At a speed of 54 km / h, electric train motors consume 900 kW of power.

At a speed of 54 km / h, electric train motors consume 900 kW of power. what is the magnitude of the resistance force during its movement?

V = 54 km / h = 15 m / s.

N = 900 kW = 900000 W.

Fsopr -?

Let us express the power of the electric train N by the formula: N = A / t, where t is the engine operation time, A is the mechanical work performed by the engine.

We express the mechanical work of the engine A by the formula: A = F * S, where F is the traction force of the electric train engine, S is the movement of the train.

The ratio of the traversed path S in the time of its passage t is called the speed V: V = S / t.

The formula for determining the power of the train will take the form: N = F * S / t = F * V.

Since the train moves uniformly, then according to Newton’s 1 law: the traction force F is equal to the resistance force Fcopr: F = Fcopr.

N = Fcopr * V.

Fcopr = N / V.

Fcopr = 900000 W / 15 m / s = 60,000 N.

Answer: when an electric train is moving, the resistance force is Fcopr = 60,000 N.