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Skier skies down slope, including friction, calculate final speed

  1. Dec 20, 2012 #1
    1. The problem statement, all variables and given/known data

    A 50Kg skier skies down a 25 degree slope. at the top of the slope her speed is 4m/s and accelerates down the hill. the coefficient of friction is 0.12 between skies and snow. ignoring air resistance calculate her speed at point that is displaced 60m downhill.


    2. Relevant equations

    E=KE+PE
    E=.5*m*V^2+m*g*h


    3. The attempt at a solution

    I believe I solved it correctly using E0=.5(50)4^2+50(9.8)(60sin(25)) and getting 12650

    then setting 12650 = .5*m*Vf^2 and I'm getting Vf=22.5 m/s

    However I'm unsure how to go about solving while taking into consideration the friction, since there is a non conservative force at work using KE0+PE0=KEf+PEf would get me close but not all the way there.
     
  2. jcsd
  3. Dec 20, 2012 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The work done by friction will reduce the total mechanical energy.
     
  4. Dec 20, 2012 #3
    Right.

    So I've just broken it up into x and y components

    Fx=Wsin25-.12(Wcos(25))=m*a

    I calculated for acceleration in the x plane, and found a=3.08 m/s^2

    Then I just did

    Vf=sqr(2ax+V0^2)
    sqr(2(3.08)(60)+4^2)= Vf= 19.6 m/s

    Does this seem correct?
     
  5. Dec 20, 2012 #4

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
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