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Work-Energy Kinematics skier

  1. May 25, 2012 #1
    1. The problem statement, all variables and given/known data
    An extreme skier, starting from rest, coasts down a mountain that makes an angle of 25.0° with the horizontal. The coefficient of kinetic friction between her skis/snow is 0.200. She coasts for a distance of 10.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.50 m below the edge. How fast is she going just before she lands?


    2. Relevant equations

    W = (F cos theta)*displacement
    W = KEf - KE0

    Kinetic friction = coefficient of kinetic friction * normal force

    Weight = mg


    3. The attempt at a solution
    I tried combining all of these, but i need the mass to do something. I simply can't figure out the final velocity with the information given.
     
  2. jcsd
  3. May 25, 2012 #2
    You can look at the problem in three main segments.
    1) Initial Point
    2) Top of Cliff
    3) Landing at Bottom

    Remember that energy is conserved. At rest, the person has some potential energy and no kinetic energy. This total energy is equal to the total energy at point 2, and the total energy at point 3. What type of energy does the person have at point 2?

    It may help you to sketch the problem.
     
  4. May 25, 2012 #3
    Please clarify. I've been stumped by this for days...
     
  5. May 25, 2012 #4

    Doc Al

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    You'll find that the mass will simply drop out of your equations. You won't need a numerical value. Just call the mass 'm' and continue to set up your equations.

    How does the total mechanical energy at the start compare to the total mechanical energy when she reaches the edge of the cliff?
     
  6. May 25, 2012 #5
    I can't use any other methods except the formulae given, unfortunately.
     
  7. May 25, 2012 #6

    Doc Al

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    No problem. Find the net force on the skier and compute the work done as she coasts.
     
  8. May 25, 2012 #7
    But to do that, I need to find the mass of the skier. Can you show me specifically how to?
     
  9. May 25, 2012 #8
    Can you show us the equations you formed?
     
  10. May 26, 2012 #9
    Please... I have no clue what to do.
     
  11. May 26, 2012 #10

    Doc Al

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    No you don't. Don't try to find a numerical value, just express it symbolically.
    What forces act on the skier?
     
  12. May 26, 2012 #11
    The normal force, which is equal to mg cos 25° but is balanced out by gravity's y component;
    gravity's x component, mg sin 25°;
    friction: 0.200 * normal force
     
  13. May 26, 2012 #12

    Doc Al

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    Excellent!

    So what's the net force on the skier? Then you can use Wnet = ΔKE to find the speed at which she reaches the cliff edge.
     
  14. May 26, 2012 #13
    So the net force is (mg sin 25°) - friction
    = (mg sin 25°) - (0.200 * mg cos 25°)
     
  15. May 26, 2012 #14

    Doc Al

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    Good. Now find the work done by that net force.
     
  16. May 26, 2012 #15
    sin 25° is around 0.423, and cos 25° is around 0.906, so:

    F = 0.423mg - 0.200*0.906mg
    = 0.423mg - 0.181mg
    = 0.242 mg

    Using W = (F cos θ)* displacement, this is what I get:

    (0.242mg cos 0°)*displacement


    = 0.242mg * displacement

    But what is the displacement?
     
  17. May 26, 2012 #16

    Doc Al

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    That was given in the problem statement.
     
  18. May 26, 2012 #17
    13.9 m?
     
  19. May 26, 2012 #18

    Doc Al

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    No. We are only dealing (for now) with the part where she coasts down the hill.
     
  20. May 26, 2012 #19
    So 10.4 m?
     
  21. May 26, 2012 #20

    Doc Al

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    Yes.
     
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