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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?

Respuesta :

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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