A train of mass 2 10^5 kg is moving up an inclined track (where = 0.01 ) with a constant speed of 15 m s ⁻¹ . The frictional resistance offered by the track is 5000 N . The power delivered by the train's engine is: (Take g = 10 m s ⁻² )

Step-by-step solution

Mass of the train, m = 2 10^5 kg Speed of the train, v = 15 m s ⁻¹ Frictional resistance, f = 5000 N Component of weight down the incline, W_ = mg W_ = (2 10^5) 10 0.01 = 20000 N Since the train is moving with a constant speed, the net force is zero. The force exerted by the engine must balance the downward forces. F = W_ + f F = 20000 + 5000 = 25000 N Power delivered by the engine is given by: P = F v P = 25000 15 = 375000 W = 375 kW Answer: 375 kW