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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

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College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

A skier starts from rest at the top of a hill that is inclined 10.5° with respect to the horizontal. The hillside is 2.00 × 102 m long, and the coefficient of friction between snow and skis is 0.075 0. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?

To determine
How far does the skier glide along the horizontal portion of the snow before coming to rest.

Explanation

Given Info:

The hill is inclined 10.5° with respect to the horizontal.

The length of the hillside is 2.00×102m .

The coefficient of friction between the snow and skis is 0.0750 .

Consider the level of the base of the hill to be at the zero gravitational potential.

Since the hill is inclined 10.5° with respect to the horizontal, the normal force on the skis when it is on the hillside is,

n1=mgcos10.5°

  • n1 is the normal force on the skier when it is on the hillside
  • m is the mass of the skier
  • g is the free fall acceleration

The normal force on the skis on the horizontal portion is,

n2=mg

In the entire trip the work done by frictional force will be:

W=μk(mgcos10.5°)(2.00×102m)μk(mg)(x)=μkmg((cos10.5°)(2.00×102m)+x)       (1)

  • x is the distance travelled in the horizontal portion
  • μk is the coefficient of friction between the snow and the ski

In the entire trip the change in the mechanical energy is:

W=(KE+PEg)f(KE+PEg)i=(0+0)(0+mg(200m)(sin10

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