Problem 5.32 - enhanced - with feedback 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 22 m/s on level rails. you may want to review (pages 138 - 142). part a if the engineer disengages the engine, how much time will it take the locomotive to coast to a stop?

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Chemistry

Joshua Wolfe

Today, 12:48

Problem 5.32 - enhanced - with feedback 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 22 m/s on level rails. you may want to review (pages 138 - 142). part a if the engineer disengages the engine, how much time will it take the locomotive to coast to a stop?

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Jovanni Douglas · Answer 1

The frictional force can be calculate as:

Ff = μr * N

where μr is the frictional constant while N is the normal force which is also equivalent to weight, hence:

Ff = 0.002 * 180,000 kg * 9.81 m/s^2

Ff = 3,531.6 N

The frictional force is also equivalent to the product of mass and acceleration, so we can find a:

Ff = m * a

a = 3,531.6 N / 180,000 kg

a = 0.01962 m/s^2 (in negative direction)

We can solve for time using the formula:

v = vi + a t

where v is final velocity = 0, vi is initial velocity = 22 m/s, t is time

0 = 22 - 0.01962 * t

t = 1,121.3 seconds