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 19 m/s on level rails. A) If the engineer disengages the engine, how much time will it take the locomotive to coast to a stop? B) How far will the locomotive move during this time?

Answer 1

M*g = 180,000 * 9.8 = 1,764,000 N. = Wt. of locomotive = Normal force(Fn).

Fr = ur*Fn = 0.002 * 1,764,000 = 3528 N. = Force of rolling friction.

A. Fap-Fr = M*a.
0-3528 = 180,000*a
a = -0.0196 m/s^2.

V = Vo+a*t.
0 = 19 - 0.0196t.
t = 969.4s.

B. V^2 = Vo^2 + 2a*d.
0 = 19^2 - 2*0.0196*d.
d = ?.