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Skier on slope - work done

  1. Oct 18, 2009 #1
    ok guys, i got one that wants me to figure out how much work is done:
    A skier of a mass 79.1 kg, starting from rest, slides down a slope at an angle of 38 degrees with the horizontal. The coefficient of kinetic friction, u, is 0.09. What is the net work (the net gain in kinetic energy) done on the skier in the first 8.9 s of decent?
    now i dont want the answer, i just need help know what equations i need when.

    I know it gives me m=79.1kg theta=38degrees coefficient of kinetic friction=.09 (not sure of unit) and t=8.9

    how do i get how much work?

    I have figured the mass in the verticle direction to be 48.69 (is this right?) by taking sin38=x/79.1 or 79.1sin38=x
     
    Last edited: Oct 18, 2009
  2. jcsd
  3. Oct 18, 2009 #2

    Doc Al

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    Figure out the acceleration. What's the net force on the skier? Use Newton's 2nd law.
     
  4. Oct 18, 2009 #3
    ok since i got 48.69 to be the mass in the y direction, i take that times -9.8 as the acceleration due to gravity and get -477.16 as the force, i think i newtons, but where does friction come in?
     
  5. Oct 18, 2009 #4
    I can complete this problem without friction, or even with friction, but in one dimention, but what confuses me is that usually gravity and friction work together to slow down an object, now gravity is working to speed it up and friction is trying to slow it down, so i'm not sure what equation to use
     
  6. Oct 18, 2009 #5

    Doc Al

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    Don't think in terms of mass in a particular direction--that doesn't make much sense. Think of the component of the weight (mg) parallel to the incline, which equals mgsinθ.

    Friction is another force acting on the skier; it will equal μN, where N is the normal force. (How do you determine the normal force?)
     
  7. Oct 18, 2009 #6

    Doc Al

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    You need the net force down the incline. Since gravity and friction act in opposite directions, they will have opposite signs.
     
  8. Oct 18, 2009 #7
    ok so N=mgcos(theta)
    N=79.1x9.8xCos(38)
    N=477.25 (i think it is a little different because of rounding earlier)


    so friction=uN
    friction=-0.09x477.25
    friction=-42.95

    so the net forces N+friction=477.25-42.95 or 434.30 so that is the net force

    now K=mv^2
    w=dK=(K-K_0)/2
    so 1/2K (because he starts at rest so v_0=0 so K_0=0)
    or (mv^2)/2
    now m=79.1kg, v=axt or because a=F/m v=434.30/79.1x8.9 or 48.87m/s
    so K=1/2(79.1)(48.87)^2
    w=94,456.35j
    (is this right and is joules the right unit?)
     
  9. Oct 18, 2009 #8

    Doc Al

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    Redo this calculation. (Cosine, not sine!)
     
  10. Oct 18, 2009 #9
    oops... my bad, i did use sin i just accidentally wrote cos
    so is it supposed to be cos?
     
    Last edited: Oct 18, 2009
  11. Oct 18, 2009 #10

    Doc Al

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    You should have used cosine, as written.
     
  12. Oct 18, 2009 #11
    ok thanks
     
  13. Oct 18, 2009 #12
    ok so N=mgcos(theta)
    N=79.1x9.8xCos(38)
    N=610.85

    so friction=uN
    friction=-0.09x610.85
    friction=-54.98

    so the net forces N+friction=610.85-54.98 or 422.27 so that is the net force

    now K=mv^2
    w=dK=(K-K_0)/2
    so 1/2K (because he starts at rest so v_0=0 so K_0=0)
    or (mv^2)/2
    now m=79.1kg, v=axt or because a=F/m v=422.27/79.1x8.9 or 47.51m/s
    so K=1/2(79.1)(47.51)^2
    w=89,272.26j
    so joules is the right unit?
     
  14. Oct 18, 2009 #13

    Doc Al

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    OK.

    The net force is the component of gravity down the incline (not N) minus the friction force.
     
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