A 0.50\ kg ball moves at 6.0\ m s^-1. Find its kinetic energy. [2 marks]

A motor delivers 2.0\ kW of useful power while consuming 2.5\ kW. Find its efficiency. [2 marks]

WJEC Eduqas Eduqas 2018-style practice: A car of mass 900\ kg travels at a constant 25\ m s^-1 along a level road against a total resistive force of 620\ N. Calculate the useful output power of the engine and, if the engine is 28 per cent efficient, the rate at which it burns fuel energy.

At constant speed the driving force equals the resistive force, 620\ N. Useful output power: P = Fv = 620 × 25 = 15\,500\ W = 15.5\ kW. Input power from efficiency: efficiency = useful outputinput, so input = 15\,5000.28 = 55\,400\ W, about 55\ kW. Markers reward equating driving and resistive forces at constant speed, P = Fv = 15.5\ kW, and the input power about 55\ kW.

WJEC Eduqas Eduqas 2020-style practice: A ball of mass 0.20\ kg is dropped from a height of 1.8\ m and rebounds to 1.2\ m. Using energy ideas, calculate the speed just before impact and the fraction of the kinetic energy lost in the bounce. Take g = 9.81\ m s^-2.

Speed before impact from energy conservation: mgh = 12mv^2, so v = sqrt(2gh) = sqrt(2 × 9.81 × 1.8) = sqrt(35.3) = 5.9\ m s^-1. The kinetic energy just after the bounce is proportional to the rebound height, so the fraction of kinetic energy retained is 1.21.8 = 0.67. The fraction lost is 1 - 0.67 = 0.33, about a third. Markers reward v = sqrt(2gh) = 5.9\ m s^-1, recognising kinetic energy is proportional to rebound height, and the fraction lost about 0.33.

EnglandPhysicsSyllabus dot point

How do we account for energy as it is transferred and conserved, and how is power related to force and velocity?

Energy concepts: work done by a force, the conservation of energy, kinetic and gravitational potential energy, power as the rate of energy transfer, the relation P = Fv, and efficiency.

A focused answer to the Eduqas A-Level Physics Component 1 energy concepts, covering work done by a force, the principle of conservation of energy, kinetic and gravitational potential energy, power as the rate of energy transfer, the relation P = Fv, and efficiency.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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What this dot point is asking

Eduqas wants you to calculate the work done by a force, apply the principle of conservation of energy, use kinetic energy $\frac{1}{2}mv^2$ and gravitational potential energy $mg\Delta h$ , define power as the rate of energy transfer, use $P = Fv$ , and calculate efficiency as the ratio of useful output to total input.

The answer

Work done by a force

Conservation of energy

Kinetic and potential energy

Power and P = Fv

Efficiency

A crane lifting a load

A crane raises a $500\ \text{kg}$ load through $12\ \text{m}$ in $8.0\ \text{s}$ at constant speed. Find the useful output power and, if the crane's motor draws $9.0\ \text{kW}$ , its efficiency. Take $g = 9.81\ \text{m s}^{-2}$ .

step 1: Useful work done

Work against gravity: $W = mg\Delta h = 500 \times 9.81 \times 12 = 58\,860\ \text{J}$ .

step 2: Useful output power

$P = \dfrac{W}{t} = \dfrac{58\,860}{8.0} = 7\,360\ \text{W}$ , about $7.4\ \text{kW}$ .

step 3: Efficiency

$\text{efficiency} = \dfrac{7\,360}{9\,000} = 0.82$ , about 82 per cent.

Examples in context

Energy accounting underpins the entire energy industry, from the efficiency of power stations and engines to the rating of electrical appliances. The $P = Fv$ relation sets a vehicle's top speed, where the engine's maximum power is reached when the driving force equals the rising air resistance. Conservation of energy lets engineers design roller coasters, pendulum rides and hydroelectric schemes, and lets athletes and coaches model the energy cost of a sprint or a jump.

Try this

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

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

Q2. A $0.50\ \text{kg}$ ball moves at $6.0\ \text{m s}^{-1}$ . Find its kinetic energy. [2 marks]

Cue. $E_k = \frac{1}{2}mv^2 = \frac{1}{2}(0.50)(6.0)^2 = 9.0\ \text{J}$ .

Q3. A motor delivers $2.0\ \text{kW}$ of useful power while consuming $2.5\ \text{kW}$ . Find its efficiency. [2 marks]

Cue. $\frac{2.0}{2.5} = 0.80$ , i.e. 80 per cent.

Exam-style practice questions

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

Eduqas 20184 marksA car of mass

900\ \text{kg}

travels at a constant

25\ \text{m s}^{-1}

along a level road against a total resistive force of

620\ \text{N}

. Calculate the useful output power of the engine and, if the engine is 28 per cent efficient, the rate at which it burns fuel energy.

Show worked answer →

At constant speed the driving force equals the resistive force, $620\ \text{N}$ .

Useful output power: $P = Fv = 620 \times 25 = 15\,500\ \text{W} = 15.5\ \text{kW}$ .

Input power from efficiency: $\text{efficiency} = \dfrac{\text{useful output}}{\text{input}}$ , so input $= \dfrac{15\,500}{0.28} = 55\,400\ \text{W}$ , about $55\ \text{kW}$ .

Markers reward equating driving and resistive forces at constant speed, $P = Fv = 15.5\ \text{kW}$ , and the input power about $55\ \text{kW}$ .

Eduqas 20204 marksA ball of mass

0.20\ \text{kg}

is dropped from a height of

1.8\ \text{m}

and rebounds to

1.2\ \text{m}

. Using energy ideas, calculate the speed just before impact and the fraction of the kinetic energy lost in the bounce. Take

g = 9.81\ \text{m s}^{-2}

Show worked answer →

Speed before impact from energy conservation: $mgh = \tfrac{1}{2}mv^2$ , so $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 1.8} = \sqrt{35.3} = 5.9\ \text{m s}^{-1}$ .

The kinetic energy just after the bounce is proportional to the rebound height, so the fraction of kinetic energy retained is $\dfrac{1.2}{1.8} = 0.67$ . The fraction lost is $1 - 0.67 = 0.33$ , about a third.

Markers reward $v = \sqrt{2gh} = 5.9\ \text{m s}^{-1}$ , recognising kinetic energy is proportional to rebound height, and the fraction lost about $0.33$ .

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)