Question

A skier slides straight down an incline of 25 degrees without using her poles. The slope itself is 96 meters long, and the skier starts from rest at the top. a. What would the velocity of the skier be at the bottom of the incline if friction can be neglected? b. What would your answer be to the previous question if the coefficient of kinetic friction between the skis and the snow is 0.13 on the incline? c. Upon reaching the bottom of the incline in part b, she reaches a flat portion, and decides to just let friction slow her to a stop. How far does the skier travel along the horizontal portion before coming to a stop, if the coefficient of kinetic friction is 0.28 on the flat portion?

Answer 1

a) acceleration of skier on incline surface, a = g*sin(25)

= 9.8*sin(25)

= 4.14 m/s^2

now use, vf^2 -vi^2 = 2*a*d

vf^2 - 0^2 = 2*4.14*96

vf = sqrt(2*4.14*96)

= 28.2 m/s

b) net force acting on the skier,

Fnet = m*g*sin(theta) - fk

m*a = m*g*sin(theta) - mue_k*m*g*cos(theta)

a = g*sin(theta) - mue_k*g*cos(theta)

= 9.8*sin(25) - 0.13*9.8*cos(25)

= 2.99 m/s

now use, vf^2 -vi^2 = 2*a*d

vf^2 - 0^2 = 2*4.14*96

vf = sqrt(2*2.99*96)

= 23.9 m/s

c) acceleration of skier on flat surface, a = -g*mue_k

= -9.8*0.28

= -2.74 m/s

now use, vf^2 -vi^2 = 2*a*d

0^2 - 23.9^2 = 2*(-2.74)*d

==> d = 23.9^2/(2*2.74)

= 104 m