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Physics: skier problem

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A 65-kg skier coasts up a snow-covered hill that makes an angle of 25° with the horizontal. The initial speed of the skier is 6.4 m/s. After coasting 1.9 m up the slope, the skier has a speed of is 3.2 m/s.


    2. Relevant equations



    3. The attempt at a solution

    Work done if force F applied over a distance d.
    W= Fd
    Also
    Ke= 0.5mV^2
    Difference in Kinetic energy dKe
    dKe=Ke1 - Ke2
    dKe= 0.5m(V1^2 - V2^2)
    dKe= W

    a) W=dKe= 0.5m(V1^2 - V2^2)
    W= 0.5x65(6.4^2 - 3.2^2)
    W= 998.4 Joules

    P=mgh
    Pe= 65 x 9.81 x 1.9 sin(25)= 511.5 J

    Wf= W- Pe= 998.4 - 511.5 = 486.9 J

    what am i doing wrong?
     
  2. jcsd
  3. Oct 13, 2009 #2

    tiny-tim

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    Hi triplel777! :smile:
    I'm confused :confused:

    what's the question? :wink:
     
  4. Oct 13, 2009 #3
    (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?
     
  5. Oct 14, 2009 #4

    tiny-tim

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    Hi triplel77! :smile:

    (try using the X2 tag just above the Reply box; and we usually write KE and PE with capitals, and joules without :wink:)
    Apart from making PE = 512.0 J, I get the same work done by friction as you do. :redface:
     
  6. Oct 14, 2009 #5
    huh... ok thanks
     
  7. Oct 15, 2009 #6
    my teacher said to use this formula
    Wnc= delta KE+ delta PE
    would this make any difference?
     
  8. Oct 15, 2009 #7
    delta PE is just your change in gravitational potential energy, delta KE is your change in kinetic energy. Does this formula seem different than the one you used?
     
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