OxfordAQA International AS and A-level Physics: Work, power, efficiency and conservation of energy (9630)

This guide is aligned to the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for International AS and A-level exams from May/June 2020 onwards. It covers sections 3.2.7 (work, energy and power) and 3.2.8 (conservation of energy). Both sections are shared International AS and A-level content, so they apply to AS and to the full A-level.

Use it with the revision notes and the practice set. The course page is OxfordAQA A-level Physics and the printable list is the checklist. To find weak spots first, try the free diagnostics. For the forces and motion that sit underneath this unit, see Motion and Newton’s laws.

What you must be able to do

Section	Outcome
3.2.7	Energy transferred W = Fs cos θ, where θ is the angle between F and s
3.2.7	Explain the significance of the area under a force-displacement graph, including for variable forces
3.2.7	Power as rate of doing work: P = ΔW/Δt = Fv
3.2.7	Efficiency = useful output power / input power, as a ratio or a percentage
3.2.8	State the principle of conservation of energy
3.2.8	Use ΔEp = mgΔh and Ek = ½mv²
3.2.8	Apply energy conservation, with and without numbers, to gravitational and elastic potential energy, kinetic energy and work done against resistive forces

Work done: W = Fs cos θ

Work is the energy transferred by a force when its point of application moves. Only the part of the force along the displacement does work. If the force F acts at angle θ to the displacement s, the component along s is F cos θ, so

W = Fs cos θ, in joules (J), where 1 J = 1 N m.

Three cases are worth knowing:

• θ = 0°: cos θ = 1, so W = Fs. The force is along the movement and the transfer is as large as it can be.
• θ = 90°: cos θ = 0, so W = 0. A force at right angles to the movement transfers no energy. The normal force on a block sliding along a table does no work.
• θ greater than 90°: cos θ is negative, so the work is negative. The force opposes the motion and takes energy from the object. Friction and drag usually do this.

Work is a scalar quantity.

Worked example 1: a rope at an angle

A sledge is pulled 25 m across level snow by a rope. The tension is 80 N and the rope makes 35° with the ground. Find the work done by the tension.

1. Only the component along the ground matters: F cos 35° = 80 × 0.819 = 65.5 N.
2. W = 65.5 × 25 = 1.6 × 10³ J (1638 J before rounding).

If the rope were horizontal the work would be 80 × 25 = 2000 J. Pulling at an angle wastes part of the force lifting the sledge instead of moving it forward.

Area under a force-displacement graph

When the force is constant, W = Fs is a rectangle of height F and width s on a force-displacement graph. The same idea works when the force varies: the work done equals the area under the graph, with F plotted along the displacement. This also gives the energy transferred in each stage of a journey.

For a graph made of straight lines, split the area into rectangles, triangles and trapeziums. For a curve, count squares or divide it into thin strips. Regions below the axis count as negative work. The graph must show the force component along the displacement.

Worked example 2: a variable push

A 5.0 kg trolley starts from rest on a smooth horizontal track. The forward force on it rises uniformly from 0 to 40 N over the first 2.0 m. It then stays at 40 N until the trolley has moved 5.0 m, and falls uniformly to 10 N at 6.0 m. Find the work done and the final speed.

1. Triangle, 0 to 2.0 m: ½ × 2.0 × 40 = 40 J.
2. Rectangle, 2.0 to 5.0 m: 3.0 × 40 = 120 J.
3. Trapezium, 5.0 to 6.0 m: ½ × (40 + 10) × 1.0 = 25 J.
4. Total work = 40 + 120 + 25 = 185 J.
5. The track is smooth, so this work becomes kinetic energy: ½ × 5.0 × v² = 185.
6. v² = 74, so v = 8.6 m s⁻¹.

Using the peak force of 40 N for the whole 6.0 m would overestimate the work.

Power: P = ΔW/Δt = Fv

Power is the rate of doing work, which is the same as the rate of energy transfer:

P = ΔW/Δt, in watts (W), where 1 W = 1 J s⁻¹.

If a constant force F moves an object at constant speed v along its own line, the work in time Δt is F × vΔt. Dividing by Δt gives

P = Fv

This is the power delivered by that force at that instant. For a vehicle at steady speed, the driving force equals the total resistive force, so the engine power needed is resistive force × speed. Doubling the speed against a constant resistive force doubles the power. Where drag grows with speed, the power rises faster than that.

Always check which force F refers to. In P = Fv it is the force doing the work, in the direction of the velocity.

Worked example 3: a van on a hill

A van of mass 1800 kg climbs a road that rises 1.0 m for every 12 m travelled along it. It travels at a steady 15 m s⁻¹ against a constant 650 N of resistive forces. Take g = 9.81 N kg⁻¹. Find the useful power at the wheels.

1. The weight component down the slope is mg × (1/12) = 1800 × 9.81 ÷ 12 = 1472 N.
2. Constant speed means zero resultant force, so the driving force is 650 + 1472 = 2122 N.
3. P = Fv = 2122 × 15 = 3.18 × 10⁴ W, or 32 kW.

Efficiency

No device turns all its input into useful output. The rest spreads into the surroundings, mostly as thermal energy.

Efficiency = useful output power / input power

The same ratio holds for energy: useful energy output / total energy input. The result is a number from 0 to 1, or a percentage once multiplied by 100. A value above 100% would break energy conservation.

Worked example 4: a hoist

A motor with an electrical input of 3.5 kW raises a 120 kg load at a steady 1.8 m s⁻¹. Find the efficiency.

1. Useful output power = Fv = mgv = 120 × 9.81 × 1.8 = 2119 W.
2. Efficiency = 2119 ÷ 3500 = 0.605, which is 61%.
3. The remaining 3500 − 2119 = 1381 W, about 1.4 kW, is dissipated.

Conservation of energy

Energy cannot be created or destroyed. In an isolated system the total stays fixed while energy moves between stores and forms.

When resistive forces act, mechanical energy (kinetic plus potential) is not conserved on its own. The missing energy is not destroyed: it becomes thermal energy in the object and its surroundings. The work done against resistive forces measures it.

A reliable method for any energy question:

1. List the energy forms at the start and at the end (kinetic, gravitational, elastic, thermal).
2. Include any work done on the system, or any work done against resistive forces.
3. Write: initial energy = final energy + energy dissipated.
4. Solve for the unknown.

Gravitational potential energy and kinetic energy

ΔEp = mgΔh is the change in gravitational potential energy for a height change Δh near the Earth’s surface, where the field strength g is uniform. Only the vertical change matters, never the path length.

Ek = ½mv² is the kinetic energy of a mass m moving at speed v. Because v is squared, doubling the speed quadruples the kinetic energy.

Worked example 5: sliding down a rough ramp

A 3.0 kg block is released from rest and slides 4.0 m down a ramp inclined at 25° to the horizontal. A constant friction force of 6.0 N acts up the ramp. Find the speed at the bottom. Take g = 9.81 N kg⁻¹.

1. Height lost: Δh = 4.0 × sin 25° = 1.69 m.
2. Potential energy lost: mgΔh = 3.0 × 9.81 × 1.69 = 49.8 J.
3. Work done against friction: 6.0 × 4.0 = 24 J.
4. Kinetic energy at the bottom: 49.8 − 24 = 25.8 J.
5. ½ × 3.0 × v² = 25.8, so v² = 17.2 and v = 4.1 m s⁻¹.

With no friction the same calculation gives √(2gΔh) = 5.8 m s⁻¹.

Elastic potential energy

A stretched or compressed spring stores elastic potential energy. For a spring obeying Hooke’s law, the force rises in proportion to the extension, so the work done in stretching it is the area of a triangle on a force-extension graph. This gives energy stored = ½FΔL, which equals ½kΔL² when F = kΔL. The spring constant k and Hooke’s law are in section 3.2.9, which has its own treatment. Here you only need to use the stored energy in energy-conservation problems.

Worked example 6: a spring launcher

A spring with k = 240 N m⁻¹ is compressed by 0.15 m. It fires a 0.060 kg ball vertically upward. The ball leaves the spring at its natural length, 0.15 m above the start. Ignore air resistance.

1. Elastic energy stored: ½ × 240 × 0.15² = 2.7 J.
2. Some of this lifts the ball through the 0.15 m while it is still in contact: 0.060 × 9.81 × 0.15 = 0.088 J.
3. Kinetic energy at release: 2.7 − 0.088 = 2.61 J, so v = √(2 × 2.61 ÷ 0.060) = 9.3 m s⁻¹.
4. Further rise: 2.61 = 0.060 × 9.81 × h, so h = 4.4 m above the release point.

Fired horizontally on a smooth surface, the same spring would give v = √(2 × 2.7 ÷ 0.060) = 9.5 m s⁻¹, since no energy goes into height.

Qualitative energy questions

You may be asked to describe energy transfers without numbers. Name every store involved, say which way each changes, and say where dissipated energy goes. For example, a child on a swing gives up gravitational potential energy as the swing falls, gains kinetic energy at the lowest point, and then converts it back. Each swing the height drops slightly because air resistance and friction at the pivot do work, transferring energy to the surroundings as thermal energy. A push from behind adds energy back.

Common errors

• Using the path length rather than the vertical height in mgΔh.
• Taking F, not F cos θ, when the force is at an angle to the displacement.
• Using the peak force instead of the area when the force varies.
• Writing P = Fv with the wrong force. For a vehicle at steady speed it is the driving force, equal to the resistive force on level ground.
• Saying energy is “gone” or “lost” rather than naming where it went.
• Not squaring v in ½mv² before multiplying by the mass.
• Giving an efficiency above 100%, or treating it as a number with a unit.
• Leaving out a negative sign for work done against a resistive force when adding energy terms.

Try the practice set when you can do the worked examples without looking.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.2.7 and 3.2.8.