IGCSE Physics Power Questions and Worked Solutions (1.7.4) | 0625

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IGCSE Physics 0625 · Topic 1.7.4

IGCSE Physics Power Questions and Worked Solutions

50 topic-wise practice questions on Power (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 Challenge130 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: Power

Power is the rate at which work is done or energy is transferred. A more powerful device transfers the same energy in less time.

Key formulas

P = W ÷ tpower P (W), work done W (J), time t (s)Core
P = ΔE ÷ tpower as energy transferred per secondCore
1 W = 1 J/s; 1 kW = 1000 WunitsCore

Prerequisite knowledge

  • Work done (1.7.2)
  • Energy (1.7.1)
  • Converting minutes and hours to seconds

Exam tips & common mistakes

  • Convert minutes to seconds before calculating
  • kJ ÷ s gives kW
  • Compare two people's power by work done ÷ time, not just speed

Key facts from the syllabus (1.7.4)

  • Power is work done per unit time (energy transferred per unit time)
  • A 60 W lamp transfers 60 J every second
  • For lifting: P = m g h ÷ t
  • For constant speed against a force: P = F v (from W/t = F d/t)

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

1
Core[1 mark]

Define power.

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Method: Rate of doing work.

Work done per unit time (energy transferred per unit time)
2
Core[1 mark]

State the unit of power.

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Method: 1 J/s.

watt (W)
3
Core[2 marks]

A motor does 4800 J of work in 12 s. Calculate its power.

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Method: P = W ÷ t.

P = 4800 ÷ 12 = 400 W
4
Core[2 marks]

A 2.0 kW heater runs for 30 minutes. Calculate the energy transferred in J.

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Method: E = Pt with t = 1800 s.

E = 2000 × 1800 = 3600000 J
5
Core[2 marks]

A lamp of power 60 W transfers 7200 J. Calculate the time it was on.

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Method: t = E ÷ P.

t = 7200 ÷ 60 = 120 s
6
Core[3 marks]

A 55 kg student runs up stairs 4.0 m high in 5.0 s. Calculate her power (g = 9.8 N/kg).

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Method: P = mgh ÷ t.

P = (55 × 9.8 × 4.0) ÷ 5.0 = 431 W
7
Core[3 marks]

A crane lifts a 5000 N load 20 m in 25 s. Calculate its useful power output.

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Method: P = Fd ÷ t.

P = (5000 × 20) ÷ 25 = 4000 W
8
Core[2 marks]

Two students climb the same stairs; one takes less time. Explain who has the greater power if they have equal weights.

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Method: Same work in less time.

The faster student: same work done in a shorter time
9
Core[2 marks]

Convert 3.5 kW and 0.25 MW into watts.

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Method: k = 10³, M = 10⁶.

3500 W and 250 000 W
10
Core[2 marks]

An electric motor transfers 36 kJ in 2.0 minutes. Calculate its power.

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Method: t = 120 s.

P = 36000 ÷ 120 = 300 W
11
Core[3 marks]

A car engine provides a driving force of 2500 N at a constant speed of 20 m/s. Calculate the useful power.

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Method: P = Fd/t = Fv.

P = 2500 × 20 = 50000 W = 50.0 kW
12
Core[2 marks]

A 1500 W microwave oven runs for 4.0 min. Calculate the energy transferred in kJ.

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Method: E = Pt.

E = 1500 × 240 = 360000 J = 360 kJ
13
Core[3 marks]

A pump lifts 300 kg of water through 8.0 m in 60 s. Calculate its useful power.

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Method: P = mgh ÷ t.

P = (300 × 9.8 × 8.0) ÷ 60 = 392 W
14
Core[2 marks]

State what is meant by a power of 100 W.

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Method: Joules per second.

100 J of energy are transferred every second
15
Core[2 marks]

A 40 W lamp and a 100 W lamp are on for the same time. Compare the energy transferred.

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Method: E = Pt.

The 100 W lamp transfers 2.5 times as much energy
16
Core[3 marks]

A lift motor has power 12 kW. Calculate how long it takes to do 300 kJ of work.

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Method: t = W ÷ P.

t = 300000 ÷ 12000 = 25.0 s
17
Core[2 marks]

A heater transfers 90 kJ in 1.5 minutes. Calculate its power in kW.

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Method: t = 90 s.

P = 90000 ÷ 90 = 1000 W = 1.0 kW
18
Core[3 marks]

A machine has input power 800 W and is 60% efficient. Calculate the useful work done in 10 s.

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Method: Useful power = 480 W.

Useful work = 480 × 10 = 4800 J
19
Core[2 marks]

A weightlifter lifts 1800 N through 2.0 m in 1.2 s. Calculate the power.

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Method: P = W ÷ t.

P = 3600 ÷ 1.2 = 3000 W
20
Core[2 marks]

Explain why a more powerful kettle boils water faster.

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Method: More energy per second.

It transfers more energy per second, so the required energy is supplied in less time
21
Core[2 marks]

A 2.4 kW kettle runs for 3.0 minutes. Calculate the energy transferred.

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Method: E = Pt.

E = 2400 × 180 = 432000 J
22
Core[2 marks]

A machine does 9000 J of work in 30 s. Calculate its power.

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Method: P = W/t.

P = 300 W
23
Core[2 marks]

A 750 W blender runs until 45 kJ has been transferred. Calculate the time.

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Method: t = E/P.

t = 45000 ÷ 750 = 60.0 s
24
Core[3 marks]

A 70 kg climber climbs 12 m up a rope in 20 s. Calculate the climber's useful power (g = 9.8 N/kg).

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Method: P = mgh/t.

P = 70 × 9.8 × 12 ÷ 20 = 412 W
25
Core[3 marks]

A motor lifts a 40 kg load 6.0 m in 8.0 s. Calculate its useful power output (g = 9.8 N/kg).

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Method: P = mgh/t.

P = 294 W
26
Core[2 marks]

Two motors lift identical loads to the same height. Motor A takes 4 s and motor B takes 6 s. State which has the greater power and why.

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Method: Same work, less time.

Motor A: same work in less time
27
Core[2 marks]

State the power of a device that transfers 1 kJ every second.

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Method: 1000 J/s.

1 kW (1000 W)
28
Core[3 marks]

A 1.2 kW heater and a 0.80 kW heater run for 2.0 hours. Calculate the total energy transferred in J.

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Method: E = Pt.

E = 2000 × 7200 = 1.44 × 10⁷ J
29
Core[2 marks]

A cyclist produces 250 W. Calculate the energy transferred in 1 hour.

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Method: E = Pt.

E = 250 × 3600 = 900000 J
30
Core[3 marks]

A car engine's useful power is 30 kW at a constant speed of 25 m/s. Calculate the driving force.

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Method: P = Fv.

F = 30000 ÷ 25 = 1200 N
31
Core[3 marks]

A 60 W bulb is 10% efficient. Calculate the useful light energy produced in 1.0 minute.

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Method: Useful power × time.

Useful = 6.0 × 60 = 360 J
32
Core[2 marks]

Express power in terms of energy and time and state its unit in base form.

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Method: P = E/t.

P = E ÷ t; 1 W = 1 J/s
33
Core[3 marks]

A water pump raises 2000 kg of water 15 m every minute. Calculate its useful power (g = 9.8 N/kg).

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Method: P = mgh/t.

P = 2000 × 9.8 × 15 ÷ 60 = 4900 W
34
Core[2 marks]

A student runs up stairs in 4.0 s doing 2000 J of work. Calculate her power.

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Method: P = W/t.

P = 500 W
35
Core[2 marks]

A microwave oven of power 800 W heats food for 90 s. Calculate the energy transferred.

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Method: E = Pt.

E = 72000 J
36
Core[3 marks]

An escalator lifts 20 people of average mass 65 kg through 8.0 m in 40 s. Calculate the useful power (g = 9.8 N/kg).

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Method: P = mgh/t.

P = 20 × 65 × 9.8 × 8.0 ÷ 40 = 2550 W
37
Core[2 marks]

A 5.0 kW electric car charger runs for 6.0 hours. Calculate the energy in MJ.

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Method: E = Pt.

E = 5000 × 21600 = 1.08 × 10⁸ J = 108 MJ
38
Core[3 marks]

A motor has input power 500 W and does 3600 J of useful work in 12 s. Calculate its useful power and efficiency.

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Method: P = W/t.

Useful power = 300 W; efficiency = 60.0%
39
Core[2 marks]

Explain why a high-power hairdryer dries hair faster than a low-power one.

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Method: More energy per second.

It transfers more energy per second to heat the air and evaporate water
40
Core[2 marks]

A 9.0 W phone charger is left on for 8.0 hours. Calculate the energy transferred in J.

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Method: E = Pt.

E = 9.0 × 28800 = 259000 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 65 kg student runs up 30 steps, each 0.18 m high, in 6.0 s (g = 9.8 N/kg). Calculate the useful power. A second student of 80 kg takes 8.0 s; who is more powerful?

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Method: P = mgh/t.

Student 1: 573 W; student 2: 529 W; student 1 is more powerful
42
ChallengeCore[4 marks]

A lift of mass 800 kg carries 6 people of 70 kg each up 30 m in 20 s at constant speed. The motor is 60% efficient. Calculate the input power (g = 9.8 N/kg).

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Method: Useful P = mgh/t.

Useful = 1220 × 9.8 × 30 ÷ 20 = 17900 W; input = 29900 W
43
ChallengeCore[4 marks]

A car engine provides a driving force of 1800 N at 30 m/s on a level road. Calculate the useful power, and the energy used in a 10-minute journey.

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Method: P = Fv.

P = 54000 W; E = 54 000 × 600 = 3.24 × 10⁷ J
44
ChallengeCore[4 marks]

A 2.0 kW kettle heats water for 3.5 minutes; a 3.0 kW kettle heats the same water in 2.5 minutes. Calculate the energy used by each and state which wastes less energy.

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Method: E = Pt.

420000 J and 450000 J; the 2.0 kW kettle used less energy for the same job
45
ChallengeCore[3 marks]

A pump lifts water at 12 kg per second through 25 m. Calculate the minimum power needed (g = 9.8 N/kg).

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Method: P = mgh/t.

P = 12 × 9.8 × 25 = 2940 W
46
ChallengeCore[4 marks]

A cyclist produces 300 W while riding at a steady 10 m/s. Calculate the total resistive force. If she doubles her power, explain why her speed does not double.

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Method: P = Fv.

F = 30.0 N; air resistance increases with speed, so more force is needed at higher speed
47
ChallengeCore[4 marks]

An escalator moves 30 people per minute, each of mass 70 kg, up a vertical height of 6.0 m. The motor's input power is 7.0 kW. Calculate the efficiency (g = 9.8 N/kg).

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Method: Useful P.

Useful = 30 × 70 × 9.8 × 6 ÷ 60 = 2060 W; efficiency = 29.4%
48
ChallengeCore[4 marks]

A crane can lift a 5000 N load at 0.40 m/s. Calculate its useful power and how long it takes to lift the load 18 m.

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Method: P = Fv.

P = 2000 W; t = 18 ÷ 0.40 = 45.0 s
49
ChallengeCore[3 marks]

Explain why a powerful sports car accelerates faster than a family car of the same mass.

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Method: Energy per second.

It can transfer more energy per second into the kinetic store, so its kinetic energy (and speed) increases faster
50
ChallengeCore[4 marks]

A heater of 1.5 kW is used to warm a room. The room loses energy at 900 W. Calculate the net energy gained by the room in 30 minutes and explain what will eventually happen to the room's temperature.

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Method: Net power.

Net = 600 × 1800 = 1080000 J; temperature rises until losses (which increase with temperature) equal 1.5 kW, then it stays constant
All IGCSE Physics topics← 1.7.3 Energy Resources1.8 Pressure →

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