Question

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/son 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?

Express your answer using two significant figures.

Δt =		s

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Part B

How far will the locomotive move during this time?

Express your answer to two significant figures and include the appropriate units.

Δx =

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Answer 1

No air resistance on our locomotive we need to find the rate at which velocity decrease, (acceleration), ' - a ' N = m g friction force = 0.002 * (m * g) acceleration This force produce: a = friction/mass = 0.002 * g = 0.0196 m/s^2 Now part A) v = u + at 12 = 0.0196 * t t = 610 seconds 10^+ mintues !! part B) v^2 = u^2 + 2as 12^2 = 2 * 0.0196 * s s = 3700 m rightarrow 3.7 km look out for unit.