Question

A 69990 kg locomotive is traveling at 9.86 m/s when its engine and brakes both fail. How far will the locomotive roll before it comes to a stop? (The coefficient of rolling friction is 0.002.)

Answer 1

Not sure, but I'll take a crack at it since no one else answered the question. KE=Force of Friction * distance. 1/2(MV^2)=mg *U*d= 1/2[(69,990Kg)(9.86m/s)^2 =(69,990Kg) (9.8m/s^2) *(0.002) *d. Masses cancel out, so the equation should be written as 1/2(9.86m/s)^2=(9.8m/s^2)*(0.002)*d. Rearranging the equation, solve for d.