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?

Question

Dulce Mckee

Physics

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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Answers (1)

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Jayleen Hancock · Answer 1

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