Skip to main content
EnglandPhysicsSyllabus dot point

How are work, energy and power related, and how does the conservation of energy constrain what is possible?

Work, energy and power: work done by a force, the principle of conservation of energy, kinetic and gravitational potential energy, power as the rate of doing work, and efficiency.

A focused answer to the OCR H556 forces and motion content on work, energy and power, covering work done by a force, the conservation of energy, kinetic and gravitational potential energy, power as the rate of doing work with the relation to force and velocity, and efficiency.

Generated by Claude Opus 4.811 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 dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

OCR wants you to calculate the work done by a force, apply the principle of conservation of energy, use the formulae for kinetic and gravitational potential energy, define power as the rate of doing work with the relation P=FvP = Fv, and calculate efficiency.

The answer

Work done

The area under a force-displacement graph also gives the work done, which is useful when the force varies, for example stretching a spring.

Conservation of energy

Kinetic and potential energy

Power and efficiency

Examples in context

Regenerative braking in electric cars recovers kinetic energy that friction brakes would waste as heat. Power-station and motor efficiencies are quoted as the useful output divided by the energy input, with the rest lost mainly as heat. Pumped-storage hydroelectricity converts electrical energy to gravitational potential energy and back. A pole-vaulter converts the kinetic energy of the run-up into elastic energy in the pole and then gravitational potential energy at the top of the vault.

Try this

Q1. State the principle of conservation of energy. [1 mark]

  • Cue. Energy cannot be created or destroyed, only transferred between stores; total energy is constant in an isolated system.

Q2. A 1200 kg1200\ \text{kg} car accelerates from rest to 15 m s−115\ \text{m s}^{-1}. Find the gain in kinetic energy. [2 marks]

  • Cue. Ek=12mv2=12(1200)(15)2=1.35×105 JE_k = \frac{1}{2}mv^2 = \frac{1}{2}(1200)(15)^2 = 1.35 \times 10^{5}\ \text{J}.

Q3. A pump does 6.0 kJ6.0\ \text{kJ} of useful work in 8.0 s8.0\ \text{s} from an input of 1.0 kW1.0\ \text{kW}. Find the efficiency. [2 marks]

  • Cue. Output power =60008.0=750 W= \frac{6000}{8.0} = 750\ \text{W}; efficiency =7501000=75%= \frac{750}{1000} = 75\%.

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 roller-coaster car of mass 480 kg480\ \text{kg} starts from rest at the top of a 35 m35\ \text{m} drop. Assuming no friction, calculate its speed at the bottom and state how the answer would change if friction were significant. Take g=9.81 m s−2g = 9.81\ \text{m s}^{-2}.
Show worked answer →

Energy conservation: the loss in gravitational potential energy becomes kinetic energy. mgh=12mv2mgh = \frac{1}{2}mv^2, so v=2gh=2×9.81×35=687=26 m s−1v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 35} = \sqrt{687} = 26\ \text{m s}^{-1}.

The mass cancels, so it does not appear in the answer.

With friction, some energy is transferred to thermal energy and sound, so the speed at the bottom would be less than 26 m s−126\ \text{m s}^{-1}. Markers reward equating potential and kinetic energy, the value about 26 m s−126\ \text{m s}^{-1}, and the statement that friction reduces the speed.

OCR 20224 marksA motor raises a 250 kg250\ \text{kg} load at a steady 0.80 m s−10.80\ \text{m s}^{-1}. The motor draws 2.6 kW2.6\ \text{kW} of electrical power. Calculate the useful output power and the efficiency of the motor. Take g=9.81 m s−2g = 9.81\ \text{m s}^{-2}.
Show worked answer →

At steady speed the useful output power lifts the load against gravity: P=Fv=mgv=250×9.81×0.80=1962 WP = Fv = mgv = 250 \times 9.81 \times 0.80 = 1962\ \text{W}, about 2.0 kW2.0\ \text{kW}.

Efficiency: useful outputtotal input=19622600=0.75\frac{\text{useful output}}{\text{total input}} = \frac{1962}{2600} = 0.75, or 75%75\%.

Markers reward using P=mgvP = mgv for the output, dividing by the input power, and the efficiency about 75%75\%.

Related dot points

Sources & how we know this