How far does the skier travel on the horizontal surface before coming to rest? Neglect air resistance

Question

A skier starts from rest at the top of a frictionless incline of height 20.0 m, as in Figure below. At the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.210.the skier’s speed at the bottom is 19.8 m/s.

Answer 1

First of all, the speed at the bottom depends on the height from which the skier started.  The skier's potential energy at the top converts to kinetic energy at the bottom.

 

mgh = (1/2)mv² 

 

v = √(2gh) = √(2*9.8*20) m/s = 19.8 m/s

 

This agrees with the given speed, which was a redundant piece of information.

 

At the bottom, the kinetic energy goes to zero as the friction does work against the moving skier.

 

x = stopping distance

 

(1/2)mv² = μ_kmgx

 

(1/2)v² = μ_kgx

 

x = v² / (2μ_kg) = (19.8)² / (2*0.210*9.8) m = 95.2 m