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Kinetic frictional force

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    A 64.9-kg skier coasts up a snow-covered hill that makes an angle of 23.4 ° with the horizontal. The initial speed of the skier is 8.95 m/s. After coasting a distance of 2.31 m up the slope, the speed of the skier is 4.40 m/s. (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

    http://edugen.wiley.com/edugen/courses/crs2216/art/qb/qu/c06/EAT_12258936060120_435296769488702.gif

    2. Relevant equations

    W=FX 2K=mv^2 a=(v^2-v0^2)/2x

    3. The attempt at a solution

    for (b) i used f=ma-mg sinθ,

    or 1/2mv^2-1/2mv0^2-mgxsinθ for (a),

    but neither was right...
     
  2. jcsd
  3. Feb 20, 2010 #2

    Delphi51

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    Homework Helper

    1/2mv^2-1/2mv0^2-mgxsinθ for (a)
    looks good to me. What did you get?
     
  4. Feb 20, 2010 #3
    I think there may be a sign error here?

    [itex]\rm \Delta E = K_f - K_i + U_{f_{grav}}-U_{i_{grav}} = \frac{1}{2}m (v_f^2-v_i^2) +mg s Sin\theta = -1390J[/itex]
     
  5. Feb 20, 2010 #4
    Correct! thank you!
     
  6. Feb 20, 2010 #5
    You're welcome. If I had $1 for every time I've made a sign error in my lifetime, well, I'd own beach-front property in South Hampton. :smile:
     
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