# What speed can a tram develop on an uphill slope of 10 °, having four electric motors of 55

**What speed can a tram develop on an uphill slope of 10 °, having four electric motors of 55 kW each, if the weight of the tram with passengers is 30 tons, and the friction coefficient is 0.05?**

∠α = 10 °.

P1 = 55 kW = 55000 W.

m = 30 t = 30,000 kg.

g = 10 m / s2.

n = 4.

μ = 0.05.

V -?

Since the tram moves uniformly, it means, according to 1 Newton’s law, the effect of forces on it is compensated.

F + Ftr + N + m * g = 0 – vector.

ОХ: F – Ftr – m * g * sinα = 0

OU: N – m * g * cosα = 0.

N = m * g * cosα.

F = Ftr + m * g * sinα.

Ftr = μ * N = μ * m * g * cosα.

F = μ * m * g * cosα + m * g * sinα = (μ * cosα + sinα) * m * g.

The power P of the tram with uniform movement is expressed by the formula: P = F * V = (μ * cosα + sinα) * m * g * V.

P = n * P1, where n is the number of electric motors, P1 is the power of one motor.

(μ * cosα + sinα) * m * g * V = n * Р1.

V = n * Р1 / (μ * cosα + sinα) * m * g.

V = 4 * 55000 W / (0.05 * cos10 ° + sin10 °) * 30000 kg * 10 m / s2 = 3.3 m / s.

Answer: the tram will move at a speed of V = 3.3 m / s.