GGebook smalllogo
Cambridge IGCSE
white cross



1.7b Work and power

Work and power are common words with many different meanings. We might describe a weightlifter as being very powerful. If you spend the day digging or moving boxes, you feel tired - you have 'worked hard'. In science, both work and power are quantities that can be measured. In this section you will learn about the difference between these two often confused terms.

Figure 1. An Olympic weightlifter
Svetlana Podobedova
| CC 2.0



When energy is transferred from one energy store to another, work is done. (See section 4.1). Therefore work is simply the quantity of energy that has been transferred. Work is therefore also measured in joules, just like energy.

For example, these electric cells (figure 2) contain chemicals that store energy. Typical cells like these might contain 10 000 J, or 10 kJ. However they are not connected in a circuit yet, so the energy stays inside the cell. Once connected, the chemical energy store is converted to electricity. The cells are now 'doing work' because energy is being transferred from one place to another.

In the last section, a ball is thrown upwards, converting kinetic energy to gravitational energy. initially, this energy came from stored chemical energy in the thrower's muscles. Therefore the person throwing the ball has done work transferring energy from one form to another.

Figure 2. Electric cells

To calculate the work done on a system, you simply need to find how much energy has been transferred. For transfers between gravitational energy and kinetic energy, you can do this by using formulas from section 4.3.

However, if mechanical work is done, there is a simple way of calculating the energy transferred:

A tractor driving in a muddy field

Figure 3. A tractor driving through mud.

Figure 3 shows a tractor driving through thick mud. It is hard work, because the mud provides a large frictional force. The tractor has to provide a forward force equal to the friction. The bigger the force required to counter the friction from the mud, the harder work it is. Also, more work needs to be done if the tractor travels a larger distance across the field.

There is a standard maths formula for calculating the work done in this type of situation, using the force applied and the distance travelled:

work done (change in energy) = force x distance moved (in direction of force)

W = ΔE = Fd

[joules] = [newtons] x [m]

Learn this formula!


The tractor shown in figure 3 produces a force of 20 kN and moves 500 m. Calculate the work done.

The force of 20 kN needs to be given in newtons, which is 20 000 N. Using the formula above, the work done is therefore:
W = F x d
W = 20 000 x 500
So W = 10 000 000 joules, or 10 MJ

Note that if the force of friction or resistance is given in a question, it can be assumed that the forward force matches this so that the object moves forward at a constant speed. As long as the force given is parallel to the distance moved, the equation above can be used.

If you are walking up hill, then you are doing work against gravity. The downwards force is your weight, and you need to produce an equal force to move upward at a constant speed. For these sorts of questions (see question 3 below), first work out the weight of the object and then use this as the force in the above equation.


1. A dolphin swims for 1 km in water, producing a forward force of 800 N to overcome the resistance from the water. Calculate the work done by the dolphin.

We know W = F x d, and 1 km is equal to 1000 m. Therefore:
W = 800 N x 1000 m
W= 800 000 J, or 800 kJ

2. A small toy boat contains a battery storing 2 kJ of energy. When moving at top speed, the resistance from the water is 12 N. Assuming the boat motor is 100 % efficient, how far could the boat travel before the batteries run out?

We know that W = F x d, and the work done will be 2 000 J assuming all the battery energy is transferred. So:
2 000 = 12 x d
rearranging the equation gives:

d = 2000

giving d = 167 m (to 3 sig. figs.)

3. An 8 kg drone is moving upwards at a constant speed to take pictures of a new building from above. It travels 600 m upwards.

a) The weight of the drone is given by:
weight = m g, and g =9.8 N/kg on Earth (see section 1.5)
So weight = 8 x 9.8
Weight = 78.4 N

b) Work done = F d,
So Work = 78.4 x 600
Work done = 47 040 J (47.0 kJ to 3 sig. figs.)

You could also solve this question by finding the gain in G.P.E. of the drone, as the work done has transferred energy from the drone batteries into G.P.E. The gain in G.P.E. is:
G.P.E. = m g h
G.P.E. = 8 x 9.8 x 600
So G.P.E. = work done = 47.0 kJ

* Take care on this type of question - both work and weight are given the letter 'W'. In our solution, we have used words, not symbols, to identify which is which!



Question 3 above considers the work done by a drone. Imagine you work for the company needing the photographs of the building site. Two drones are available for hire. The data for each drone is given in figure 4 below:

Comparison of two drones

Figure 4. Which drone is best?

Which of the drones here would you use? Which one does the most work getting to a height of 600m, and so will use the most energy from the batteries?

The answer is that both drones do the same work. They both have the same weight so are working against the same downward force, and they both need to travel the same distance. However, drone B does this work in less time. We say drone B is more powerful than drone A.
The power output of each drone is the work done per second, and is measured in watts (W). One watt of power means one joule of work has been done each second. To calculate this we need to do the following maths:

Power (watts) = work done (or energy transferred) (J)
time taken (s)
P = W (or ΔE)
Learn this formula!

Drone B shown in figure 4 does 47 000 J of work climbing 600m. It does this in 60 seconds. Calculate the power output of drone B.


p = W and therefore:- P = 47 000
t 60

So P = 783 W. (The power output of drone A is lower at 588 watts.)

Note: How is the work done of 47 000 J calculated?
See question 3 above for how this is worked out.

Power can be defined in two ways:

When we use 'the rate of' in this context, we mean how much per unit time (per second). We already know that work is done when energy is transferred, so we can replace 'work' with 'energy transfer'.

Have a go at the following practice questions to check you understand this section:


4. A phone battery needs 28 kJ of electrical energy to be transferred to fully charge the phone. If the charger supplies 4 W of electrical power, how long will it take to charge the phone?

We know that:

p = W
4 = 28 000

rearranging this equation gives:

t = 28 000

t = 7 000 seconds (117 minutes or 1 hr 57 mins)

5. Amy is timed running up some stairs. She weighs 550 N, and takes 12.5 seconds to complete the climb. The stairs have a vertical height of 9 m.


a) Amy's weight is the downward force of gravity (F) on her. The work done is given by:
W = F d
So W = 550 x 9
W= 4 950 J

b) Her power output is given by:

p = W and therefore:- P = 4 950
t 12.5

Her power output is P = 396 W
(Note that this is her average power, as she may have been faster at the bottom of the climb, and slower at the top).






Please choose a tier of entry

Extended Tier (Core and Supplementary content, Grades A* to G)
Core Tier (Core content only, grades C to G)
Remember my choice