With what speed does a train weighing 600 tons evenly rise along a slope with a slope of 0.005

With what speed does a train weighing 600 tons evenly rise along a slope with a slope of 0.005, if the friction force is 10 kN and the engine power of the locomotive is 800 kW?

m = 600 t = 600,000 kg.

g = 10 N / kg.

sinα = 0.005.

Ftr = 100 kN = 100000 N.

N = 800 kW = 800000 W.

V -?

We express the power of the train N by the formula: N = F * V, where F is the traction force of the train, V is the speed of its movement.

V = N / F.

Since the train rises evenly, it means that the action of all forces on it is compensated:

F + Ftr + N + m * g = 0.

ОХ: F – Ftr – m * g * sinα = 0.

OU: N – m * g * cosα = 0.

F = Ftr + m * g * sinα.

N = m * g * cosα.

F = 800,000 N + 600,000 kg * 10 N / kg * 0.005 = 830,000 N.

V = 800,000 W / 830,000 N = 0.96 m / s.

Answer: the train rises at a speed of V = 0.96 m / s.