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10 September, 02:24

The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typically μr=0.002. suppose a 180,000 kg locomotive is rolling at 12 m/s on level rails. if the engineer disengages the engine, how much time will it take the locomotive to coast to a stop?

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  1. 10 September, 02:49
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    M = 180,000 kg, the mass of the locomotive.

    The normal reaction is

    N = mg

    = (9.8 m/s²) * (180,000 kg)

    = 1764,000 N

    The frictional resistance is

    F = μN

    = 0.002*1764,000

    = 3528 N

    The deceleration, a, caused by the frictional force is

    a = (3528 N) / (180,000 kg)

    = 0.0196 m/s²

    The initial velocity of the train is 12 m/s.

    The time, t, required for the train to coast to a stop is

    (12 m/s) - (0.0196 m/s²) * (t s) = 0

    t = 612.24 s

    Answer: 612.2 s
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