Question

A 50,000 kg locomotive is traveling at 10.0 when its engine and brakes both fail. How far will the locomotive roll before it comes to a stop?

Answer 1

m = 50,000 kg
u =coefficient of frictionfor steel = 0.002
v = 10 m/s
perconservation of energy:
KE = E(due to friction)
KE = 1/2*m*v^2
E(due to friction) = F * d = u * N = m * g
hence:
1/2*m*v^2 = u*m*g*d
1/2 * v^2 = u * g * d
d = v^2 / 2*u*g
= 100 / [2 * .002 * 9.81)
= 2548 .43meters

Answer 2

a = de acceleration = friction/m = mu*m*g/m = mu*g............. distance = V^2/(2*a ) = 100/(2*mu*g) = 5.1/(mu) ............... where mu = coefficient of friction