IGCSE Physics 0625 · Topic 1.7.1

IGCSE Physics Energy Questions and Worked Solutions

50 topic-wise practice questions on Energy (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 questions18 Core22 Extended10 Challenge139 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: Energy

Energy is stored in different ways and transferred between stores. It is never created or destroyed. This topic covers energy stores, transfer pathways, Sankey diagrams and (Extended) calculations with kinetic and gravitational potential energy.

Key formulas

Ek = ½ m v²	kinetic energy (J), mass (kg), speed (m/s)	Extended
ΔEp = m g Δh	change in gravitational potential energy (J)	Extended
total energy in = total energy out	conservation of energy	Core

Prerequisite knowledge

Mass and weight (1.3)
Speed (1.2)
Squaring and square roots

Exam tips & common mistakes

Square the speed BEFORE multiplying in ½mv²
Mass cancels in mgh = ½mv², so speed after falling does not depend on mass
Name the store AND the pathway in transfer questions

Key facts from the syllabus (1.7.1)

Energy stores: kinetic, gravitational potential, chemical, elastic (strain), nuclear, electrostatic, internal (thermal)
Energy is transferred by forces (mechanical work), electrical currents (electrical work), heating, and by waves
Energy cannot be created or destroyed, only transferred between stores (conservation of energy)
Sankey diagrams show energy transfers; arrow widths are proportional to the energy
Wasted energy usually ends up in the internal (thermal) store of the surroundings
Extended for a falling object with no air resistance, loss of Ep = gain in Ek

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

Core[2 marks]

Name four energy stores.

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Method: From the syllabus list.

Any four: kinetic, gravitational potential, chemical, elastic (strain), nuclear, electrostatic, internal (thermal)

Core[1 mark]

State the energy store that increases when a spring is stretched.

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Method: Elastic store.

Elastic (strain) store

Core[2 marks]

A ball is thrown upwards. Describe the energy transfer as it rises.

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Method: KE decreases, GPE increases.

Kinetic store → gravitational potential store

Core[2 marks]

State four ways energy can be transferred between stores.

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Method: Mechanical work, electrical work, heating, waves.

By forces (mechanical work), electric currents (electrical work), heating, and waves (electromagnetic/sound)

Core[1 mark]

State the principle of conservation of energy.

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Method: Energy cannot be created or destroyed.

Energy cannot be created or destroyed, only transferred between stores

Core[2 marks]

Describe the energy transfers when a battery-powered torch is switched on.

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Method: Chemical → electrical work → light and thermal.

Chemical store in the battery → electrical work → light (and internal/thermal store of surroundings)

Core[2 marks]

Describe the useful and wasted energy outputs of an electric motor.

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Method: Kinetic useful; heat and sound wasted.

Useful: kinetic; wasted: internal (thermal) and sound

Core[3 marks]

A device receives 500 J of energy and transfers 350 J usefully. Calculate the energy wasted.

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Method: Conservation of energy.

Wasted = 500 − 350 = 150 J

Core[2 marks]

Describe the energy transfers in a car braking to a stop.

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Method: Kinetic → internal store of brakes.

Kinetic store → internal (thermal) store of brakes and surroundings by friction

Core[2 marks]

Describe the energy transfers when a person climbs a flight of stairs.

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Method: Chemical → GPE + thermal.

Chemical store (food) → gravitational potential store, with some to the internal (thermal) store

Core[2 marks]

Describe the energy transfers when a bow fires an arrow.

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Method: Elastic → kinetic.

Elastic store of the bow → kinetic store of the arrow

Core[2 marks]

Describe the energy transfers in a hydroelectric power station.

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Method: GPE → kinetic → electrical.

Gravitational potential store of water → kinetic store → electrical work by the generator

Core[2 marks]

Describe the energy transfers when a falling ball hits the ground and stops.

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Method: Kinetic → thermal and sound.

Kinetic store → internal (thermal) store of ball and ground, and sound

Core[2 marks]

A device has an input of 1200 J and wastes 900 J. Calculate the useful output.

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Method: Conservation.

Useful = 1200 − 900 = 300 J

Core[2 marks]

State the energy store in a charged battery.

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Method: Chemical.

Chemical store

Core[2 marks]

State the energy store in a cup of hot tea.

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Method: Internal.

Internal (thermal) store

Core[2 marks]

Describe the energy transfers in a solar-powered calculator.

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Method: Light → electrical.

Light (electromagnetic waves) → electrical work → light and thermal from the display

Core[2 marks]

Explain why energy is always 'wasted' in real devices.

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Method: Dissipated heat.

Some energy is always transferred to the internal store of the surroundings (e.g. by friction or electrical resistance), which is not useful

Extended[2 marks]

Calculate the kinetic energy of a 0.15 kg ball moving at 20 m/s.

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Method: Ek = ½mv².

Ek = ½ × 0.15 × 20² = 30.0 J

Extended[2 marks]

Calculate the kinetic energy of a 1200 kg car moving at 25 m/s.

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Method: Ek = ½mv².

Ek = ½ × 1200 × 25² = 375000 J

Extended[3 marks]

A 2.0 kg object has 64 J of kinetic energy. Calculate its speed.

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Method: v = √(2Ek/m).

v = √(2 × 64 ÷ 2.0) = 8.00 m/s

Extended[2 marks]

Calculate the gain in gravitational potential energy when a 50 kg student climbs 4.0 m (g = 9.8 N/kg).

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Method: ΔEp = mgΔh.

ΔEp = 50 × 9.8 × 4.0 = 1960 J

Extended[3 marks]

A 0.50 kg ball is dropped from 5.0 m. Calculate its speed just before it hits the ground, ignoring air resistance.

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Method: mgh = ½mv².

v = √(2 × 9.8 × 5.0) = 9.90 m/s

Extended[3 marks]

A ball is thrown upwards at 12 m/s. Calculate the maximum height it reaches (no air resistance).

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Method: h = v² ÷ 2g.

h = 144 ÷ 19.6 = 7.35 m

Extended[2 marks]

The speed of a car doubles. State the factor by which its kinetic energy changes.

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Method: Ek ∝ v².

It increases by a factor of 4

Extended[4 marks]

A 60 kg skier starts from rest at the top of a slope 20 m high and reaches the bottom at 15 m/s. Calculate the energy wasted.

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Method: Wasted = ΔEp − Ek.

ΔEp = 11760 J; Ek = 6750 J; wasted = 5010 J

Extended[3 marks]

A Sankey diagram shows 1000 J input to a lamp, with 150 J as light. Calculate the thermal energy output and describe how the arrows would look.

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Method: Total output = input.

Thermal = 850 J; a wide input arrow splitting into a thin light arrow and a wide thermal arrow

Extended[4 marks]

A pendulum bob of mass 0.20 kg is released from 0.30 m above its lowest point. Calculate its maximum speed.

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Method: All Ep → Ek at the lowest point.

v = √(2 × 9.8 × 0.30) = 2.42 m/s

Extended[3 marks]

A 0.030 kg arrow is given 110 J of energy. Calculate its initial speed.

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Method: v = √(2Ek/m).

v = √(2 × 110 ÷ 0.030) = 85.6 m/s

Extended[2 marks]

Calculate the kinetic energy of a 70 kg runner at 8.0 m/s.

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Method: Ek = ½mv².

Ek = ½ × 70 × 64 = 2240 J

Extended[2 marks]

Calculate the gain in GPE when a 12 kg box is lifted 1.5 m (g = 9.8 N/kg).

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Method: mgh.

ΔEp = 12 × 9.8 × 1.5 = 176 J

Extended[3 marks]

A 0.40 kg ball has 20 J of kinetic energy. Calculate its speed.

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Method: v = √(2Ek/m).

v = √(100) = 10.0 m/s

Extended[3 marks]

A 900 kg roller-coaster car is at rest at the top of a 45 m drop. Calculate its maximum possible speed at the bottom.

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Method: v = √(2gh).

v = √(2 × 9.8 × 45) = 29.7 m/s

Extended[3 marks]

A 2.0 kg object gains 196 J of GPE. Calculate the height it was raised (g = 9.8 N/kg).

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Method: h = Ep/(mg).

h = 196 ÷ 19.6 = 10.0 m

Extended[3 marks]

A 0.10 kg ball is thrown upwards at 8.0 m/s and rises 2.8 m. Calculate the energy transferred to the surroundings.

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Method: Ek − Ep.

Ek = 3.20 J; Ep = 0.10 × 9.8 × 2.8 = 2.74 J; wasted = 0.456 J

Extended[2 marks]

The mass of a moving object is halved and its speed doubled. State the change in kinetic energy.

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Method: ½mv².

Kinetic energy doubles (½ × 4 = 2)

Extended[3 marks]

A cyclist of total mass 80 kg slows from 12 m/s to 4.0 m/s. Calculate the kinetic energy lost.

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Method: ΔEk.

ΔEk = ½ × 80 × (144 − 16) = 5120 J

Extended[3 marks]

A 1.5 kg book falls 2.0 m from a shelf. Calculate its speed just before landing using conservation of energy.

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Method: v = √(2gh).

v = √(2 × 9.8 × 2.0) = 6.26 m/s

Extended[2 marks]

Draw a description of a Sankey diagram for a TV: 200 J in, 120 J light and sound, 80 J thermal.

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Method: Arrow widths.

One 200-unit-wide input arrow splitting into a 120-unit useful arrow (light and sound) and an 80-unit arrow bending down (thermal)

Extended[4 marks]

A pole-vaulter of mass 65 kg runs at 9.0 m/s. Assuming all kinetic energy becomes GPE, calculate the maximum rise in height of her centre of mass.

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Method: ½mv² = mgh.

h = 81 ÷ 19.6 = 4.13 m

Challenge questions

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

ChallengeCore[4 marks]

A battery-powered toy car lifts itself up a ramp. Describe all the energy transfers from the battery to the top of the ramp, including wasted energy.

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Method: Stores and pathways.

Chemical store → electrical work → kinetic store of motor/car → gravitational potential store; wasted to internal (thermal) stores by friction and electrical resistance, and as sound

ChallengeCore[3 marks]

A Sankey diagram for a car engine shows 100 kJ in, 25 kJ useful kinetic, and the rest as thermal and sound with thermal 70 kJ. Calculate the sound energy and the efficiency.

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Method: Conservation.

Sound = 5 kJ; efficiency 25%

ChallengeExtended[5 marks]

A 0.20 kg ball is dropped from 2.0 m and rebounds to 1.4 m (g = 9.8 N/kg). Calculate (a) the speed just before impact, (b) the speed just after impact, (c) the energy transferred during the bounce.

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Method: Energy conservation.

(a) 6.26 m/s (b) 5.24 m/s (c) 0.20 × 9.8 × 0.6 = 1.18 J

ChallengeExtended[5 marks]

A 500 kg roller-coaster car starts from rest 30 m above the ground, goes down to 5.0 m, then up a hill 20 m high. Ignoring friction, calculate its speed at the bottom and at the top of the second hill.

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Method: Ep → Ek.

Bottom: √(2×9.8×25) = 22.1 m/s; top: √(2×9.8×10) = 14.0 m/s

ChallengeExtended[4 marks]

A 1500 kg car at 25 m/s brakes to 15 m/s. Calculate the kinetic energy transferred to the brakes. Explain why stopping from 30 m/s needs more than twice the energy needed from 15 m/s.

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Method: ΔEk; Ek ∝ v².

ΔEk = ½ × 1500 × (625 − 225) = 300000 J; doubling speed quadruples kinetic energy

ChallengeExtended[5 marks]

A pendulum bob of mass 0.30 kg swings with a speed of 2.4 m/s at the bottom. Calculate the maximum height it rises above the bottom, then explain why each swing gets lower.

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Method: ½mv² = mgh.

h = 2.4² ÷ (2 × 9.8) = 0.294 m; energy is transferred to thermal stores by air resistance and friction at the pivot

ChallengeExtended[4 marks]

An 80 kg skier slides from rest down a 50 m high slope and reaches 25 m/s. Calculate the percentage of the gravitational potential energy wasted.

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Method: Compare Ep and Ek.

Ep = 39200 J; Ek = 25000 J; wasted = 36.2%

ChallengeExtended[4 marks]

A spring stores 0.90 J of elastic energy and launches a 0.050 kg ball vertically. Assuming no losses, calculate the launch speed and the height reached.

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Method: Elastic → Ek → Ep.

v = √(2 × 0.90 ÷ 0.050) = 6.00 m/s; h = 0.90 ÷ (0.050 × 9.8) = 1.84 m

ChallengeExtended[4 marks]

A hydroelectric station drops 500 kg of water per second through 80 m. The water leaves the turbine at 10 m/s. Calculate the energy per second transferred to the turbine.

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Method: ΔEp − remaining Ek per second.

Ep = 392000 J; Ek left = 25000 J; to turbine = 367000 J per second

ChallengeExtended[5 marks]

Two balls of mass 1.0 kg and 2.0 kg are dropped from the same height of 10 m. Compare their speeds at the ground and their kinetic energies, explaining your answers.

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Method: v independent of mass.

Same speed: 14.0 m/s (gh does not depend on m); Ek = 98 J and 196 J — the heavier has twice the kinetic energy

All IGCSE Physics topics ← 1.6 Momentum 1.7.2 Work →

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