An electric locomotive with a traction force of 180 kN. train weight 1200t.

An electric locomotive with a traction force of 180 kN. train weight 1200t. On which section of the track the train speed increases from 18 km / h to 36 km / h.

F = 180 kN = 180,000 N.

m = 1200 t = 1200000 kg.

V1 = 18 km / h = 5 m / s.

V2 = 36 km / h = 10 m / s.

S -?

The traversed path S of the electric locomotive during acceleration is expressed by the formula: S = V1 * t + a * t2 / 2, where V1 is the speed of the electric locomotive at the beginning of acceleration, t is the acceleration time, a is the acceleration during acceleration.

The acceleration of the electric locomotive a is found from 2 Newton’s laws: a = F / m, where F is the traction force of the electric locomotive, m is the mass of the electric locomotive.

a = 180,000 N / 1,200,000 kg = 0.15 m / s2.

The acceleration time t is found by the formula: t = (V2 – V1) / a.

t = (10 m / s – 5 m / s) / 0.15 m / s2 = 33.3 s.

S = 5 m / s * 33.3 s + 0.15 m / s2 * (33.3 s) 2/2 = 250 m.

Answer: the speed of the electric locomotive will increase at a distance of S = 250 m.