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Work and velocity problem

  1. Feb 25, 2008 #1
    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?
     
    Last edited: Feb 25, 2008
  2. jcsd
  3. Feb 25, 2008 #2

    Dick

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    Homework Helper

    [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?
     
  4. Feb 25, 2008 #3
    in which part
     
  5. Feb 25, 2008 #4

    Dick

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    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.
     
  6. Feb 25, 2008 #5
    o ive messed my formula up =/
    i got correct answer now, thanks
     
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