A train of mass 2 × 10^5 kg is moving up an inclined track (where θ...

Mass of the train, $m = 2 \times 10^5 \text{ kg}$

Speed of the train, $v = 15 \text{ m s}^{-1}$

Frictional resistance, $f = 5000 \text{ N}$

Component of weight down the incline, $W_{\parallel} = mg \sin\theta$

$W_{\parallel} = (2 \times 10^5) \times 10 \times 0.01 = 20000 \text{ N}$

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.

$F = W_{\parallel} + f$

$F = 20000 + 5000 = 25000 \text{ N}$

Power delivered by the engine is given by:

$P = F \times v$

$P = 25000 \times 15 = 375000 \text{ W} = 375 \text{ kW}$

Answer: $375 \text{ kW}$