# Calculate the work done by the kinetic frictional force

## 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.

WFN=0 ----> makes a 90deg angle with the direction.
Wmgcos(theta)=0

Wmgsin(theta)=Fd
=(mgsin25.2)(2.09m)
=(62.7kg)(9.8m/s^2)(0.426)(2.09m)
=548J
WFk= ???

WTOT= 0+0+548+WFk = 1/2mVf^2-1/2mVi^2

=> 548J+WFk=321J-2120J
=> WFk=-1800J-548J=-2350J

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

## Answers and Replies

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tiny-tim
Homework Helper
Hi Student3.41!

(your equations would be easier to read ifyouleftsomespaces)
WTOT= 0+0+548+WFk = 1/2mVf^2-1/2mVi^2
no, your LHS is positive, so your RHS should be positive also …

interchange Vf and Vi

Hi Student3.41!

(your equations would be easier to read ifyouleftsomespaces)

no, your LHS is positive, so your RHS should be positive also …

interchange Vf and Vi
OK, so interchange the values so;

545J+WK= 1/2mVi^2-1/2mVf^2

= 2120J-322J
=1780J

WK= 1780J-545J = 1240J

But this is wrong. Unless I am not understanding what you mean. Only have 3 more trys :O

tiny-tim
Homework Helper
WK= 1780J-545J = 1240J

But this is wrong. Unless I am not understanding what you mean. Only have 3 more trys :O
isn't it 548, not 545?

isn't it 548, not 545?
I re calculated my answer so I didn't take everything by 3 significant digits. Either way the answer comes out to be wrong.

tiny-tim
Homework Helper
I re calculated my answer so I didn't take everything by 3 significant digits. Either way the answer comes out to be wrong.
Sorry, then I can't see what's wrong.

(btw, you should keep everything to at least 4 sig figs until the very end, and only then round off to 3 sig figs; and I made it 547.4)

Sorry, then I can't see what's wrong.

(btw, you should keep everything to at least 4 sig figs until the very end, and only then round off to 3 sig figs; and I made it 547.4)
I am not sure if that will affect the answer, my prof usually gives us some room for error.

I am not sure either, maybe an error in the answer.

Isn't the Work of friction always -fk because Wfk = fkcos180d = fk(-1)d = -fkd???

Also, is there a need to move the 547.7 to the RHS???

I am also posting another problem if you can help with that! I have done all the calculations and its been driving me crazy.. Thanks for the help :)

tiny-tim