---
title: "Power of a car moving uphill"
subject: "Engineering Mechanics"
chapter: 3
key: "semester_3/engineering_mechanics:car-power-uphill-efficiency"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/car-power-uphill-efficiency
site: BCT Engineering
---

# Power of a car moving uphill

Engineering Mechanics · Chapter 3: Kinetics of Particles: Work Energy Principles

- Asked in: year not given, 5 marks
- Video: [Physics 8  Work, Energy, and Power (16 of 37) Driving Up A Hill](https://www.youtube.com/watch?v=WEGjFjCOWB8) — Michel van Biezen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/car-power-uphill-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-3

## Question

A car of mass 1500 kg moves up a road with a gradient of 1 in 20 (1 vertical to 20 along the road) at a constant speed of 54 km/h. The total resistance to motion is 400 N. Determine the power developed at the driving wheels. If the transmission efficiency from engine to wheels is 80 %, find the power output required from the engine.

## Short answer

Tractive force balances resistance plus weight component: F = 400 + 1500×9.81×0.05 = 1135.75 N. Power at wheels is Fv = 1135.75×15 = 17.04 kW. Engine power required at 80% transmission efficiency is 17.04/0.80 = 21.30 kW.

## Answer

**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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. At constant speed up the gradient, the driving force equals:
   A. Resistance only
   B. Resistance + m g sin(theta)
   C. Resistance - m g sin(theta)
   D. m g cos(theta)
   Answer: B. Net force is zero, so F = 400 + 735.75 = 1135.75 N. (Key point: Driving force)

2. The speed 54 km/h in m/s is:
   A. 54 m/s
   B. 19.4 m/s
   C. 15 m/s
   D. 5.4 m/s
   Answer: C. 54/3.6 = 15 m/s. (Key point: Speed conversion)

3. The power at the driving wheels is:
   A. F x v = 1135.75 x 15 = 17.04 kW
   B. 400 x 15 = 6 kW
   C. 735.75 x 15 = 11.0 kW
   D. 1135.75 / 15 = 75.7 W
   Answer: A. P = F v = 17036 W. (Key point: Wheel power)

4. With transmission efficiency 80%, the engine power is:
   A. 17.04 x 0.8 = 13.6 kW
   B. 17.04 / 0.8 = 21.30 kW
   C. 17.04 + 0.8 kW
   D. 17.04 kW
   Answer: B. Efficiency = P_wheels / P_engine, so the engine must produce more than the wheel power. (Key point: Engine power)

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