A 2.0 kg ball moves at 6.0 m s^-1. Calculate its kinetic energy. [1 mark]

AQA AQA 2018-style practice: A motor lifts a 250 kg load at a steady 0.40 m s^-1. The motor draws 1.5 kW of electrical power. Calculate the useful output power and the efficiency. Take g = 9.8 m s^-2.

The useful output is the rate of gain of gravitational potential energy, which equals the lifting force times the velocity. Lifting force = mg = 250 × 9.8 = 2450 N. Useful power = Fv = 2450 × 0.40 = 980 W. Efficiency = useful outputtotal input = 9801500 = 0.65, or 65 percent. Markers reward using P = Fv for the output power and the correct efficiency ratio (a value less than 1).

EnglandPhysicsSyllabus dot point

How do we measure energy transferred by a force and the rate of that transfer?

Work done by a force including a force at an angle, the relationship between power, work and velocity, kinetic and gravitational potential energy, and efficiency as the ratio of useful output to total input.

A focused answer to AQA A-Level Physics 3.4.1.7 and 3.4.1.8, covering work done by a force including forces at an angle, the relationship between power, work and velocity, kinetic and gravitational potential energy, and the definition of efficiency.

Generated by Claude Opus 4.89 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
Work done by a force
Power
Kinetic and gravitational potential energy
Efficiency
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What this dot point is asking

AQA specification points 3.4.1.7 and 3.4.1.8 want you to calculate work done by a force (including a force at an angle to the displacement), link power to work and to force times velocity, use the equations for kinetic and gravitational potential energy, and define and calculate efficiency.

Work done by a force

Work is done only when a force causes a displacement in its own direction; it transfers energy from one store to another. No work is done when the force is perpendicular to the motion ( $\cos 90^{\circ} = 0$ ), which is why the tension in a string does no work on an object moving in a circle, and why carrying a bag horizontally at constant height does no work against gravity.

Power

The form $P = Fv$ follows because $P = \dfrac{W}{t} = \dfrac{Fs}{t} = F\dfrac{s}{t} = Fv$ . It is especially useful for vehicles and lifts moving at steady speed, where the driving force balances the resistive forces.

Kinetic and gravitational potential energy

The work done by a resultant force on an object equals its change in kinetic energy; this is the work-energy principle, which links the force and energy descriptions of motion. Note that kinetic energy depends on the square of the speed, so doubling the speed quadruples the kinetic energy, which is why braking distance grows so quickly with speed.

Efficiency

Efficiency can be calculated using either energies or powers, since power is energy per unit time and the time cancels in the ratio. To express it as a percentage, multiply the ratio by $100$ . No real machine reaches 100 percent because friction, air resistance and electrical resistance always transfer some energy to thermal stores that cannot be fully recovered, a consequence of the second law of thermodynamics. Improving efficiency means reducing these dissipative transfers, for example by lubricating moving parts or streamlining a vehicle.

Try this

Q1. State the equation linking power, force and velocity. [1 mark]

Cue. $P = Fv$ .

Q2. A $2.0 \text{ kg}$ ball moves at $6.0 \text{ m s}^{-1}$ . Calculate its kinetic energy. [1 mark]

Cue. $E_k = \tfrac{1}{2}(2.0)(6.0)^2 = 36 \text{ J}$ .

Q3. State why the efficiency of a real machine is always less than 100 percent. [1 mark]

Cue. Some energy is always dissipated, usually as heat.

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 20184 marksA motor lifts a

250 \text{ kg}

load at a steady

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

. The motor draws

1.5 \text{ kW}

of electrical power. Calculate the useful output power and the efficiency. Take

g = 9.8 \text{ m s}^{-2}

Show worked answer →

The useful output is the rate of gain of gravitational potential energy, which equals the lifting force times the velocity.

Lifting force $= mg = 250 \times 9.8 = 2450 \text{ N}$ . Useful power $= Fv = 2450 \times 0.40 = 980 \text{ W}$ .

Efficiency $= \dfrac{\text{useful output}}{\text{total input}} = \dfrac{980}{1500} = 0.65$ , or $65$ percent.

Markers reward using $P = Fv$ for the output power and the correct efficiency ratio (a value less than 1).

AQA 20214 marksA force of

40 \text{ N}

pulls a sledge a distance of

10 \text{ m}

along level ground, with the rope at an angle of

30^{\circ}

above the horizontal. Calculate the work done by the force, and explain why the vertical component does no work.

Show worked answer →

Work done is $W = Fs\cos\theta = 40 \times 10 \times \cos 30^{\circ} = 400 \times 0.866 = 346 \text{ J}$ .

Only the component of the force along the displacement does work. The vertical component $F\sin\theta$ acts at right angles to the horizontal motion, and since the sledge does not move vertically, that component does no work ( $\cos 90^{\circ} = 0$ ).

Markers reward the correct work calculation using $\cos\theta$ and explaining that a force perpendicular to the displacement does no work.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)