No topics match this filter.
Practise 27 exam-style questions on Power for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
An electric kettle transfers 180 000 J of energy in 60 s.
Calculate the power of the kettle.
P = E ÷ t
P = 180 000 ÷ 60
P = 3000 W
Power is the energy transferred divided by the time taken. With energy in joules and time in seconds, the answer is in watts; 3000 W is the same as 3 kW.
P = 180 000 ÷ 60
an answer of 3000 (W) scores 2 marks
3000 (W)
allow 3 kW
Kettle X has a power of 3.0 kW. Kettle Y has a power of 2.0 kW. Each kettle is used to boil the same mass of water, starting at the same temperature.
Explain which kettle boils the water faster. Assume both kettles waste the same amount of energy.
Kettle X boils the water faster. Both kettles must transfer the same amount of energy to the water, but kettle X has a higher power, so it transfers more energy every second and takes less time.
Time = energy ÷ power. For the same energy, a higher power means a shorter time: kettle X takes two-thirds of the time kettle Y takes.
kettle X
the same energy is needed, and X transfers energy at a faster rate / transfers more energy each second
allow t = E ÷ P, so a larger power gives a shorter time
A power station has a power output of 1.2 GW.
Calculate the energy transferred by the power station in one day.
Give your answer in standard form.
P = 1.2 GW = 1.2 × 10⁹ W
t = 24 × 60 × 60 = 86 400 s
E = P × t = 1.2 × 10⁹ × 86 400
E = 1.04 × 10¹⁴ J
Giga means 10⁹, and a day contains 24 × 60 × 60 = 86 400 seconds. Large answers like this are best written in standard form.
1.2 GW = 1.2 × 10⁹ W and 1 day = 86 400 s
both needed for this mark; an answer of 1.04 × 10¹⁴ (J) scores 3 marks
E = 1.2 × 10⁹ × 86 400
allow ecf from incorrect conversions
1.04 × 10¹⁴ (J)
allow 1.0 × 10¹⁴ (J); answer must be in standard form; allow ecf
Car A and car B each have a mass of 1000 kg. Starting from rest, each car accelerates to a speed of 20 m/s. Car A takes 8.0 s and car B takes 10 s.
Calculate the average useful power output of each car, and explain why they are different.
Kinetic energy gained by each car: Eₖ = 0.5 × 1000 × 20² = 200 000 J
Car A: P = E ÷ t = 200 000 ÷ 8.0 = 25 000 W
Car B: P = 200 000 ÷ 10 = 20 000 W
Both cars gain the same kinetic energy, but car A does it in less time, so its power output is greater.
The useful energy transferred is the same for both cars, so the difference in power comes only from the difference in time. A more powerful car can accelerate to the same speed more quickly.
Eₖ = 0.5 × 1000 × 20² = 200 000 (J)
car A: P = 200 000 ÷ 8.0 = 25 000 (W)
allow 25 kW; allow ecf
car B: P = 200 000 ÷ 10 = 20 000 (W)
allow 20 kW; allow ecf
both cars transfer the same energy (to the kinetic store) but car A does it in less time
Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.
Hi! Stuck? Ask me for a hint and I’ll help you work it out. Once you’ve checked your answer I can explain exactly where the marks are.
