Question

The rolling resistance for steel on steel is quite low; the coefficient of rolling friction is typicallyµ,,= 0.002. Suppose a 180,000 kg locomotive is rolling at 10 mis (just over 20 mph) on level rails. If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? How far will the locomotive move during this time?

Answer 1

Step-by-step explanation:

Given

coefficient of kinetic friction
`\mu _k=0.002`

mass of locomotive=180,000 kg

v=10 m/s

deceleration provided by friction
`a=\mu _kg`

`a=0.002* 9.8=0.0196 m/s^2`

Time taken to stop completely is

v=u+at

where v=final velocity

u=initial velocity

a=acceleration

`0=10-0.0196* t`

t=510.20 s

Distance traveled during this point

`v^2-u^2=2 a s`

`0-10^2=2(-0.0196)s`

`s=(100)/(0.0392)`

s=2551.02 m