Question

A skier slides horizontally along the snow for a distance of 12 m before coming to rest. the coefficient of kinetic friction between the skier and the snow is μk = 0.060. initially, how fast was the skier going?

Answer 1

We can use work-energy to solve this question. The work done by friction on the skier will be equal in magnitude to the skier's initial kinetic energy. KE = Work (1/2)mv^2 = mg ÎĽk d v^2 = 2g ÎĽk d v = sqrt{ 2g ÎĽk d } v = sqrt{ (2) (9.80 m/s^2) (0.060) (12 m) } v = 3.76 m/s Initially the skier was moving at a speed of 3.76 m/s