# Locomotive traction forces F = 240 kN. The locomotive, moving evenly in 14.4 minutes, covers the distance

Locomotive traction forces F = 240 kN. The locomotive, moving evenly in 14.4 minutes, covers the distance between the two stations equal to 10.8 km. Determine the work done by the traction force of the locomotive and its power.

Initial data: F (locomotive traction force) = 240 kN (240 * 10 ^ 3 N); the locomotive movement is uniform; t (duration of the locomotive movement) = 14.4 min (864 s); S (distance between two stations) = 10.8 km (10.8 * 10 ^ 3 m).

1) The work performed by the traction force of the locomotive: A = F * S = 240 * 10 ^ 3 * 10.8 * 103 = 2592 * 10 ^ 6 J (2592 MJ).

2) Locomotive power: N = A / t = 2592 * 10 ^ 6/864 = 3 * 10 ^ 6 W (3 MW).

Answer: The locomotive has completed work of 2592 MJ, the power of the locomotive is 3 MW.

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