An ice skater of mass m is given a shove on a frozen pond. After the shove she has a speed of 2m/s. Assuming that the only horizontal force that acts on her is a slight frictional force between the blades of the skates and the ice:

Question

a.) Identify the horizontal force and two vertical forces that act on her.

b.) Use the work-energy theorem to find the distance the skater moves before coming to rest. Assume that the coefficient of kinetic friction between the blades of the skates and ice is .12

Answer 1

horizontal friction

vertical gravity down, force up from ice equal and opposite

initial Ke = (1/2) m v^2 = .5*4 m = 2 m Jloules
final Ke = 0

normal force at ice = 9.81 * m
so
friction force = .12 (9.81) m

work done = Force * distance
= .12 * 9.81* m * d = change in ke = 2 m
so
d = .12*9.81/2 meters

Answer 2

d=2/(.12*9.81)