With an emergency stop of a train moving at a speed of 72 km / h, the stopping distance was 100 m.

With an emergency stop of a train moving at a speed of 72 km / h, the stopping distance was 100 m. what is the coefficient of friction between the wheels of the train and the rails?

V0 = 72 km / h = 20 m / s.

V = 0 m / s.

S = 100 m.

g = 10 m / s2.

μ -?

Emergency braking of the train occurs under the action of only the friction force Ffr. For the horizontal direction, we write 2 Newton’s law: Ftr = m * a, where m is the mass of the train, a is the acceleration of the train during emergency braking.

The friction force Ftr, which acts on the train, is expressed by the formula: Ftr = μ * m * g, where μ is the friction coefficient, m * g is the train’s gravity.

m * a = μ * m * g.

a = μ * g.

We express the acceleration by the formula: a = V0 ^ 2/2 * S.

μ * g = V0 ^ 2/2 * S.

μ = V0 ^ 2/2 * S * g.

μ = (20 m / s) ^ 2/2 * 100 m * 10 m / s2 = 0.2.

Answer: the coefficient of friction between the wheels of the train and the rails is μ = 0.2.