OCR GCSE Physics A: Energy stores, transfers and energy calculations (J249) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018, topic P7.1 Work done (P7.1a to P7.1e). Every statement is assessed in both tiers, so no question here is labelled Higher Tier only. Use g = 9.8 N/kg where needed.

Links: study guide, revision notes, OCR GCSE Physics hub, printable checklist and free diagnostics.

Questions

1. State the law of conservation of energy for a closed system, and explain what happens to the energy when a bouncing ball finally comes to rest. [3]

2. For each situation, describe which energy stores change.

(a) A cyclist brakes and comes to a stop. [2] (b) An electric kettle brings water to the boil. [2]

3. A runner of mass 62 kg moves at 7.5 m/s.

(a) Calculate the runner’s kinetic energy. Give your answer to 2 significant figures. [2] (b) The runner speeds up to 9.0 m/s. Calculate the increase in kinetic energy, to 2 significant figures. [2]

4. A parcel of mass 3.5 kg is lifted from the floor onto a shelf 1.8 m high.

(a) Calculate how much energy is transferred to the parcel’s gravitational store. [2] (b) The parcel falls off the shelf. Ignoring air resistance, find how fast it is moving as it reaches the floor. [3]

5. A spring of spring constant 120 N/m is compressed by 0.060 m.

(a) Calculate the energy stored in the spring. [2] (b) The spring is released and all this energy is transferred to a dart of mass 0.020 kg. Calculate the speed of the dart. [3]

6. A block of iron of mass 0.75 kg (specific heat capacity 450 J/kg °C) is heated from 15 °C to 95 °C.

(a) Calculate the energy transferred to the block. [2] (b) The heater has a power of 150 W and all its energy goes to the block. Calculate the time taken. [2] (c) Explain why the time in a real experiment would be longer. [1]

7. A 9.0 V supply drives a current of 0.50 A through a motor for 40 s. The motor lifts a 0.80 kg load through 5.0 m in this time.

(a) Find the total charge that passes through the motor. [2] (b) Calculate the energy transferred by the supply. [2] (c) Calculate the gravitational potential energy gained by the load and state how much energy goes to other stores. [3]

8. An immersion heater has a power of 3.0 kW and is switched on for 40 minutes. Electricity costs 30 p per kW h.

(a) Calculate the energy transferred in kW h. [2] (b) Calculate the cost. [1] (c) Convert your answer to (a) into joules. [2]

9. A ball of mass 0.060 kg is dropped from a height of 2.0 m and rebounds to 1.3 m.

(a) Calculate the energy dissipated during the bounce. [2] (b) State where this energy goes. [1] (c) Ignoring air resistance, find the ball’s speed on reaching the floor. [2] (d) Suggest one way to make the measurement of the rebound height more reliable. [1]

10. A car of mass 1200 kg travels at 25 m/s and brakes to a stop.

(a) Calculate the kinetic energy of the car. [2] (b) Describe the changes in energy stores as the car stops. [2] (c) The braking force is a constant 7500 N. Calculate the braking distance. [2] (d) The car now travels at 50 m/s and brakes with the same force. Use calculation to show how the braking distance changes. [2]

11. A crate of mass 40 kg is pushed up a ramp 6.0 m long by a constant force of 120 N acting along the ramp. The crate rises 1.5 m vertically.

(a) Calculate the work done by the pushing force. [2] (b) Find how much the crate’s gravitational potential energy increases. [2] (c) Calculate the energy transferred to the thermal store and explain why it is not zero. [3] (d) The crate is released at the top and slides back down. Suppose all its gravitational potential energy became kinetic energy. Calculate its speed at the bottom. [3]

Answers

1. Energy is never created or destroyed; it is only moved from one store to another [1]. In a closed system the total energy does not change [1]. The ball’s kinetic and gravitational energy are transferred to the thermal stores of the ball, floor and air (and sound), so the energy is spread out, not lost [1]. [3] Examiner insight: An answer that says energy is “lost” or “used up” contradicts the law. Naming the thermal store of the surroundings is the point that separates a full answer from a partial one.

2. (a) The kinetic store of the cyclist and bike decreases [1]. The thermal store of the brakes, wheels and surroundings increases [1]. [2] (b) Work is done by the current, so energy is transferred from the electrical supply [1] to the thermal store of the water (and some to the kettle and air) [1]. [2] Examiner insight: Name a store that goes down and a store that goes up. Writing only “kinetic energy turns into heat” without saying what is heated is easy to under-credit.

3. (a) E = ½mv² = ½ × 62 × 7.5² [1] = 1743.75 J = 1700 J (2 s.f.) [1]. [2] (b) E = ½ × 62 × 9.0² = 2511 J [1]. Increase = 2511 − 1744 = 767 J = 770 J (2 s.f.) [1]. [2] Examiner insight: Write out your substitution, not just the final value. Squaring only the 7.5 and not the whole speed is the usual slip, so write v² clearly.

4. (a) E = mgh = 3.5 × 9.8 × 1.8 [1] = 61.74 J = 62 J (2 s.f.) [1]. [2] (b) The gravitational potential energy becomes kinetic energy: ½mv² = 61.74 [1]. v² = 2 × 61.74 / 3.5 = 35.28 [1]. v = 5.94 = 5.9 m/s [1]. [3] Examiner insight: State the energy link (gravitational potential becomes kinetic) in words. It shows your method even if a calculation slip follows.

5. (a) E = ½kx² = ½ × 120 × 0.060² [1] = 0.216 J [1]. [2] (b) ½mv² = 0.216 [1]. v² = 2 × 0.216 / 0.020 = 21.6 [1]. v = 4.65 = 4.6 m/s (2 s.f.) [1]. [3] Examiner insight: Convert the extension into metres before squaring. Using 6.0 for centimetres would give an energy 10 000 times too large.

6. (a) ΔE = mcΔθ = 0.75 × 450 × (95 − 15) [1] = 27 000 J [1]. [2] (b) t = E / P = 27 000 / 150 [1] = 180 s [1]. [2] (c) Some energy heats the surroundings (or the heater itself) rather than the block [1]. [1] Examiner insight: The temperature change is a difference (80 °C), not 95 °C. For the explanation, say where the extra energy goes, not just “heat is lost”.

7. (a) Q = It = 0.50 × 40 [1] = 20 C [1]. [2] (b) E = QV = 20 × 9.0 [1] = 180 J [1]. [2] (c) E = mgh = 0.80 × 9.8 × 5.0 [1] = 39.2 J [1]. Energy to other stores = 180 − 39.2 = 140.8 J (about 140 J), mostly the thermal store of the motor and surroundings [1]. [3] Examiner insight: Show how each part feeds the next. Using your own answer from (b) in (c) is good practice, and writing the equation each time makes your method easy to follow.

8. (a) t = 40 / 60 = 0.667 h [1]. E = Pt = 3.0 × 0.667 = 2.0 kW h [1]. [2] (b) Cost = 2.0 × 30 = 60 p [1]. [1] (c) 1 kW h = 3 600 000 J [1], so E = 2.0 × 3 600 000 = 7 200 000 J (7.2 × 10⁶ J) [1]. [2] Examiner insight: Keep kW with hours. Using 40 for the time with kW gives a figure 60 times too large, and mixing minutes into the equation is a common slip.

9. (a) Energy dissipated = loss in gravitational potential energy = mg × change in height = 0.060 × 9.8 × (2.0 − 1.3) [1] = 0.4116 J = 0.41 J [1]. [2] (b) It goes to the thermal stores of the ball, the floor and the air, with some as sound [1]. [1] (c) mgh = ½mv², so v = √(2gh) = √(2 × 9.8 × 2.0) [1] = 6.26 = 6.3 m/s [1]. [2] (d) Repeat the drop several times and calculate a mean rebound height (or read the height at eye level against a fixed rule) [1]. [1] Examiner insight: For (d), give a specific action that improves reliability. “Be more careful” does not say what you would do.

10. (a) E = ½mv² = ½ × 1200 × 25² [1] = 375 000 J [1]. [2] (b) The kinetic store decreases [1]. The thermal store of the brakes, tyres and road increases by the same amount [1]. [2] (c) Work done = Fs, so s = 375 000 / 7500 [1] = 50 m [1]. [2] (d) At 50 m/s, E = ½ × 1200 × 50² = 1 500 000 J [1]. s = 1 500 000 / 7500 = 200 m, four times as far [1]. [2] Examiner insight: A “show the effect” question wants figures from a calculation. Saying “it takes longer to stop” without the 200 m comparison is not a calculation.

11. (a) W = Fs = 120 × 6.0 [1] = 720 J [1]. [2] (b) E = mgh = 40 × 9.8 × 1.5 [1] = 588 J [1]. [2] (c) Energy to thermal store = 720 − 588 = 132 J [1]. It is not zero because friction acts between the crate and the ramp [1], so some of the work done heats the crate and ramp [1]. [3] (d) ½mv² = 588 [1], v² = 2 × 588 / 40 = 29.4 [1], v = 5.42 = 5.4 m/s [1]. [3] Examiner insight: In (b) use the vertical height of 1.5 m, not the 6.0 m slope length. Using the slope length is the most common error here.

Where marks are usually lost

• Writing that energy is “used up” or “lost” instead of naming the store it moves to.
• Saying “heat” or “energy” with no named store.
• Not squaring the speed or extension, or squaring only part of the product.
• Leaving extension in centimetres or mass in grams.
• Putting the length along a ramp into mgh in place of the vertical rise.
• Mixing kW and seconds, or W and hours, in E = Pt.
• Giving the final temperature rather than the temperature change in mcΔθ.
• Rounding in the middle of a calculation, or giving too many significant figures at the end.
• Leaving out units, or giving joules when kW h is asked for.

Next steps

• Check your recall with the revision notes.
• Re-read the full explanations in the study guide.
• Open the OCR GCSE Physics hub and the printable checklist.
• Find weak topics with the free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

This page is based on the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018, published by OCR (Oxford Cambridge and RSA). Topic P7.1 Work done.