What work does an electric locomotive do in 10 minutes, moving a train with a mass of 3000 tons

What work does an electric locomotive do in 10 minutes, moving a train with a mass of 3000 tons along a horizontal path at a constant speed of 72 km / h, if the friction coefficient is 0.005?

t = 10 min = 600 s.

m = 3000 t = 3000000 kg.

g = 10 m / s2.

V = 72 km / h = 20 m / s.

μ = 0.005.

А – ?

Since the movement of the electric locomotive S coincides with the direction of its traction force Ft, its mechanical work A will be determined by the formula: A = Ft * S.

Since the electric train moves uniformly, the traction force of the train Ft is equal to the friction force Ftr: Ft = Ftr.

The friction force Ffr is expressed by the formula: Ffr = μ * m * g.

We express the displacement S by the formula: S = V * t.

A = μ * m * g * V * t.

A = 0.005 * 300000 kg * 10 m / s2 * 20 m / s * 600 s = 180,000,000 J = 180 MJ.

Answer: when moving the train, the electric train performs work A = 180 MJ.