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## Homework Statement

I have a few questions I need help with on my assignment. I cannot wrap my head around this work and energy chapter.

Anyway here it is;

A 62.7-kg skier coasts up a snow-covered hill that makes an angle of 25.2° with the horizontal. The initial speed of the skier is 8.22 m/s. After coasting 2.09 m up the slope, the skier has a speed of 3.20 m/s. Calculate the work done by the kinetic frictional force that acts on the skis

## Homework Equations

W=Fd

Wtotal=1/2mvf^2-1/2mvi^2

## The Attempt at a Solution

FBD. Fn in the positive y-direction, mgcos(theta) in the negative y-direction, Fk and mgsin(theta) in the negative y-diretion.

W

_{FN}=0 ----> makes a 90deg angle with the direction.

W

_{mgcos(theta)}=0

W

_{mgsin(theta)}=Fd

=(mgsin25.2)(2.09m)

=(62.7kg)(9.8m/s^2)(0.426)(2.09m)

=548J

W

_{Fk}= ???

W

_{TOT}= 0+0+548+W

_{Fk}= 1/2mVf^2-1/2mVi^2

=> 548J+W

_{Fk}=321J-2120J

=> W

_{Fk}=-1800J-548J=-2350J

I am using three sig digits for my calculations. The answer is wrong.