physics coefficiant of friction problem-help??

Question

A 2480-Newton bobsled leaves a horizontal track with a speed of 26.9 m/s. It then moves onto a horizontal track lightly sprinkled with sawdust and comes to a stop in a distance of 147 meters. What is the coefficient of kinetic friction between the runners of the sled and the sawdust-covered surface?

Answer 1

Work done to slow the bobsled down equal to force of friction times the path and is also equal to the initial kinetic energy just before breaking.

W=f S
f=uN=uw=umg
W=Ke=(1/2) mV^2

umgS = (1/2) mV^2
u= V^2/(2gS)
u=(26.9)^2 / (2 x 9.81x 147)
u=0.25