A motor car of mass 1100 kg travels at a uniform speed of 100 km.hr-1. It is subjected to two forces: The frictional resistance offered by the road surface, -Froad = 0.013 mg, where m is the mass of the car. The force due to the air resistance, -Fair = 0.35 (N.s2.m-2) v2, where v is the velocity of the car.

What power does the engine put out? (g = 9.80 m.s^-2.)

Solution:

A velocity of 100 km.hr^-1 = 100000 (m)/3600 (s) = 27.8 m.s^-1.

Force required to overcome friction = F_road = 0.013 x 1100 (kg) x 9.80 (m.s^-2) = 140 N

Force required to overcome air resistance = F_air = 0.35 (N.s².m^-2) x 27.8² (m.s^-1)² = 270 N

The total force overcome by the engine is therefore F = 140 + 270 = 410 N.

The average power is P = Fv = 410 (N) x 27.8 (m.s^-1) = 11400 W = 11.4 kW.