A 1500\,W motor lifts a 30\,kg load. Find the time to raise it 4.0\,m, assuming 100 per cent efficiency. Take g = 9.81\,m s^-2.

WJEC WJEC 2019-style practice: A pump raises 0.50\,m^3 of water per second through a vertical height of 12\,m. The density of water is 1000\,kg m^-3. Calculate the useful output power and, if the pump draws 75\,kW from the supply, determine its efficiency.

First find the mass of water lifted each second, then the gravitational potential energy gained per second, which is the useful output power. Mass per second: m = V = 1000 × 0.50 = 500\,kg s^-1. Useful output power equals potential energy gained per second: P_out = mgh per second = 500 × 9.81 × 12 = 58\,860\,W ≈ 58.9\,kW. Efficiency: P_outP_in = 58.975 = 0.785, or about 79\%. Markers reward computing mass flow from density and volume flow, using mgh per second as the output power, and the efficiency ratio expressed as a percentage.

WJEC WJEC 2021-style practice: A ball of mass 0.15\,kg is dropped from rest and reaches the ground at 8.0\,m s^-1 after falling 4.0\,m. Show that energy is not conserved as kinetic energy alone, and calculate the energy transferred to other forms.

Compare the gravitational potential energy lost with the kinetic energy gained. Potential energy lost: E_p = mgh = 0.15 × 9.81 × 4.0 = 5.89\,J. Kinetic energy gained: E_k = 12mv^2 = 12(0.15)(8.0)^2 = 4.8\,J. The kinetic energy gained (4.8\,J) is less than the potential energy lost (5.89\,J), so not all of it became kinetic energy. Energy transferred to other forms (air resistance, heating the air): 5.89 - 4.8 = 1.1\,J. Total energy is still conserved overall. Markers reward both energy calculations and identifying the difference as energy lost to air resistance.

WalesPhysicsSyllabus dot point

How do work, energy, power and efficiency describe energy transfer?

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

A focused answer to WJEC A-Level Physics Unit 1 energy concepts, covering work done by a force, kinetic and gravitational potential energy, the principle of conservation of energy, power as the rate of doing work, and efficiency.

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

WJEC wants you to calculate work done by a force, use the formulae for kinetic and gravitational potential energy, apply conservation of energy, and define and calculate power and efficiency. Energy is the universal accounting tool of physics, and the examiners often reward a clean energy argument where a forces-and-motion approach would be far longer.

The answer

Work

When the force is perpendicular to the motion ( $\theta = 90^\circ$ ), no work is done. This is why the tension in a string holding a conical pendulum, or the magnetic force on a moving charge, does no work even though a force clearly acts.

Kinetic and potential energy

Kinetic energy is the energy of a moving body, $E_k = \frac{1}{2}mv^2$ . Gravitational potential energy gained when a mass is raised a height $h$ near the Earth's surface is $E_p = mgh$ . The kinetic energy formula follows from the work-energy theorem: the work done by a resultant force equals the change in kinetic energy of the body.

Conservation of energy

Power and efficiency

Power is the rate of transferring energy or doing work, $P = \dfrac{W}{t}$ . For a force moving at velocity $v$ , $P = Fv$ , which comes from $P = W/t = Fx/t = Fv$ .

Efficiency is a ratio (or percentage) and is always less than 1 because some energy transfers to less useful forms such as heat.

Examples in context

Example 1. Regenerative braking. An electric car of mass $1500\,\text{kg}$ slowing from $20\,\text{m s}^{-1}$ to rest has $\tfrac{1}{2}(1500)(20)^2 = 300\,\text{kJ}$ of kinetic energy. If the regenerative system is $70\%$ efficient, it returns $0.70 \times 300 = 210\,\text{kJ}$ to the battery, with the remaining $90\,\text{kJ}$ lost as heat in the motor windings and tyres. This is why electric cars are far more efficient than petrol cars in stop-start traffic.

Example 2. A hydroelectric turbine. Water falls $90\,\text{m}$ at $200\,\text{kg s}^{-1}$ , delivering $mgh = 200 \times 9.81 \times 90 = 176\,\text{kW}$ of gravitational power. If the turbine and generator together are $85\%$ efficient, the electrical output is $0.85 \times 176 = 150\,\text{kW}$ . Scaling this up to thousands of cubic metres per second is how stations like Dinorwig in Snowdonia generate hundreds of megawatts.

Try this

Q1. A $1500\,\text{W}$ motor lifts a $30\,\text{kg}$ load. Find the time to raise it $4.0\,\text{m}$ , assuming 100 per cent efficiency. Take $g = 9.81\,\text{m s}^{-2}$ . [3 marks]

Cue. $E_p = mgh = 30\times9.81\times4.0 \approx 1177\,\text{J}$ ; $t = \frac{E}{P} = \frac{1177}{1500} \approx 0.78\,\text{s}$ .

Q2. Define the efficiency of a device. [1 mark]

Cue. Useful output energy (or power) divided by total input energy (or power).

Exam-style practice questions

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

WJEC 20195 marksA pump raises

0.50\,\text{m}^3

of water per second through a vertical height of

12\,\text{m}

. The density of water is

1000\,\text{kg m}^{-3}

. Calculate the useful output power and, if the pump draws

75\,\text{kW}

from the supply, determine its efficiency.

Show worked answer →

First find the mass of water lifted each second, then the gravitational potential energy gained per second, which is the useful output power.

Mass per second: $m = \rho V = 1000 \times 0.50 = 500\,\text{kg s}^{-1}$ .

Useful output power equals potential energy gained per second:

$P_{\text{out}} = mgh \text{ per second} = 500 \times 9.81 \times 12 = 58\,860\,\text{W} \approx 58.9\,\text{kW}$ .

Efficiency: $\dfrac{P_{\text{out}}}{P_{\text{in}}} = \dfrac{58.9}{75} = 0.785$ , or about $79\%$ .

Markers reward computing mass flow from density and volume flow, using $mgh$ per second as the output power, and the efficiency ratio expressed as a percentage.

WJEC 20213 marksA ball of mass

0.15\,\text{kg}

is dropped from rest and reaches the ground at

8.0\,\text{m s}^{-1}

after falling

4.0\,\text{m}

. Show that energy is not conserved as kinetic energy alone, and calculate the energy transferred to other forms.

Show worked answer →

Compare the gravitational potential energy lost with the kinetic energy gained.

Potential energy lost: $E_p = mgh = 0.15 \times 9.81 \times 4.0 = 5.89\,\text{J}$ .

Kinetic energy gained: $E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(0.15)(8.0)^2 = 4.8\,\text{J}$ .

The kinetic energy gained ( $4.8\,\text{J}$ ) is less than the potential energy lost ( $5.89\,\text{J}$ ), so not all of it became kinetic energy.

Energy transferred to other forms (air resistance, heating the air): $5.89 - 4.8 = 1.1\,\text{J}$ . Total energy is still conserved overall. Markers reward both energy calculations and identifying the difference as energy lost to air resistance.

Related dot points

Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)