Question

An engine pulls a car of mass 1500 kg on a level road at a constant velocity of 5ms^-1. If the frictional force is 500 N, what power does the engine generate? What extra power must the engine develop to maintain the same speed up a plane inclined to have gradient in 10?

Answer 1

Since the car moves with a constant velocity, the power generated by the engine is just sufficient to overcome the frictional force.

P = F × V = 500 × 5

= 2500 W

2.5 kW

When the car moves over an inclined road, in addition to the friction, the engine must over come the component of the weight of the car along the inclined road. The component of the weight along the AB is, F = mg Sinθ

= 1500 × 9.8 × \(\frac {1}{10}\)(since Sinθ =\(\frac {1}{10}\))

∴ Extra power required = F × v

= 1500 ×\(\frac{9.8 \times 5}{10}\)

= 7350 W = 7.35 kW.