IGCSE Physics Work Questions and Worked Solutions (1.7.2) | 0625

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IGCSE Physics 0625 · Topic 1.7.2

IGCSE Physics Work Questions and Worked Solutions

50 topic-wise practice questions on Work (Unit 1: Motion, Forces and Energy), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions40 Core0 Extended10 Challenge121 marksTextbook §1.7 · pp. 60–77
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Work

Work is done when a force moves an object through a distance. The work done equals the energy transferred, so work and energy share the same unit, the joule.

Key formulas

W = F dwork done W (J), force F (N), distance moved in direction of force d (m)Core
work done = energy transferred1 J = 1 N mCore

Prerequisite knowledge

  • Forces (1.5.1)
  • Energy stores and transfers (1.7.1)

Exam tips & common mistakes

  • Use the distance moved IN THE DIRECTION of the force
  • Convert km to m and kN to N
  • Holding a heavy object still does no work on it

Key facts from the syllabus (1.7.2)

  • Mechanical work is done by a force moving through a distance; electrical work is done by a current
  • If there is no movement in the direction of the force, no work is done
  • Work done against friction transfers energy to the internal (thermal) store
  • Work done lifting an object = weight × height = gain in gravitational potential energy
  • Braking work = kinetic energy lost

Textbook: §1.7 Energy, work and power, pp. 60–77

1
Core[1 mark]

State the relationship between work done and energy transferred.

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Method: They are equal.

Work done = energy transferred
2
Core[1 mark]

State the equation for mechanical work done.

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Method: W = Fd.

Work done = force × distance moved in the direction of the force
3
Core[2 marks]

A force of 40 N pushes a box 3.0 m along the floor. Calculate the work done.

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Method: W = Fd.

W = 40 × 3.0 = 120 J
4
Core[2 marks]

A crane lifts a 2000 N load 15 m. Calculate the work done on the load.

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Method: W = Fd.

W = 2000 × 15 = 30000 J
5
Core[2 marks]

110 J of work is done pulling a bow string back 0.45 m. Calculate the average force.

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Method: F = W ÷ d.

F = 110 ÷ 0.45 = 244 N
6
Core[2 marks]

A student does 600 J of work pushing a trolley with a force of 150 N. Calculate the distance moved.

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Method: d = W ÷ F.

d = 600 ÷ 150 = 4.00 m
7
Core[1 mark]

State the unit of work.

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Method: Same as energy.

joule (J)
8
Core[2 marks]

A person holds a heavy box still. Explain why no mechanical work is done on the box.

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Method: No distance moved.

The box does not move, so distance in the direction of the force is zero
9
Core[3 marks]

A 65 kg student climbs stairs of vertical height 3.5 m. Calculate the work done against gravity (g = 9.8 N/kg).

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Method: W = weight × height.

W = 65 × 9.8 × 3.5 = 2230 J
10
Core[2 marks]

Explain why a cyclist does work against friction when riding at constant speed.

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Method: Friction/drag act against motion over a distance.

The driving force moves through a distance against resistive forces, transferring energy to thermal stores
11
Core[2 marks]

A lift does 54 000 J of work raising people 12 m. Calculate the force it exerted.

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Method: F = W ÷ d.

F = 54000 ÷ 12 = 4500 N
12
Core[2 marks]

Calculate the work done when a 7.5 N force moves an object 80 cm.

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Method: Convert cm to m.

W = 7.5 × 0.80 = 6.00 J
13
Core[2 marks]

State the energy transfer when work is done pushing a block across a rough floor at constant speed.

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Method: Friction heats surfaces.

Energy is transferred to the internal (thermal) store of the block and floor
14
Core[3 marks]

A car engine provides a driving force of 3000 N over 2.5 km. Calculate the work done in MJ.

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Method: W = Fd with d = 2500 m.

W = 3000 × 2500 = 7500000 J = 7.50 MJ
15
Core[1 mark]

A force is perpendicular to the direction of motion. State the work done by this force.

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Method: No distance in the direction of the force.

Zero
16
Core[2 marks]

A weightlifter raises a 1500 N bar by 2.0 m. Calculate the energy transferred to the bar's gravitational potential store.

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Method: W = Fd = ΔE.

ΔE = 1500 × 2.0 = 3000 J
17
Core[2 marks]

Electrical work of 1800 J is done by a motor. State the total energy transferred by the motor.

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Method: Work done = energy transferred.

1800 J
18
Core[2 marks]

A 20 N force pulls a sledge 15 m, and friction is 20 N. Calculate the work done against friction.

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Method: W = Ffriction × d.

W = 20 × 15 = 300 J
19
Core[2 marks]

A bow is pulled back with an average force of 240 N through 0.45 m. Calculate the work done.

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Method: W = Fd.

W = 240 × 0.45 = 108 J
20
Core[3 marks]

A box is pushed 6.0 m with a force of 50 N, then 4.0 m with 30 N. Calculate the total work done.

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Method: Add the two amounts.

W = 300 + 120 = 420 J
21
Core[2 marks]

A 25 N force moves a sledge 12 m. Calculate the work done.

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Method: W = Fd.

W = 300 J
22
Core[2 marks]

A person does 2400 J of work pushing a car 8.0 m. Calculate the average force.

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Method: F = W ÷ d.

F = 300 N
23
Core[2 marks]

A lift motor does 36 000 J of work with a force of 6000 N. Calculate the height raised.

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Method: d = W ÷ F.

d = 6.00 m
24
Core[3 marks]

A 50 kg box is lifted 1.2 m onto a shelf (g = 9.8 N/kg). Calculate the work done.

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Method: W = mgh.

W = 50 × 9.8 × 1.2 = 588 J
25
Core[2 marks]

State what happens to the energy when work is done against friction.

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Method: Thermal.

It is transferred to the internal (thermal) store of the surfaces and surroundings
26
Core[2 marks]

A trolley is pulled with a force of 15 N for 4.0 m, then 10 N for 6.0 m. Calculate the total work done.

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Method: Sum.

W = 60 + 60 = 120 J
27
Core[2 marks]

Calculate the work done when a 0.50 N force moves through 40 cm.

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Method: Convert cm to m.

W = 0.50 × 0.40 = 0.200 J
28
Core[2 marks]

State the work done by gravity on a satellite moving in a circular orbit at constant height.

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Method: Force perpendicular to motion.

Zero
29
Core[2 marks]

A car brakes with a force of 6000 N and stops in 25 m. Calculate the work done by the brakes.

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Method: W = Fd.

W = 150000 J
30
Core[3 marks]

A car of mass 1200 kg moving at 20 m/s brakes to rest. Using energy, calculate the work done by the brakes.

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Method: Work = Ek lost.

W = ½ × 1200 × 400 = 240000 J
31
Core[3 marks]

For the car in the previous question, the braking force is 8000 N. Calculate the braking distance.

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Method: d = W ÷ F.

d = 240 000 ÷ 8000 = 30.0 m
32
Core[2 marks]

Explain why a cyclist going uphill at constant speed does more work than on the flat.

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Method: Against gravity too.

Work is done against gravity (increasing GPE) as well as against friction and air resistance
33
Core[2 marks]

State the name of the unit equal to 1 N m of work.

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Method: Joule.

joule (J)
34
Core[2 marks]

Express 1 joule in terms of newtons and metres.

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Method: W = Fd.

1 J = 1 N × 1 m
35
Core[2 marks]

A crane lifts a 12 kN load through 25 m. Calculate the work done in kJ.

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Method: W = Fd.

W = 12 × 25 = 300 kJ
36
Core[2 marks]

A child pulls a toy 5.0 m with a 4.0 N force; the toy speeds up. State whether all the work increases the toy's kinetic energy.

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Method: Friction.

No, some work is done against friction and becomes thermal energy
37
Core[3 marks]

A 60 kg climber climbs a 300 m hill (g = 9.8 N/kg). Calculate the work done against gravity in kJ.

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Method: W = mgh.

W = 60 × 9.8 × 300 = 176400 J ≈ 176 kJ
38
Core[2 marks]

A student pushes against a wall with a force of 200 N for 30 s. State the work done on the wall and explain.

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Method: No movement.

Zero, because the wall does not move
39
Core[3 marks]

A box is dragged 8.0 m across a floor with a horizontal force of 45 N at constant speed. State the friction force and calculate the work done against friction.

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Method: Constant speed: balanced.

Friction = 45 N; W = 45 × 8.0 = 360 J
40
Core[2 marks]

An electric motor lifts a 30 N weight through 2.0 m. Calculate the useful work done.

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Method: W = Fd.

W = 60 J

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

A man pushes a 40 kg crate 12 m up a ramp, raising it 3.0 m. He pushes with 140 N along the ramp. Calculate (a) work done by the man, (b) gain in GPE (g = 9.8 N/kg), (c) work done against friction.

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Method: W = Fd; Ep = mgh.

(a) 1680 J (b) 1180 J (c) 504 J
42
ChallengeCore[4 marks]

A car of mass 1000 kg moving at 20 m/s brakes to rest. The brakes do 160 000 J of work and the rest is done by air resistance and rolling friction. Calculate the work done by other resistive forces.

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Method: Ek lost = total work.

Ek = 200000 J; other = 40000 J
43
ChallengeCore[4 marks]

A crane lifts a 2.0 t load at constant speed through 18 m (g = 9.8 N/kg). The motor uses 600 kJ. Calculate the useful work and the efficiency.

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Method: W = mgh.

W = 2000 × 9.8 × 18 = 353000 J; efficiency = 58.8%
44
ChallengeCore[3 marks]

A box is pushed 5.0 m across a floor at constant speed against 60 N of friction, then lifted 1.2 m onto a table (weight 150 N). Calculate the total work done.

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Method: Add.

W = 300 + 180 = 480 J
45
ChallengeCore[4 marks]

A 70 kg cyclist (including bike) freewheels down a hill dropping 15 m and reaches the bottom at 12 m/s from rest. Calculate the work done against resistive forces (g = 9.8 N/kg).

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Method: Ep − Ek.

Ep = 10300 J; Ek = 5040 J; work against resistance = 5250 J
46
ChallengeCore[4 marks]

A force–distance graph for stretching a bow rises steadily from 0 N to 300 N over 0.50 m. Calculate the work done (area under the graph).

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Method: Area of triangle.

W = ½ × 0.50 × 300 = 75.0 J
47
ChallengeCore[3 marks]

Explain why a satellite in a circular orbit has no work done on it by gravity, although gravity acts on it all the time.

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Method: Force perpendicular to motion.

Gravity acts towards the centre, at right angles to the motion, so no distance is moved in the direction of the force and no work is done
48
ChallengeCore[4 marks]

A 60 kg student does 20 pull-ups, raising their body 0.45 m each time, in 1.0 minute. Calculate the total work done against gravity and the average power.

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Method: W = mgh × 20.

W = 60 × 9.8 × 0.45 × 20 = 5290 J; P = 88.2 W
49
ChallengeCore[3 marks]

A train's engine provides 50 kN for 2.0 km while the train travels at constant speed. State the total resistive force and calculate the work done against it.

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Method: Constant speed: balanced.

50 kN; W = 50 000 × 2000 = 1.0 × 10⁸ J
50
ChallengeCore[4 marks]

Two people move identical boxes onto a 1.0 m high truck. One lifts vertically; the other pushes up a frictionless 4.0 m ramp. Compare the work done and the force needed.

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Method: Same Ep gain.

Same work (equal to the gain in GPE); the ramp needs a quarter of the force but over four times the distance
All IGCSE Physics topics← 1.7.1 Energy1.7.3 Energy Resources →

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