Practise Work done and energy transfer. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
When a force causes an object to move through a distance, work is done on the object and energy is transferred. You must recall and use W = F s, convert between joules and newton-metres, and explain why doing work against friction makes objects warm up.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
State that work done is energy transferredWhen a force moves an object through a distance, energy is transferred and work is done.
4
Calculate work done using W = F sFor example, 50 N × 3.0 m = 150 J.
5
Rearrange W = F s for force or distanceF = W ÷ s and s = W ÷ F.
5
Convert between joules and newton-metres1 J = 1 N m, so 250 N m of work is 250 J.
6
Explain heating by work done against frictionWork done against friction transfers energy to the thermal store, so the temperature of the object rises.
7
Link work done to changes in energy storesWork done by brakes equals the kinetic energy lost; work done lifting an object equals its gain in gravitational potential energy.
Notes
Doing work
Work is done when a force causes an object to move through a distance.
Work done = energy transferred. Both are measured in joules (J).
work done = force × distance: W = F s
W in joules (J), F in newtons (N), s in metres (m). The distance must be moved along the line of action of the force.
If the object does not move, no work is done, however hard you push.
Joules and newton-metres
One joule of work is done when a force of one newton causes a displacement of one metre.
So 1 joule = 1 newton-metre (1 J = 1 N m). A value in N m is the same number in J.
Convert before calculating: kJ to J (× 1000), kN to N (× 1000), cm to m (÷ 100).
Work done against friction
When an object slides against friction, or moves through air, work is done against the frictional force.
This transfers energy to the thermal energy store of the object (and its surroundings), so its temperature rises.
Examples: rubbing your hands together warms them; brake pads get hot when a car slows down; a sliding box warms the floor slightly.
Work and energy stores grade 7+
Lifting an object at a steady speed: the force needed equals its weight, so the work done = m g h, which equals its gain in gravitational potential energy.
Braking: the work done by the braking force equals the kinetic energy the vehicle loses.
Cheatsheet
W = F s (work done = force × distance moved along the line of action of the force)
Units: W in J, F in N, s in m
1 J = 1 N m
Work done = energy transferred
Work done against friction transfers energy to the thermal store, so the temperature rises
No movement along the line of the force means no work is done
Work done lifting at steady speed = weight × height = m g h = gain in gravitational potential energy grade 7+
How to answer each type of question
Calculate work done
2 marks4
Check the force is in N and the distance in m.
Write W = F s and substitute.
Give the answer in J.
Example. A student pushes a box 4.5 m across a floor with a constant horizontal force of 120 N. Calculate the work done by the student.
Show the model answer
W = 120 × 4.5 (1) W = 540 J (1)
Rearrange to find force or distance
3 marks5
Convert units first (kJ to J).
Substitute into W = F s.
Rearrange and calculate, with a unit.
Example. A crane does 36 kJ of work lifting a load a vertical height of 15 m at a steady speed. Calculate the force the crane exerts on the load.
Show the model answer
36 kJ = 36 000 J (1) 36 000 = F × 15 (1) F = 2400 N (1)
Explain heating caused by friction
2 to 3 marks6
Identify the friction.
Say that work is done against friction.
Say that energy is transferred to the thermal store, so the temperature rises.
Example. A child slides down a playground slide. Afterwards, the surface of the slide and the child's clothes are warmer. Explain why.
Show the model answer
There is friction between the child and the slide (1). Work is done against the friction (1) so energy is transferred to the thermal energy store of the slide and the clothes, and their temperature increases (1).
Multi-step: work done and kinetic energy
4 marks7
Calculate the energy in the relevant store (e.g. Ek = ½ m v2).
Set the work done equal to the energy transferred.
Use W = F s to find the unknown.
Example. A car of mass 900 kg is travelling at 20 m/s. The driver brakes and the car stops. The average braking force is 6000 N. Calculate the braking distance.
Show the model answer
Ek = 0.5 × 900 × 202 (1) Ek = 180 000 J (1) work done by the brakes = kinetic energy lost, so 180 000 = 6000 × s (1) s = 30 m (1)
Shortcuts and memory tricks
Work done = energy transferred, so many energy questions are really 'work done' questions in disguise.
N × m = N m = J. If you get N m in a work question, you can write J.
Formula triangle: W on top, F and s underneath.
Three different W's: W for work done (J), W for weight (N), and W for watts (the unit of power). Read carefully.
Where marks are lost
Using a distance in cm, or a force in kN, without converting.
Using the wrong distance: it must be the distance moved in the direction of the force (e.g. the vertical height when lifting).
Saying friction 'creates heat energy'. Say that work done against friction transfers energy to the thermal store and the temperature rises.
Confusing work done (in J) with power (in W). Power is the rate of doing work.
Exam technique
W = F s must be recalled: it is not on the equations sheet.
Show the equation, the substitution and the answer with a unit. Each can earn a mark.
In 'explain' questions about warming, use the phrase 'work done against friction'.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
A removal worker pushes a box 6.0 m along a corridor with a horizontal force of 150 N. Calculate the work done by the worker. Use the equation: work done = force × distance
900 J
Write down the equation that links distance (s), force (F) and work done (W).
W = F s
The lift took 1.4 s. Calculate the average useful power output of the weightlifter during the lift.
2100 W (2058 W) W
Write down the equation that links distance moved along the line of action of the force (s), force (F) and work done (W).
W = F s
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy5 marks
(a) A removal worker pushes a box 6.0 m along a corridor with a horizontal force of 150 N. Calculate the work done by the worker. Use the equation: work done = force × distance[2]
(b) Which unit is equivalent to the joule? Tick (✓) one box.[1]
N/m
N m
N/m2
kg m/s
(c) Friction acts between the box and the floor. What happens to the temperature of the box and the floor as the box is pushed?[1]
(d) Complete the sentence. When a force causes an object to move through a distance, ............ is done on the object.[1]
Show the answer and mark scheme
(a)Answer: 900 J
W = 150 × 6.0
900 (J)
(b)Answer: N m
(c)Answer: It increases.
it increases / they get warmer
(d)Answer: work
work
Question 2Medium7 marks
A crane lifts a steel beam of mass 850 kg vertically upwards through a height of 24 m at a constant speed. gravitational field strength = 9.8 N/kg
(a) Calculate the weight of the beam.[2]
(b) Calculate the work done by the crane on the beam. Give your answer in kilojoules.[3]
(c) Describe the energy transfer that takes place as the crane lifts the beam.[2]
Show the answer and mark scheme
(a)Answer: 8330 N
W = 850 × 9.8
8330 (N)
(b)Answer: 200 kJ (199.92 kJ) kJ
W = 8330 × 24
199 920 (J)
200 (kJ) / 199.92 (kJ)
(c)Answer: The crane does work on the beam, transferring energy to the beam’s gravitational potential energy store.
work is done on the beam by the (tension in the) cable / crane
energy is transferred to the gravitational potential energy store of the beam
Question 3Hard7 marks
A shop assistant lifts a box of mass 8.0 kg from the floor onto a shelf 1.5 m above the floor. gravitational field strength = 9.8 N/kg
(a) Calculate the work done on the box as it is lifted onto the shelf.[3]
(b) The assistant then carries the box 12 m across the shop, keeping it at the same height and moving at a constant speed. A student says: ‘The assistant does 78.4 × 12 = 941 J of work on the box while carrying it.’ Explain why the student is wrong.[2]
(c) The assistant then pushes a trolley 25 m across the shop. The assistant pushes on the handle at an angle to the horizontal. The horizontal component of the push is 48 N. Calculate the work done by the push.[2]
Show the answer and mark scheme
(a)Answer: 118 J (117.6 J) J
weight = 8.0 × 9.8 = 78.4 (N)
W = 78.4 × 1.5
118 (J) / 117.6 (J)
(b)Answer: The upward force is perpendicular to the horizontal motion, so the box moves no distance along the line of action of the force and no work is done by it.
the upward force on the box is at right angles to its (horizontal) movement / the box does not move in the direction of the upward force
so the distance moved along the line of action of the force is zero / no work is done by this force / the gravitational potential energy of the box does not change