Askier with a mass of 63 kg starts from rest and skis down an icy (frictionless) slope that has a length of 50 m at an angleof 32° with respect to the horizontal. at the bottom of the slope. the path levels out and becomes horizontal, the snowbecomes less icy, and the skier begins to slow down, coming to rest in a distance of 140 m along the horizontal path, (a)what is the speed of the skier at the bottom of the slope? (b) what is the coefficient of kinetic friction between the skierand the horizontal surface?

Explanation:

When the skier reaches the bottom of the slope , height lost by it

h = 50 sin32 m

= 26.5 m

potential energy lost

= mgh

Gain of kinetic energy

= 1/2 mv²

mgh = 1/2 mv²

v = √ 2gh

= √ (2x9.8 x 26.5)

= 22.8 m /s

b )

Let μ be the coefficient of kinetic friction required.

friction force acting

= μmg

work done by friction in displacement of  d (40 m ) on horizontal surface

- μmg x d

This negative work will be equal to positive kinetic energy of the skier on horizontal surface .

=  μmg x d  = (1/2) m v²

μ = v² / (2 gd)

= 519.4 / (2 x 9.8 x 140 )

= .19

i would say the answer is food.

hope this !

c) up

hope this