Question

A skier 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?

Answer 1

Answer:

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

Answer 2

a. The speed of the skier at the bottom of the slope should be 22.8 m/s

b. the coefficient of kinetic friction between the skier and the horizontal surface should be 0.19.

Calculation of the speed and coefficient:

Since

h = 50 sin32 m

= 26.5 m

Now  

potential energy lost should be

= mgh

Now Gain of kinetic energy

= 1/2 mv²

So,

mgh = 1/2 mv²

v = √ 2gh

= √ (2x9.8 x 26.5)

= 22.8 m /s

b )

here we assume μ be the coefficient of kinetic friction required.

So,

friction force acting

= μmg

And,

work done by friction in displacement of  d (40 m ) on horizontal surface is - μmg x d

So,

=  μmg x d  = (1/2) m v²

μ = v² / (2 gd)

= 519.4 / (2 x 9.8 x 140 )

= .19

learn more about speed here: https://brainly.com/question/3902437