A motor lifts a 50\ kg load through 4.0\ m in 8.0\ s. Calculate the useful output power. Take g = 9.8\ m s^-2.

A ball is dropped from rest at a height of 1.8\ m. Using energy conservation, find its speed just before it lands. Take g = 9.8\ m s^-2.

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How are work, energy and power linked, and when is energy conserved?

Work done by a force, kinetic and gravitational potential energy, the principle of conservation of energy, power, and efficiency.

A CCEA A-Level Physics answer on work done by a force, kinetic and gravitational potential energy, the principle of conservation of energy, the definition of power, and how efficiency is calculated.

Generated by Claude Opus 4.811 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
The answer
Examples in context
Try this

What this dot point is asking

CCEA wants you to define and calculate work done by a force, use the expressions for kinetic and gravitational potential energy, apply the principle of conservation of energy, define power, and calculate efficiency. Conservation-of-energy problems and efficiency calculations recur in every paper.

The answer

Work and energy

The two mechanical energy stores at this level are kinetic energy, $E_k = \tfrac{1}{2}mv^2$ , and gravitational potential energy, $E_p = mgh$ . Work done on a body transfers energy to it; work done by a body transfers energy away.

Conservation of energy

Power and efficiency

Worked example: pumped storage

Energy and power of a pumped-storage scheme

A pumped-storage station raises $2.0 \times 10^{6}\ \text{kg}$ of water through a height of $300\ \text{m}$ in $1.0$ hour. Take $g = 9.81\ \text{m s}^{-2}$ . Find the energy stored and the average input power if the pumping is $80\%$ efficient.

step Find the gravitational potential energy stored

E_p = mgh = 2.0 \times 10^{6} \times 9.81 \times 300 = 5.89 \times 10^{12}\ \text{J}.

step Find the input energy from the efficiency

\text{efficiency} = \frac{E_{\text{useful}}}{E_{\text{in}}} \Rightarrow E_{\text{in}} = \frac{5.89 \times 10^{12}}{0.80} = 7.36 \times 10^{12}\ \text{J}.

step Find the average input power

P = \frac{E_{\text{in}}}{t} = \frac{7.36 \times 10^{12}}{3600} = 2.0 \times 10^{9}\ \text{W} = 2.0\ \text{GW}.

This is comparable to the output of a large power station, which is why such schemes store off-peak energy for later release.

Examples in context

Example 1. A roller coaster. At the top of the first drop the cars have maximum gravitational potential energy and almost no kinetic energy. As they fall, $mgh$ converts to $\tfrac{1}{2}mv^2$ , reaching maximum speed at the bottom. Because some energy is lost to friction and air resistance, each successive hill must be lower than the last, an everyday demonstration of conservation of energy with dissipation.

Example 2. Rating a kettle. A $2.0\ \text{kW}$ kettle transfers $2000\ \text{J}$ every second. To heat water needing $2.5 \times 10^{5}\ \text{J}$ takes $t = E/P = 2.5 \times 10^{5} / 2000 = 125\ \text{s}$ , ignoring losses. The efficiency is below $100\%$ because some heat warms the kettle body and the surroundings rather than the water.

Try this

Q1. A motor lifts a $50\ \text{kg}$ load through $4.0\ \text{m}$ in $8.0\ \text{s}$ . Calculate the useful output power. Take $g = 9.8\ \text{m s}^{-2}$ . [3 marks]

Cue. $E_p = 50 \times 9.8 \times 4.0 = 1960\ \text{J}$ , so $P = \frac{1960}{8.0} = 245\ \text{W}$ .

Q2. State the principle of conservation of energy. [1 mark]

Cue. Energy cannot be created or destroyed, only transferred from one store to another.

Q3. A ball is dropped from rest at a height of $1.8\ \text{m}$ . Using energy conservation, find its speed just before it lands. Take $g = 9.8\ \text{m s}^{-2}$ . [2 marks]

Cue. $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.8} = \sqrt{35.3} = 5.9\ \text{m s}^{-1}$ .

Exam-style practice questions

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

CCEA 20206 marksA 60 kg skier starts from rest at the top of a slope 25 m high. At the bottom her speed is 18 m per second. Calculate the gravitational potential energy lost, the kinetic energy gained, and the energy dissipated by friction and air resistance. Take g as 9.81.

Show worked answer →

Gravitational potential energy lost:

$E_p = mgh = 60 \times 9.81 \times 25 = 1.47 \times 10^{4}$ J.

Kinetic energy gained:

$E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 60 \times 18^2 = \tfrac{1}{2} \times 60 \times 324 = 9.72 \times 10^{3}$ J.

Energy dissipated equals the difference, by conservation of energy:

$E_{\text{lost}} = 1.47 \times 10^{4} - 9.72 \times 10^{3} = 5.0 \times 10^{3}$ J.

Markers reward both energy stores, the conservation-of-energy step, and a positive dissipated value (the missing energy has become heat and sound).

CCEA 20185 marksAn electric motor with an input power of 750 W raises a 40 kg load vertically at a steady speed of 1.2 m per second. Calculate the useful output power and the efficiency of the motor. Explain what happens to the energy that is not usefully transferred. Take g as 9.81.

Show worked answer →

At steady speed the lifting force equals the weight, $F = mg = 40 \times 9.81 = 392$ N.

The useful output power is

$P_{\text{out}} = Fv = 392 \times 1.2 = 471$ W.

The efficiency is

$\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{471}{750} = 0.63 = 63\%$ .

The remaining $750 - 471 = 279$ W is dissipated, mostly as heat in the motor windings (resistive heating) and in friction in the bearings, plus a little sound.

Markers reward $P = Fv$ at constant speed, the efficiency ratio, and naming heat from resistance and friction as the wasted energy.

Related dot points

Sources & how we know this

CCEA GCE Physics specification — CCEA (2016)