Power and Efficiency Revision Notes | A Level AQA Physics

1. Why Power and Efficiency Matter

A car engine and a bicycle both move a person from A to B — but they do it at very different rates. Power is the quantity that captures how fast energy is being transferred, and efficiency tells you how much of that energy is actually doing something useful.

These two ideas sit at the heart of engineering, transport, and energy policy. Every AQA mechanics question that involves a moving object under a driving force will almost certainly call on one or both of them.

2. Power — The Rate of Doing Work

Think about lifting a heavy box. You could lift it slowly over ten seconds, or quickly in two seconds. The work done against gravity is identical in both cases — but the second scenario demands a much higher rate of energy transfer. That rate is power.

The definition makes the unit transparent: one watt means one joule of energy transferred every second. A 60 W light bulb transfers 60 J of energy to its surroundings every second — mostly as heat, which is why efficiency matters so much here.

There is a second, equally important form of the power equation. When a constant force $F$ acts on an object moving at constant velocity $v$, the work done per second is $F \times v$. This follows directly from substituting $\Delta W = F \Delta s$ into $P = \frac{\Delta W}{\Delta t}$ and recognising that $\frac{\Delta s}{\Delta t} = v$.

Notice that $P = Fv$ applies at any instant, not just over a long interval. If the velocity is changing, you can still use it to find the instantaneous power at a particular moment — a subtlety that AQA questions exploit regularly.

3. Worked Calculations Using Both Power Equations

The two power equations are complementary tools. Some questions give you energy and time; others give you force and speed. Recognising which form to reach for first saves time under exam conditions.

Worked example: Calculating power from work done and time

A motor lifts a 40 kg crate through a vertical height of 6.0 m in 8.0 s. Calculate the power output of the motor. Take $g = 9.8$ N kg⁻¹.

1. $\Delta W = mg\Delta h = 40 \times 9.8 \times 6.0$ — Work done against gravity equals the gain in gravitational potential energy. [M1] for correct formula and substitution.
2. $40 \times 9.8 = 392$ N (weight), then $392 \times 6.0 = 2352$ J — Carry out the arithmetic in stages to avoid errors. $\Delta W = 2352$ J.
3. $P = \frac{\Delta W}{\Delta t} = \frac{2352}{8.0}$ — [M1] for correct power equation with values substituted.
4. $P = 294$ W — [A1] for correct answer with unit. Round to 3 significant figures to match the data given: $P = 294$ W.

Answer: $P = 294$ W

Worked example: Calculating power from force and velocity

A car travels at a constant speed of 25 m s⁻¹ against a total resistive force of 800 N. Calculate the power output of the engine.

1. At constant speed, driving force = resistive force = 800 N — Newton's first law: zero net force at constant velocity, so the engine force exactly balances resistance. [B1] for this reasoning.
2. $P = Fv = 800 \times 25$ — [M1] for correct equation and substitution.
3. $P = 800 \times 25 = 20\,000$ W — [A1] for correct answer. This may also be written as 20 kW.

Answer: $P = 20\,000$ W (or 20 kW)

4. Efficiency — How Much Power Is Actually Useful?

Every real machine wastes some energy. A car engine generates heat, sound, and vibration alongside the kinetic energy you actually want. Efficiency quantifies how well a system converts input power into useful output power.

Because power is energy per unit time, the same ratio holds for energy: $\text{efficiency} = \frac{\text{useful output energy}}{\text{total input energy}}$. Both forms are valid and AQA mark schemes accept either, provided you are consistent with the quantities you substitute.

A key physical constraint: efficiency can never exceed 1 (or 100%). If your calculation gives a value greater than 1, you have either inverted the fraction or used the wrong power in the numerator. This is a reliable self-check.

5. Worked Calculations Using Efficiency

Efficiency questions often combine with power calculations — you may need to find input power first using $P = Fv$ or $P = \frac{\Delta W}{\Delta t}$, then apply the efficiency ratio. Work through the chain carefully and keep track of which power is 'useful' and which is 'total input'.

Worked example: Calculating efficiency and expressing as a percentage

An electric motor has an input power of 500 W. It raises a 20 kg mass through 4.0 m in 5.0 s. Calculate the efficiency of the motor and express it as a percentage. Take $g = 9.8$ N kg⁻¹.

1. Useful work done: $\Delta W = mg\Delta h = 20 \times 9.8 \times 4.0 = 784$ J — The useful output is the gravitational potential energy gained by the mass. $20 \times 9.8 = 196$ N, then $196 \times 4.0 = 784$ J. [M1]
2. $P_{\text{useful out}} = \frac{\Delta W}{\Delta t} = \frac{784}{5.0} = 156.8$ W — [M1] for dividing energy by time correctly.
3. $\text{efficiency} = \frac{P_{\text{useful out}}}{P_{\text{in}}} = \frac{156.8}{500}$ — [M1] for correct ratio, with useful power in the numerator and total input in the denominator.
4. $\text{efficiency} = \frac{156.8}{500} = 0.3136$ — Carry the unrounded value forward to avoid compounding rounding errors.
5. $\text{efficiency} = 0.3136 \times 100 = 31.4\%$ (3 s.f.) — [A1] for correct percentage with appropriate significant figures. The motor converts about 31% of its electrical input into useful mechanical work; the remaining 69% is dissipated as heat in the motor windings and bearings.

Answer: Efficiency = 0.314 (or 31.4%)

Worked example: Finding input power from efficiency and useful output

A pump has an efficiency of 60% and delivers a useful output power of 1800 W. Calculate the input power to the pump.

1. $\text{efficiency} = \frac{P_{\text{useful out}}}{P_{\text{in}}}$ — State the equation before rearranging. [M1]
2. $P_{\text{in}} = \frac{P_{\text{useful out}}}{\text{efficiency}}$ — Multiply both sides by $P_{\text{in}}$ and divide both sides by efficiency.
3. $\text{efficiency} = 60\% = 0.60$ — Always convert to a decimal before substituting. Leaving it as 60 would give an answer 100 times too small.
4. $P_{\text{in}} = \frac{1800}{0.60} = 3000$ W — [A1] for correct answer with unit. $1800 \div 0.60 = 3000$ W. As a sanity check: 60% of 3000 = 1800 W. ✓

Answer: $P_{\text{in}} = 3000$ W

6. Connecting Power and Efficiency to Real Systems

Consider a car travelling at constant velocity on a motorway. The engine provides a driving force to overcome air resistance and friction. Using $P = Fv$, you can find the engine's mechanical output power. The efficiency then tells you how much fuel energy per second is actually converted into that mechanical output — the rest heats the engine, exhaust gases, and surroundings.

This chain of reasoning — from force and velocity to power, then from power to efficiency — is exactly the multi-step structure AQA uses in higher-tariff questions. Practising it as a sequence, rather than as isolated formula recalls, is what separates a grade 7 response from a grade 9 one.

The chart above illustrates why efficiency matters in context. A petrol engine converts only around 25% of the chemical energy in fuel into useful kinetic energy — the rest is lost as heat and sound. An electric motor, by contrast, can exceed 85% efficiency, which is one reason electric vehicles transfer significantly less energy to waste heat per kilometre travelled.

7. Exam Technique and Common Mistakes

AQA mark schemes for power and efficiency questions reward method as well as accuracy. Even if you make an arithmetic slip, you can still collect method marks — but only if your working is clearly laid out. Always write the equation, then substitute with units, then evaluate.

For extended-response questions that ask you to evaluate the efficiency of a system, structure your answer as: state the efficiency value → identify what the wasted power represents physically → compare to an ideal system → reach a conclusion about whether the system is fit for purpose. This mirrors the AQA level-of-response mark scheme structure.

Power and efficiency are two of the most frequently examined ideas in the mechanics section of AQA A Level Physics. Mastering the two power equations — $P = \frac{\Delta W}{\Delta t}$ and $P = Fv$ — and the efficiency ratio $\frac{P_{\text{useful out}}}{P_{\text{in}}}$ gives you the toolkit to tackle everything from single-step recall questions to multi-stage calculation and evaluation tasks.