A car of mass 800kg moves up an incline of 1 in 20 (1 in 20 means for every 20m along the road the car gains 1m in height) at a constant speed of 20m/s. The frictional force opposing motion is 500N. How much work has been done by the engine after the car has moved 50m?

Answer:

27) D

Height of car after 50m = 50/20 = 2.5

PE = mgh
PE = 800 x 10 x 2.5
PE = 8000 x 2.5 = 20,000J

Force of engine against friction = 500N
Work done against friction = Force x distance = 500 x 50 = 25,000J

Total work done = 20,000 + 25,000 = 45,000J = 45 kJ