A motor lifts a 50\ kg load at constant speed 0.40\ m s^-1. The motor draws 300\ W. Find its efficiency.

SEAB Original-style practice: A 1500\ kg car climbs a hill, rising 40\ m vertically, while travelling at constant speed against a constant resistive force of 600\ N over a road distance of 500\ m. Take g = 9.81\ m s^-2. (a) Find the work done against gravity. (b) Find the total useful work done by the engine. (c) If the climb takes 25\ s, find the average output power.

(a) Work against gravity equals the gain in gravitational potential energy: W_g = mgh = 1500 × 9.81 × 40 = 5.89 × 10^5\ J. (b) Work against resistance: W_r = F d = 600 × 500 = 3.00 × 10^5\ J. Since speed is constant, kinetic energy is unchanged, so total work by the engine = W_g + W_r = 5.89 × 10^5 + 3.00 × 10^5 = 8.89 × 10^5\ J. (c) Power = worktime = 8.89 × 10^525 = 3.56 × 10^4\ W = 35.6\ kW. Markers reward mgh for the gravity term, Fd for the resistance term, recognising constant speed means no kinetic energy change, and power as work over time.

SEAB Original-style practice: A pump raises 0.50\ m^3 of water (density 1000\ kg m^-3) per minute to a height of 12\ m. The pump's input power is 1.5\ kW. Take g = 9.81\ m s^-2. Calculate the efficiency of the pump.

Mass of water per minute: m = V = 1000 × 0.50 = 500\ kg. Useful energy output per minute: E = mgh = 500 × 9.81 × 12 = 5.886 × 10^4\ J. Useful output power: P_out = 5.886 × 10^460 = 981\ W. Efficiency: P_outP_in × 100 = 9811500 × 100 = 65.4\%. Markers reward finding the mass from density and volume, the useful output energy from mgh, conversion to power over the minute, and the efficiency as a ratio of output to input power.

SingaporePhysicsSyllabus dot point

How do work, energy and power describe energy transfer, and how does the work-energy theorem connect force to a change in kinetic energy?

Define work, kinetic and potential energy and power, apply the work-energy theorem and conservation of energy, and calculate efficiency

A focused answer to the H2 Physics learning outcome on work, energy and power. Work as force times displacement, the work-energy theorem, kinetic and gravitational potential energy, conservation of energy, power and efficiency.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

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

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What this dot point is asking

SEAB wants you to define work, kinetic energy, gravitational potential energy and power, to apply the work-energy theorem and conservation of energy, and to calculate efficiency. These ideas tie force directly to motion through energy and underpin much of the rest of the syllabus.

The answer

Work done by a force

Work is done when a force moves its point of application. For a constant force $F$ and displacement $s$ at angle $\theta$ between them:

W = Fs\cos\theta

Work is a scalar measured in joules. Only the component of force along the displacement does work; a force perpendicular to motion (such as the centripetal force in circular motion) does no work.

For a variable force, the work done is the area under the force-displacement graph.

Kinetic and potential energy

Kinetic energy is the energy of motion:

E_k = \tfrac{1}{2}mv^2

Gravitational potential energy near the Earth's surface (relative to a chosen reference level):

E_p = mgh

The work-energy theorem

The net work done on a body equals its change in kinetic energy:

W_{\text{net}} = \Delta E_k = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2

This is a direct route from force to speed change without needing the suvat equations, and it works even when the force varies.

Conservation of energy

Energy cannot be created or destroyed, only transferred. For a system with only conservative forces (gravity), mechanical energy is conserved:

E_k + E_p = \text{constant}

When resistive forces act, mechanical energy decreases and the difference appears as heat and sound, but the total energy is still conserved.

Power and efficiency

Power is the rate of doing work or transferring energy:

P = \frac{W}{t}, \qquad P = Fv

The second form (force times velocity) is useful for a body moving at constant speed against resistance. Efficiency compares useful output to total input:

\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\%

Efficiency is always less than $100\%$ for a real machine, because some energy is always transferred to unwanted forms.

Worked example

A $20\ \text{kg}$ crate is pushed $8.0\ \text{m}$ across a floor by a horizontal force of $60\ \text{N}$ against a friction force of $25\ \text{N}$ . The crate starts from rest. Find its final speed using the work-energy theorem.

Step 1: Find the net force

F_{\text{net}} = 60 - 25 = 35\ \text{N}

Step 2: Find the net work done

W_{\text{net}} = F_{\text{net}} \times s = 35 \times 8.0 = 280\ \text{J}

Step 3: Apply the work-energy theorem

Starting from rest, $W_{\text{net}} = \tfrac{1}{2}mv^2 - 0$ .

280 = \tfrac{1}{2}(20)v^2 = 10v^2 \implies v^2 = 28 \implies v = 5.3\ \text{m s}^{-1}

Step 4: State the result

The crate reaches $5.3\ \text{m s}^{-1}$ after $8.0\ \text{m}$ . Note that only the net force (not the applied force alone) does work that changes kinetic energy.

Examples in context

Example 1. A roller coaster. Neglecting friction, a car at the top of a $30\ \text{m}$ drop converts gravitational potential energy entirely to kinetic energy: $mgh = \tfrac{1}{2}mv^2$ , giving $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 30} = 24.3\ \text{m s}^{-1}$ at the bottom, independent of the car's mass. In practice friction reduces this, with the lost energy appearing as heat in the wheels and track.

Example 2. A car at top speed. A car cruising at constant speed has zero net force, so all the engine's output power goes into overcoming resistance: $P = Fv$ where $F$ is the total resistive force. This is why top speed scales with available power and air resistance, and why doubling speed demands far more than double the power.

Try this

Q1. Define work done by a force and state the condition under which a force does no work. [2 marks]

Cue. $W = Fs\cos\theta$ ; a force does no work when it is perpendicular to the displacement ( $\theta = 90^\circ$ ).

Q2. A $0.20\ \text{kg}$ ball is dropped from $2.0\ \text{m}$ . Using energy conservation, find its speed just before hitting the ground (ignore air resistance). [2 marks]

Cue. $mgh = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 2.0} = 6.3\ \text{m s}^{-1}$ .

Q3. A motor lifts a $50\ \text{kg}$ load at constant speed $0.40\ \text{m s}^{-1}$ . The motor draws $300\ \text{W}$ . Find its efficiency. [3 marks]

Cue. Useful power $= mgv = 50 \times 9.81 \times 0.40 = 196\ \text{W}$ ; efficiency $= \dfrac{196}{300} \times 100 = 65\%$ .

Exam-style practice questions

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

Original5 marksA

1500\ \text{kg}

car climbs a hill, rising

40\ \text{m}

vertically, while travelling at constant speed against a constant resistive force of

600\ \text{N}

over a road distance of

500\ \text{m}

. Take

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

. (a) Find the work done against gravity. (b) Find the total useful work done by the engine. (c) If the climb takes

25\ \text{s}

, find the average output power.

Show worked answer →

(a) Work against gravity equals the gain in gravitational potential energy: $W_g = mgh = 1500 \times 9.81 \times 40 = 5.89 \times 10^5\ \text{J}$ .

(b) Work against resistance: $W_r = F d = 600 \times 500 = 3.00 \times 10^5\ \text{J}$ . Since speed is constant, kinetic energy is unchanged, so total work by the engine $= W_g + W_r = 5.89 \times 10^5 + 3.00 \times 10^5 = 8.89 \times 10^5\ \text{J}$ .

(c) Power $= \dfrac{\text{work}}{\text{time}} = \dfrac{8.89 \times 10^5}{25} = 3.56 \times 10^4\ \text{W} = 35.6\ \text{kW}$ .

Markers reward $mgh$ for the gravity term, $Fd$ for the resistance term, recognising constant speed means no kinetic energy change, and power as work over time.

Original4 marksA pump raises

0.50\ \text{m}^3

of water (density

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

) per minute to a height of

12\ \text{m}

. The pump's input power is

1.5\ \text{kW}

. Take

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

. Calculate the efficiency of the pump.

Show worked answer →

Mass of water per minute: $m = \rho V = 1000 \times 0.50 = 500\ \text{kg}$ .

Useful energy output per minute: $E = mgh = 500 \times 9.81 \times 12 = 5.886 \times 10^4\ \text{J}$ .

Useful output power: $P_{\text{out}} = \dfrac{5.886 \times 10^4}{60} = 981\ \text{W}$ .

Efficiency: $\dfrac{P_{\text{out}}}{P_{\text{in}}} \times 100 = \dfrac{981}{1500} \times 100 = 65.4\%$ .

Markers reward finding the mass from density and volume, the useful output energy from $mgh$ , conversion to power over the minute, and the efficiency as a ratio of output to input power.