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Work problemby rsala
Tags: work 
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#1
Feb2508, 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? 


#2
Feb2508, 06:04 PM

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HW Helper
Thanks
P: 25,235

[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? 


#3
Feb2508, 06:07 PM

P: 40

in which part



#4
Feb2508, 06:09 PM

Sci Advisor
HW Helper
Thanks
P: 25,235

Work problem



#5
Feb2508, 06:11 PM

P: 40

o ive messed my formula up =/
i got correct answer now, thanks 


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