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

These notes are aligned to the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.2.7 (work, energy and power) and 3.2.8 (conservation of energy). Both are shared International AS and A-level content. For full explanations, start with the study guide. Test yourself with the practice set, keep the A-level Physics course page and the checklist to hand, and use the free diagnostics to find gaps. Related reading: Motion and Newton’s laws.

Key definitions

• Work done (energy transferred) by a force is the force multiplied by the displacement in the direction of the force.
• Power is the rate of doing work, which is the rate of energy transfer.
• Efficiency is useful output power divided by input power (or useful energy output divided by total energy input).
• Principle of conservation of energy: energy is never created or destroyed. In an isolated system the total stays the same while energy moves between forms and stores.

Formulae (3.2.7 and 3.2.8)

Quantity	Relationship	Unit and notes
Work done	W = Fs cos θ	J. θ is the angle between F and s
Work from a graph	W = area under force-displacement graph	J. Below the axis counts as negative
Power	P = ΔW/Δt	W (J s⁻¹)
Power and velocity	P = Fv	W. F is the force doing the work, along v
Efficiency	useful output power / input power	No unit. Multiply by 100 for a percentage
Gravitational potential energy	ΔEp = mgΔh	J. Vertical height only. Uniform field
Kinetic energy	Ek = ½mv²	J. Speed is squared
Elastic energy (spring)	½FΔL = ½kΔL²	J. Area under force-extension graph. Hooke’s law is in 3.2.9

Work and the angle

Angle between F and s	cos θ	Work
0°	1	Maximum positive, W = Fs
60°	0.5	Half of Fs
90°	0	Zero (for example normal force on a sliding block, centripetal force on a circling object)
180°	−1	Negative, W = −Fs (friction or drag opposing motion)

Method for work from a graph

1. Check the axes: force (not mass or time) against displacement.
2. Split the shape into rectangles, triangles and trapeziums.
3. Find each area and add. Subtract any area below the axis.
4. Give the answer in joules and, where asked, link it to the kinetic energy gained.

Worked reminder. A 120 N force acts at 60° to an 8.0 m displacement: W = 120 × 8.0 × cos 60° = 480 J.

Power and efficiency

• Steady speed on level ground: driving force = resistive force, so P = Fv with F the resistive force.
• Steady speed up a slope: F = resistive force + weight component along the slope (mg sin θ).
• If power is fixed and speed rises, the driving force falls. At maximum speed, driving force = resistive force and the acceleration is zero.
• Resultant force = driving force − resistive force. Use this with F = ma when the speed is not steady.
• Average power over a time: energy transferred ÷ time.

Worked reminder (power). An 18 kW engine drives at a steady 30 m s⁻¹: F = P ÷ v = 18 000 ÷ 30 = 600 N.

Worked reminder (efficiency). A machine is 85% efficient and gives 2.4 kW of useful power. Input power = 2400 ÷ 0.85 = 2.8 kW (2824 W). Wasted power = 2824 − 2400 = 424 W.

Efficiency pointers

• Efficiency compares useful output with total input, for the same time or the same energy.
• Never above 100%. If you get that, you have inverted the ratio or left out a loss.
• “Energy wasted” means spread out into the surroundings, mostly as thermal energy. It still exists.

Conservation of energy: method in steps

1. Pick two points: a start and an end.
2. List the energy at each: Ek, gravitational, elastic.
3. Add any work done on the system (a motor, a push) at the start side.
4. Add any work done against resistive forces at the end side.
5. Equate: initial energy + work in = final energy + work against resistance.
6. Solve. Convert height changes with Δh = (distance along slope) × sin(angle).

Worked reminder (free fall). A ball is dropped from rest through 1.8 m. Ignoring air resistance, mgΔh = ½mv², so v = √(2gΔh) = √(2 × 9.81 × 1.8) = 5.9 m s⁻¹. The mass cancels.

Worked reminder (spring). A spring with k = 800 N m⁻¹ is compressed 0.060 m. Energy stored = ½ × 800 × 0.060² = 1.44 J.

Worked reminder (resistance). If a 1.0 kg object slides down 2.0 m vertically and reaches 5.0 m s⁻¹ (g = 9.81 N kg⁻¹), the kinetic energy is ½ × 1.0 × 25 = 12.5 J, the potential energy lost is 19.62 J, and the work against resistive forces is 19.62 − 12.5 = 7.1 J.

Must-know distinctions

Compare	Key difference
Work and power	Work is energy transferred (J). Power is the rate of transfer (W)
Energy lost and dissipated	Energy is never lost. It spreads to the surroundings, mostly as thermal energy
Height and distance	mgΔh uses vertical height. Work against friction uses distance along the path
F and F cos θ	Only the component along the displacement does work
Efficiency of power and energy	Same ratio. Use the same time interval for both
Ek and momentum	Ek depends on v², is a scalar and is never negative. Momentum depends on v, is a vector
Gravitational and elastic energy	mgΔh needs a height change. ½kΔL² needs an extension or compression

Qualitative energy chains

Write each as a chain with the direction of change and the dissipation:

• Bouncing ball: gravitational potential to kinetic as it falls. In contact with the floor, kinetic goes to elastic energy in the ball, then back to kinetic. Each bounce is lower because energy is dissipated as thermal energy and sound in the ball and floor, and by air resistance.
• Pendulum: gravitational potential at the ends, kinetic at the lowest point. Air resistance slowly reduces the amplitude.
• Catapult or bow: the person does work to store elastic energy. This transfers to kinetic energy of the projectile.
• Vehicle braking: kinetic energy goes to thermal energy in the brakes and the tyres, and to the surroundings.

Where marks are usually lost

• Using F instead of F cos θ, or sin instead of cos, when the angle is between force and displacement. Check for a right-angled triangle.
• Taking the slope length as Δh. Resolve with sin of the angle.
• Reading a graph’s area wrongly: a trapezium is ½(a + b) × width, a triangle is ½ × base × height.
• Using P = Fv when the speed is not steady without stating which force is meant.
• Writing that energy is “gone” or “lost”. Name where it went.
• Dropping units, or not converting kW to W and kJ to J.
• Giving efficiency as a percentage greater than 100, or writing it with a unit.
• Not squaring speed in ½mv², or squaring the whole of ½m.
• Quoting too many significant figures, or rounding a figure part-way through. Keep extra digits until the end.

Quick self-test

1. A 50 N force acts at 40° to the displacement of 4.0 m. Find the work done.
2. Find the kinetic energy of a 0.40 kg ball travelling at 6.0 m s⁻¹.
3. A 2.5 kg mass is raised 3.2 m. Find its gain in gravitational potential energy (g = 9.81 N kg⁻¹).
4. A machine does 900 J of work in 15 s. What is its power?
5. A force rises uniformly from 0 to 30 N over 0.50 m. Find the work done.
6. A 450 N driving force moves a vehicle at a steady 8.0 m s⁻¹. What is the power?
7. A device takes in 1500 J and gives out 600 J of useful energy. State its efficiency.
8. An 18 J kinetic energy belongs to a 0.25 kg object. Find its speed.
9. Ignoring air resistance, what speed does a stone reach after falling 5.0 m from rest? (g = 9.81 m s⁻²)
10. By what factor does kinetic energy change if the speed is doubled?
11. What work is done by a 20 N force that stays at right angles to the displacement?
12. A ball is thrown up at 12 m s⁻¹. Ignoring air resistance, find the maximum height.

Answers

1. W = 50 × 4.0 × cos 40° = 153 J.
2. Ek = ½ × 0.40 × 6.0² = 7.2 J.
3. ΔEp = 2.5 × 9.81 × 3.2 = 78 J.
4. P = 900 ÷ 15 = 60 W.
5. Area of triangle = ½ × 0.50 × 30 = 7.5 J.
6. P = Fv = 450 × 8.0 = 3600 W (3.6 kW).
7. 600 ÷ 1500 = 0.40, so 40%.
8. v = √(2 × 18 ÷ 0.25) = 12 m s⁻¹.
9. v = √(2 × 9.81 × 5.0) = 9.9 m s⁻¹.
10. Ek is proportional to v², so it rises by a factor of 4.
11. cos 90° = 0, so the work done is zero.
12. h = v² ÷ 2g = 144 ÷ 19.62 = 7.3 m.

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.