work problem


by rsala
Tags: work
rsala
rsala is offline
#1
Feb25-08, 05:50 PM
P: 40
1. The problem statement, all variables and given/known data
A skier moving at 5.00 m/s encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. the patch is 2.9 meters long.
what is her velocity after passing the patch?

problem must be solved by using work, avoid newtons
2. Relevant equations
work = fs
work = [tex]K_{final} - K_{initial} =- \frac{1}{2}m v^{2}_{final}- \frac{1}{2}m v^{2}_{initial}[/tex]
3. The attempt at a solution
let s be displacement (2.9m)

[tex] -\mu_{k}mg*s = - \frac{1}{2}m v^{2}_{final}- \frac{1}{2} m v^{2}_{initial}
[/tex]

m is irrelevant, factor it out and cancel.

[tex]
-\mu_{k}g*s = - \frac{1}{2} v^{2}_{final}- \frac{1}{2} v^{2}_{initial}[/tex]

solve for [tex] v_{final} [/tex]

[tex]\frac{-\mu_{k}gs+.5v^{2}_{initial}}{-.5} = v^{2}_{final}
[/tex]

[tex]
\sqrt{-2(-\mu_{k}gs+\frac{1}{2}v^{2}_{initial})} = v_{final}
[/tex]

[tex]
\sqrt{ -12.4952}
[/tex]
cannot take sqrt of negative number.
i can't go beyond this part, is there a solution?
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Dick
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#2
Feb25-08, 06:04 PM
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P: 25,175
[tex]
\mu_{k}g*s = \frac{1}{2} v^{2}_{initial}- \frac{1}{2} v^{2}_{final}[/tex]

Where are you finding all the extra minus signs?
rsala
rsala is offline
#3
Feb25-08, 06:07 PM
P: 40
in which part

Dick
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#4
Feb25-08, 06:09 PM
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P: 25,175

work problem


Quote Quote by rsala View Post
in which part
All over. Look at your first expression for K_(final)-K(initial). That's supposed to be the difference of two kinetic energies, not the negative of their sum.
rsala
rsala is offline
#5
Feb25-08, 06:11 PM
P: 40
o ive messed my formula up =/
i got correct answer now, thanks


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