The electric locomotive, when moving off the train station, develops a maximum traction force of 650 kN.

The electric locomotive, when moving off the train station, develops a maximum traction force of 650 kN. What acceleration it will impart to the train with a mass of 3250 tons, if the drag coefficient is 0.005.

Ft = 650 kN = 650,000 N.

m = 3250 t = 3250000 kg.

g = 9.8 m / s ^ 2.

μ = 0.005.

a -?

Let’s write 2 Newton’s law in vector form: m * a = Ft + Ftr + N + m * g.

Let’s find expressions 2 of Newton’s law in projections on the axis.

ОХ: m * a = Fт – Fтр.

OU: 0 = N – m * g.

a = (Ft – Ftr) / m.

N = m * g.

The friction force is determined by the formula: Ftr = μ * N = μ * m * g.

We find the acceleration of the train by the formula: a = (Fт – μ * m * g) / m.

a = (650,000 N – 0.005 * 3250,000 kg * 9.8 m / s ^ 2) / 3250,000 kg = 0.151 m / s ^ 2.

Answer: the train will have an acceleration a = 0.151 m / s ^ 2.