) The initial acceleration when the car reaches level road at the top of the hill, assuming that
the power remains constant.