# The freight train moves on a flat section of the road at a speed of 60 km / h, while the electric locomotive

**The freight train moves on a flat section of the road at a speed of 60 km / h, while the electric locomotive develops a net power of 100 kW. At what speed (in km / h) it is necessary to climb the section with a slope of 1 m by 200 m of the path so that the developed power is equal to 120 kW? The drag force is 0.01 times the gravity of the train.**

V1 = 60 km / h = 16.6 m / s.

N1 = 100 kW = 100000 W.

h = 1 m.

L = 200 m.

Fcopr = 0.01 * m * g.

N2 = 120 kW = 120,000 W.

V2 -?

With uniform movement, the power is determined by the formula: N = Ft * V, where Ft is the traction force of the train, V is the speed of its movement.

Since the train moves uniformly, the traction force Fт1 is equal to the resistance force Fcop: Ft1 = Fcop.

The resistance force is determined by the formula: Fcopr = 0.01 * m * g.

N1 = 0.01 * m * g * V1.

m * g = N1 / 0.01 * V1.

When lifting, the traction force is equal to Fт2 = Fсopr + m * g * sinα, where ∠α is the angle of inclination of the road section.

sinα = h / L.

Ft2 = 0.01 * m * g + m * g * h / L.

N2 = (0.01 * m * g + m * g * h / L) * V2.

V2 = N2 / (0.01 * m * g + m * g * h / L).

V2 = N2 / (N1 / V1 + N1 * h / V1 * 0.01 * L).

V2 = 120,000 W / (100,000 W / 16.6 m / s + 100,000 W * 1 m / 16.6 m / s * 0.01 * 200 m) = 13.2 m / s.

Answer: the train must move at a speed V2 = 13.2 m / s.