A freight electric locomotive, when starting off, develops a maximum traction force of 380 kN.

| August 13, 2021 | education

A freight electric locomotive, when starting off, develops a maximum traction force of 380 kN. What acceleration will it impart to a composition with a mass of 500 tons, if the resistance force is 250 kN?

These tasks: Ft (maximum traction force of the considered freight electric locomotive) = 380 kN (380 * 10 ^ 3 N); m (mass of the composition) = 500 t (in the SI system m = 500 * 10 ^ 3 kg); Fcopr (resistance force) = 250 kN (250 * 10 ^ 3 N).

The acceleration that the considered freight electric locomotive will report to the train can be expressed from the formula: m * a = Ft – Fcopr, whence a = (Ft – Fcopr) / m.

Calculation: a = (380 * 10 ^ 3 – 250 * 10 ^ 3) / (500 * 10 ^ 3) = 0.26 m / s2.

Answer: The considered freight electric locomotive will give the train an acceleration of 0.26 m / s2.



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