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3 years ago

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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 25 m/s on level rails. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. Part A If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop?

1 answer:

Irina18 [472]3 years ago

7 0

Answer:

t = 0.354 hours

Explanation:

given,

coefficient of rolling friction μr=0.002

mass of locomotive = 180,000 Kg

rolling speed = 25 m/s

The force of friction = μ mg

= (.002) x (180000) x (9.8)

= 3528 N

F = m a

now,

m a = 3528 N

180000 x a = 3528

a = 0.0196 m/s²

Then apply

v = u + at

0 = 25 - 0.0196 x t

t = 1275.51 sec

t = 1275.61/3600 hours

t = 0.354 hours

time taken by the locomotive to stop = t = 0.354 hours

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