Physics 9702 · Paper 2

Power & Speed: P = Fv

For anything moving at a steady speed against a force, there is a faster route to the power than counting joules.
The derivation is two lines, and the exam asks for it in full.

The two-line derivation

Worked example

Show that P = Fv

A force F moves an object through displacement x in time t at constant velocity v, in the direction of the force. Use the definition of power to show that P = Fv. (9702/21/M/J/22 Q3(a)(ii))

Work done: $W = Fx$ .
Power: $P = \dfrac{W}{t} = \dfrac{Fx}{t} = Fv$ , because $v = x/t$ .

Answer

P = Fv

The same derivation was set twice in the 2025 May/June session. The step the mark scheme wants named is $x = vt$ (9702/12/F/M/25 Q14).

At constant velocity the resultant force is zero (lesson 3.01), so the driving force exactly equals the resistive force. That is why P = Fv so often uses the drag value as F (9702/12/F/M/24 Q16).
Lifting at constant speed is the vertical version: the force is the weight, so $P = mgv$ . A 120 kg load rising at 2.5 m s⁻¹ needs $120 \times 9.81 \times 2.5 = 2.9 \times 10^{3} \text{ W}$ (9702/11/M/J/25 Q12).

Using it: cars and slopes

Worked example

Exam version: an electric car

An electric car travels at a constant 35 m s⁻¹ on a level road against a total resistive force of 1750 N. Find the power delivered to the wheels, and the useful work done over 17 km. (9702/21/M/J/25 Q3(b)(i)-(ii))

Constant speed: driving force = 1750 N. $P = Fv = 1750 \times 35 = 6.1 \times 10^{4} \text{ W}$ .
$W = Fs = 1750 \times 17\,000 = 3.0 \times 10^{7} \text{ J}$ .

Answer

P = 6.1 \times 10^{4} \text{ W}, \; W = 3.0 \times 10^{7} \text{ J}

Worked example

One change: climbing a slope

A 1500 kg car climbs a 6.0° slope at a constant 30 m s⁻¹ against a resistive force of 1600 N. Show that it gains about 46 000 J of GPE each second, then find the engine's output power. (9702/23/M/J/25 Q3(b))

Height gained per second: $30\sin 6.0^\circ$ m, so $\dfrac{\Delta E_p}{t} = 1500 \times 9.81 \times 30\sin 6.0^\circ = 46\,000 \text{ J s}^{-1}$ .
The engine also pushes against resistance: $P_{\text{drag}} = Fv = 1600 \times 30 = 48\,000 \text{ W}$ .
Total: $P = 48\,000 + 46\,000 = 9.4 \times 10^{4} \text{ W}$ .

Answer

P = 9.4 \times 10^{4} \text{ W}

Your turn— tap to reveal the worked answer (9702/21/M/J/25 Q3(c))

The same car now goes down a slope at the same speed. State the effect on the air resistance, and on the current drawn by the motor. (9702/21/M/J/25 Q3(c))

Answer: air resistance is unchanged (same speed). The motor supplies less power, because the fall in GPE now helps drive the car, so the current is smaller.

When drag grows with speed: P grows like v³

For cars and aircraft the drag is roughly proportional to $v^2$ . If drag is 800 N at 20 m s⁻¹, at 40 m s⁻¹ it is $800 \times 4 = 3200 \text{ N}$ , and the power needed is $3200 \times 40 = 128 \text{ kW}$ : sixteen times the force-power product at half the speed would suggest. Twice the speed needs eight times the power (9702/12/M/J/24 Q14).

See “constant speed” in a power question? First write the force balance, then multiply the balancing force by v. Add the rate of GPE gain if the road climbs.

PreviousPower NextEfficiency

Power & Speed: P = Fv

For anything moving at a steady speed against a force, there is a faster route to the power than counting joules.
The derivation is two lines, and the exam asks for it in full.

The two-line derivation

Worked example

Show that P = Fv

A force F moves an object through displacement x in time t at constant velocity v, in the direction of the force. Use the definition of power to show that P = Fv. (9702/21/M/J/22 Q3(a)(ii))

Work done: $W = Fx$ .
Power: $P = \dfrac{W}{t} = \dfrac{Fx}{t} = Fv$ , because $v = x/t$ .

Answer

P = Fv

The same derivation was set twice in the 2025 May/June session. The step the mark scheme wants named is $x = vt$ (9702/12/F/M/25 Q14).

At constant velocity the resultant force is zero (lesson 3.01), so the driving force exactly equals the resistive force. That is why P = Fv so often uses the drag value as F (9702/12/F/M/24 Q16).
Lifting at constant speed is the vertical version: the force is the weight, so $P = mgv$ . A 120 kg load rising at 2.5 m s⁻¹ needs $120 \times 9.81 \times 2.5 = 2.9 \times 10^{3} \text{ W}$ (9702/11/M/J/25 Q12).

Using it: cars and slopes

Worked example

Exam version: an electric car

Constant speed: driving force = 1750 N. $P = Fv = 1750 \times 35 = 6.1 \times 10^{4} \text{ W}$ .
$W = Fs = 1750 \times 17\,000 = 3.0 \times 10^{7} \text{ J}$ .

Answer

P = 6.1 \times 10^{4} \text{ W}, \; W = 3.0 \times 10^{7} \text{ J}

Worked example

One change: climbing a slope

Height gained per second: $30\sin 6.0^\circ$ m, so $\dfrac{\Delta E_p}{t} = 1500 \times 9.81 \times 30\sin 6.0^\circ = 46\,000 \text{ J s}^{-1}$ .
The engine also pushes against resistance: $P_{\text{drag}} = Fv = 1600 \times 30 = 48\,000 \text{ W}$ .
Total: $P = 48\,000 + 46\,000 = 9.4 \times 10^{4} \text{ W}$ .

Answer

P = 9.4 \times 10^{4} \text{ W}

Your turn— tap to reveal the worked answer (9702/21/M/J/25 Q3(c))

The same car now goes down a slope at the same speed. State the effect on the air resistance, and on the current drawn by the motor. (9702/21/M/J/25 Q3(c))

Answer: air resistance is unchanged (same speed). The motor supplies less power, because the fall in GPE now helps drive the car, so the current is smaller.

When drag grows with speed: P grows like v³

See “constant speed” in a power question? First write the force balance, then multiply the balancing force by v. Add the rate of GPE gain if the road climbs.