A train weighing 1000 tons moves along a horizontal track. The locomotive’s traction force is 1500 kh

A train weighing 1000 tons moves along a horizontal track. The locomotive’s traction force is 1500 kh, the friction coefficient is 0.005. With what acceleration is the train moving?

m = 1000 t = 1,000,000 kg.

g = 10 m / s2.

Ft = 1500 kN = 1500000 N.

μ = 0.005.

a -?

For the movement of a train Let us write 2 Newton’s law in vector form: m * a = Ft + m * g + N + Ftr, where Ft is the traction force of a steam locomotive, m * g is the force of gravity, N is the reaction force of the rail surface, Ftr is the friction force …

ОХ: m * a = Fт – Fтр.

OU: 0 = N – m * g.

a = (Ft – Ftr) / m.

N = m * g.

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

a = (Ft – μ * m * g) / m.

a = (1,500,000 N – 0.005 * 1,000,000 kg * 10 m / s2) / 1,000,000 kg = 1.45 m / s2.

Answer: the train is moving with an acceleration a = 1.45 m / s2.