A polar bear cub slides (starting from rest) 31 m toward his mother down a snow-covered 37^\circ slope starting from rest. At the end of the 31 m, the cub’s speed is 4.7 m/s. Find the coefficient of friction between the cub and the snow-covered surface.

Answer 1

How far down did he go
31 sin 37 = 31 * 0.602 = 18.7 meters down
so
loss of potential energy = m g * 18.7
gain in kinetic energy = (1/2) m v^2 = m * 4.7^2 / 2 = 11 m
so energy wasted on friction = m (18.7 g - 11 )
that is the work done by the friction force
mu m g cos 37 * 31 = m (18.7 g - 11)
say g = 9.81
mu * 9.81 * 31 * cos 37 = 18.7 * 9.81 - 11
243 mu = 172
mu = 0.71