Power of a car moving uphill

Given: $m = 1500$ kg, gradient 1 in 20 (take $\sin\theta = 1/20 = 0.05$), $v = 54$ km/h $= 54/3.6 = 15$ m/s, resistance $R = 400$ N, $\eta = 0.80$.

At constant speed the driving force along the road balances the resistance and the weight component down the slope.

$$\begin{aligned} mg\sin\theta &= 1500 \times 9.81 \times 0.05 = 735.75\ \text{N} \\ F &= R + mg\sin\theta = 400 + 735.75 = 1135.75\ \text{N} \end{aligned}$$

Power at the driving wheels

$$P_{wheels} = Fv = 1135.75 \times 15 = 17036\ \text{W} = 17.04\ \text{kW}$$

Engine power

Efficiency $= P_{wheels}/P_{engine}$:

$$P_{engine} = \frac{17.04}{0.80} = 21.30\ \text{kW} \;(\approx 28.5\ \text{hp})$$

Answer: Power at wheels = 17.04 kW; engine power = 21.30 kW.