EnglandCombined ScienceSyllabus dot point

How is energy stored, transferred and conserved, and how can we use it efficiently?

Energy stores and systems, calculating kinetic, gravitational and elastic energy, the conservation and dissipation of energy, efficiency and reducing unwanted transfers, and national and global energy resources.

A focused answer to the AQA GCSE Combined Science: Trilogy Energy topic, covering energy stores and systems, kinetic, gravitational and elastic energy calculations, conservation and dissipation, efficiency, and national and global energy resources.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Energy stores and transfers
Energy calculations
Conservation, dissipation and efficiency
Energy resources

What this topic is asking

AQA wants you to describe energy stores and systems, calculate kinetic, gravitational potential and elastic energy, explain conservation and dissipation of energy, calculate efficiency and describe how to reduce unwanted transfers, and compare energy resources.

Energy stores and transfers

When a system changes, energy is transferred between stores, but the total energy stays the same. For example, a falling ball transfers energy from its gravitational potential store to its kinetic store; a moving car braking transfers energy from its kinetic store to the thermal store of the brakes and surroundings. AQA wants you to describe these changes in terms of named stores and pathways rather than vague "types of energy".

Energy calculations

Because energy is conserved, you can equate stores to solve problems. For a falling object (ignoring air resistance), the gravitational potential energy lost equals the kinetic energy gained, $mgh = \tfrac{1}{2}mv^2$ , so the speed at the bottom does not depend on the mass. This kind of energy-conservation argument is a favourite higher-tariff question.

Conservation, dissipation and efficiency

Efficiency $= \dfrac{\text{useful output energy transfer}}{\text{total input energy transfer}}$ , and can also be written using power as $\dfrac{\text{useful power output}}{\text{total power input}}$ . Multiply by 100 to give a percentage. Unwanted (wasted) transfers can be reduced by lubrication (reducing friction between moving parts) and thermal insulation (reducing the rate of energy transfer by conduction, for example loft insulation, cavity walls and thick walls with a low thermal conductivity). No device can be more than 100 percent efficient, because some energy is always dissipated.

Energy resources

Renewable resources are replenished as fast as they are used and include wind, solar, hydroelectric, tidal, wave, geothermal and biofuels. Non-renewable resources are finite and will run out: the fossil fuels (coal, oil and natural gas) and nuclear fuel. Fossil fuels are reliable and provide energy on demand, but burning them releases carbon dioxide (a greenhouse gas) and other pollutants; renewables are cleaner and do not run out, but many (wind, solar, tidal) are less reliable because they depend on the weather or time. The choice of resource for electricity generation, transport and heating balances reliability, environmental impact and cost, and political and ethical factors influence which are used.

Using energy conservation to find a falling speed

A ball of mass 0.5 kg is dropped from a height of 1.8 m. Use energy conservation to find its speed just before it hits the ground. (Take $g = 9.8$ N/kg and ignore air resistance.)

step 1: Calculate the gravitational potential energy lost

$E_p = mgh = 0.5 \times 9.8 \times 1.8 = 8.82$ J.

step 2: Equate it to the kinetic energy gained

By conservation of energy, all the gravitational potential energy becomes kinetic energy, so $\tfrac{1}{2}mv^2 = 8.82$ J.

step 3: Rearrange for the speed

$v^2 = \dfrac{2 \times 8.82}{m} = \dfrac{17.64}{0.5} = 35.28$ , so $v = \sqrt{35.28} \approx 5.9$ m/s.

step 4: Note what cancels

The mass cancels in the algebra ( $mgh = \tfrac{1}{2}mv^2$ gives $v^2 = 2gh$ ), so any object dropped from 1.8 m reaches the same speed in the absence of air resistance. This is a useful check on the method.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksA car of mass 1200 kg accelerates from rest to a speed of 20 m/s. Calculate the increase in the kinetic energy of the car.

Show worked answer →

A Physics Paper 1 calculation. Method: kinetic energy is $E_k = \dfrac{1}{2}mv^2$ . At the start the car is at rest, so its kinetic energy is 0 J. At 20 m/s, $E_k = \dfrac{1}{2} \times 1200 \times 20^2 = \dfrac{1}{2} \times 1200 \times 400 = 240000$ J (240 kJ). The increase is therefore 240000 J. Markers award the correct substitution, squaring the speed (the most common error is using 20 rather than 400), and the final value with a unit. State the increase equals the final value because the initial kinetic energy was zero.

AQA 20214 marksAn electric kettle transfers 360000 J of energy from the mains. Only 306000 J is usefully transferred to heat the water. Calculate the efficiency of the kettle, and suggest where the wasted energy goes.

Show worked answer →

A Physics Paper 1 efficiency calculation with application. Method: efficiency $= \dfrac{\text{useful output energy transfer}}{\text{total input energy transfer}} = \dfrac{306000}{360000} = 0.85$ , which is 85 percent. For the wasted energy: it is dissipated to the surroundings as thermal energy, for example heating the kettle body and the air, and a little as sound. Markers credit the correct ratio, the value (as a decimal or percentage), and a sensible destination for the wasted energy. A common error is to invert the fraction and get an efficiency above 1, which is impossible.

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Sources & how we know this

AQA GCSE Combined Science: Trilogy (8464) specification — AQA (2016)