EFFICIENCY OF A DEVICE

In a machine, the useful work done (work output) is always less than the work input. Total
energy is conserved but some of the energy input is wasted in form of heat due to friction in
moving parts, sound etc.

No real machine can have an efficiency of 100%.

In general, when one form of energy is converted into another form, some energy is wasted.

When an electric bulb, for example, is in use the useful energy output is light. The heat energy
produced is wasted energy. Hence the useful energy output (in the form of light) is less than
the energy input (in the form of electrical energy).

In the motor car engine, we supply the energy in the form of chemical energy in the petrol. The
useful energy that is given out is mechanical energy. But some energy is wasted as heat in the
radiator and exhaust system, more heat is produced by friction in moving parts of the car and
some energy is wasted as sound.

In general when one form of energy is converted into another form, some energy is wasted.
Wasted energy

- Heat
- Sound
- Unburnt fuel
- Sound energy

Defn: The efficiency of a machine or device is the ratio of the useful work (energy) output to
the total work (energy) input.

The efficiency of a device, being a ratio, has no units. Its value will always be less than 1. To
express efficiency as a percentage, multiply by 100.
More precisely,
𝐄𝐧𝐞𝐫𝐠𝐲 𝐨𝐮𝐭𝐩𝐮𝐭
Efficiency = x100%
𝐄𝐧𝐞𝐫𝐠𝐲 𝐢𝐧𝐩𝐮𝐭
Efficiency is usually less than 1 or less than 100%. It is important to cut down energy wasted so
that the useful work obtained from the device is as high as possible.
Example
1. Find the efficiency % of an electric motor that is capable of pulling a 50 kg mass
through a height of 15 m after consuming 30 kJ of electrical energy.
Solution
Work output = mgh = (50 x 10 x 15) J = 7 500 J
Work (Energy) input = 30 kJ = 30 000 J
Work output 7 500 J
∴ Efficiency % = x 100% = x 100% = 25%
Work input 30 000J
2. A man uses a pulley system to lift a car of weight 2 000 N a height of 1 metre. He pulls
the rope a distance of 8 m. A force of 300 N is used. Calculate the efficiency of the
pulley system.
Solution
Useful work output = Fs = 2 000 N x 1 m = 2 000 J
Work input = Fs = 300 N x 8 = 2 400 J
Work output 2 000 J
∴ Efficiency % = x 100% = x 100% = 83%
Work input 2 400J
3. The following is a diagram showing a system of energy conversion in a moving car.

Calculate:
(a) The total energy lost from the system as the car move.
(b) The efficiency of the car engine.
Solution
(a) Total energy lost = chemical energy – kinetic energy
= (300 – 250)J
= 50 J
 useful energy output
(b) Efficiency = x 100%
Total energy input
250 J
= x 100%= 83.33%
300 J

4. The input energy of a motor is 800 J. calculate its efficiency if its output energy over the
same time interval is 680 J.
Solution
useful energy output
Efficiency = x 100%
Total energy input
680 J
= x 100%
800 J
= 85%
POWER
When we speak of power we mean how quickly work is done.
Power is the work done per second, or the rate at which work is done or
Definition the amount of energy transferred per second.

work done energy converted

Hence, power = =
time taken time taken
Equation W 𝐸 Where P: power
P= =
t t W: work done (in J)
E: energy converted (in J)
t: time taken (in s)

- SI unit of power: joule per second (J s-1) or watt (W)

- 1W=1
Relationship between power and velocity
Consider a body being moved at a steady rate by a force, F,
Work done = force x distance moved in the direction of force
work done
Power =
time taken
 force x distance moved in the direction of force
∴ Power =
time taken

distance moved in the direction

But, = velocity
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛

∴ Power = force x velocity at which point of application of force is moving

P = Fv. The velocity v may be uniform or average

Examples
1. When a 300 N force is applied to abox weighing 600 N, the box moves 3.0 m
horizontally in 20 s.

What is the average power?

A 45 W C 900 W
B 90 W D 1800 W
work done 300 N x 3 m
[Ans] A P= == = 45 W
time 20 s
2. A windmill is used to raise water from a well. The depth of the well is 5 m. the windmill
raises 200 kg of water every day. What is the useful power extracted from the wind?
(g= 10N/kg)
200 x 10 x 4 200 𝑥 10
A W C W
5 5 x 24
200 x 10 x 5 200 x 5
B W D = W
24 x 60 x 60 24 x 60
E
[Ans] B Power P =
t
Fxs
P=
t
𝟐𝟎𝟎 𝐱 𝟏𝟎 𝐱 𝟓
= W
𝟐𝟒 𝐱 𝟔𝟎 𝐱 𝟔𝟎
 Exercise
3. A man lifts 4 books of 2kg each through 5 stairways in 30s. If the stairs are of the same
length and the man’s mass is 50kg.

Determine the power of the man in climbing all the 5 stairs.

4. A student of mass 45kg runs up a flight of 40 steps in a stair case each 15 cm in 12

seconds. Find the power of the student.
5. A car engine developed 24kW while travelling along a level road. If there was a
resistance of 800 N due to friction, calculate the maximum speed attained.
6. A student of mass 60 kg runs up a flight of stairs of height 4.0m in a time of 3.0 s.
calculate assuming that the gravitational force on a mass of 1.00 kg is 10.0N,
(a) the student’s gain in potential energy,
(b) the useful power developed by the student in climbing the stairs.
7. The figure below shows a wind turbine used to produce electricity. The turbine blades
are turned by the wind and are connected to an electrical generator.

Fig. 7.1 Fig. 7.2

(a) Energy can be converted from one form into another.

(i) State the useful energy transformation that occur during the operation of a
wind turbine.
 (ii) Describe briefly why more energy is produced per second if
1. the wind blows faster,
2. the turbine blades are longer,
3. the turbine is more efficient.
(b) In one revolution the blades sweep out a circle, as shown in Fig. 7.2. in 60 s a
volume of 540 000 m3 of air travelling at a speed of 6.0 m/s is incident at right
angles on that circle. The density of air is 1.2 kg/m3.
Calculate
(i) the mass of air that passes through this circle in 60 s,
(ii) the initial kinetic energy of this mass of air,
(iii) the maximum input power available to the wind turbine.

8. The graph shows the variation with time of the speed of a car as it travels along a level
road. The car brakes when the time t=20 s, and comes to rest when t= 24 s.
The car has a mass of 800 kg and the forward driving force on the wheels is 1200 N.

(a) For the first 20 s of the motion shown in the graph

Calculate
(i) the distance travelled,
(ii) the work done by the driving force,
 (iii) the power supplied by the driving force.
(b) (i) Calculate the kinetic energy of the car while it is travelling at the constant top
speed.
(iii) When the car is travelling at a constant speed there is no change in its kinetic
energy.
1. Explain why the car engine is needed to maintain this constant speed.
2. Suggest what happens to the energy that is provided by the engine.
(c) During braking the speed of the car decreases uniformly. The engine no longer
provides a driving force.
(i) Calculate the deceleration of the car between t=20 s and t = 24 s.
(ii) Calculate the total braking force acting on the car during this period.
(iii) Explain why the power dissipated in the brakes to slow the car is greater at
the beginning of the breaking period than at the end.
9. The Figure below is a diagram showing a bowling ball as it rolls down an incline for 5.0
seconds before reaching the horizontal ground.

The changes in gravitational potential energy of the bowling ball as it rolls down to the
horizontal ground are shown in table below.
Time elapsed in seconds 0 1 2 3 4 5
Potential energy in joules 9 7.2 5.4 3.6 1.8 0

(a) State the law of conservation of energy.

(b) Using the values in the table draw a graph of potential energy against time.
(c) Calculate the height, h of the inclined plane, when the potential energy was 9J, if the
mass of the ball was 900g.
(d) Using the graph, estimate the kinetic energy of the bowling ball at time
(i) 2.5 seonds,
(ii) 4.0 seconds.