Skip to main content
EnglandCombined ScienceSyllabus dot point

How is energy stored and transferred, and why is the total energy always conserved?

Energy stores and transfers, the conservation of energy, work done as energy transferred by a force, kinetic and gravitational potential energy, and the calculation of these energy stores.

A focused answer to the OCR Gateway GCSE Combined Science A topic P5 on energy stores and transfers, covering the main energy stores, conservation of energy, work done by a force, and calculating kinetic and gravitational potential energy.

Generated by Claude Opus 4.89 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. Energy stores, transfers and conservation
  3. Work done
  4. Kinetic and gravitational potential energy

What this topic is asking

OCR wants you to describe energy stores and transfers, state the principle of conservation of energy, calculate the work done by a force, and calculate kinetic and gravitational potential energy.

Energy stores, transfers and conservation

The central principle is the conservation of energy: energy cannot be created or destroyed, only transferred from one store to another or one object to another. In any real process some energy is dissipated to the surroundings, usually to the thermal energy store (for example by friction or air resistance), where it is spread out and becomes less useful. So the total energy stays the same, but the amount of useful energy decreases. Describing a process means identifying the store the energy starts in, the transfer, and the store it ends in, for example a falling ball transfers energy from the gravitational potential store to the kinetic store.

Work done

Work done is just energy transferred by a force, so it is measured in joules, the same unit as energy. When you lift an object you do work against gravity and the energy goes to the gravitational potential store; when you push an object along a rough surface you do work against friction and the energy goes to the thermal store.

Kinetic and gravitational potential energy

Two energy stores have equations you must use confidently.

The kinetic energy of a moving object is:

Ek=12mv2E_k = \frac{1}{2}mv^2

(one half times mass in kg times the speed in m/s squared), so doubling the speed quadruples the kinetic energy, which is why high speed is so dangerous in a crash.

The gravitational potential energy gained when an object is raised is:

Ep=mghE_p = mgh

(mass times gravitational field strength times height), where gg is about 9.89.8 N/kg on Earth. By conservation of energy, when an object falls (ignoring air resistance) the gravitational potential energy lost equals the kinetic energy gained, which lets you find the speed of a falling object.

Exam-style practice questions

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

OCR 20194 marksA ball of mass 0.5 kg is dropped from a height of 8 m. Calculate the gravitational potential energy it had at the top, and state how much kinetic energy it has just before hitting the ground (ignore air resistance). Take g as 9.8 N/kg.
Show worked answer →

A Physics Paper 6 energy calculation. Method: gravitational potential energy Ep=mgh=0.5×9.8×8=39.2E_p = mgh = 0.5 \times 9.8 \times 8 = 39.2 J. By conservation of energy, with air resistance ignored, all of this gravitational potential energy is transferred to the kinetic energy store as the ball falls, so just before impact the kinetic energy is also 39.239.2 J. Markers credit the correct use of Ep=mghE_p = mgh with units, and the statement that the kinetic energy equals the lost gravitational potential energy (39.239.2 J) because energy is conserved. A common slip is to forget that the two stores are equal only when other transfers (such as to air resistance) are ignored.

OCR 20214 marksCalculate the kinetic energy of a car of mass 1000 kg travelling at 12 m/s, and explain what happens to this energy when the car brakes to a stop.
Show worked answer →

A P5 calculation plus explanation. Method: kinetic energy Ek=12mv2=12×1000×122=12×1000×144=72000E_k = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 1000 \times 12^2 = \dfrac{1}{2} \times 1000 \times 144 = 72000 J (7272 kJ). Markers credit squaring the speed before multiplying, and the answer with units. The explanation: when the car brakes, the kinetic energy is transferred mainly to the thermal energy store of the brakes (and tyres and road) through friction, so the brakes heat up; energy is conserved, just transferred to a less useful (dissipated) form. Markers want the correct kinetic energy and the transfer to thermal energy by friction in the brakes.

Related dot points

Sources & how we know this