Work problem

  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. Dick

    Dick 25,853
    Science Advisor
    Homework Helper

    Chegg
    [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. in which part
     
  5. Dick

    Dick 25,853
    Science Advisor
    Homework Helper

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