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Work and Energy of an extreme skier

  1. Nov 24, 2007 #1
    Hey,

    I have a question that I can't seem to work out...can anybody please help me?!

    Question:

    An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25 degrees with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts down 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?


    I have set the problem up and come to realize that this will take 2 seperate analyses: one before she flies off the cliff and one after to get the final velocity of the second analysis, where v final of the 1st analysis will equal v initial of the 2nd.

    What has me stumped is the fact that I was given no mass. In order to to find my kinetic energy (KE = 1/2 mv (initial)^2 + W) I need a mass.

    Can anybody please help me?! I am a lost cause at the moment. Your help is MUCH appreciated!

    Thank you!!
     
    Last edited: Nov 24, 2007
  2. jcsd
  3. Nov 24, 2007 #2

    Doc Al

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    You won't need the actual mass--just call it "m" and keep going. Hint: What's the force of friction?
     
  4. Nov 24, 2007 #3
    So I got to a point where I think the masses cancel.... And Im sorry, I don't quite understand what you mean by: what's the force of friction? I think they gave me a number (0.200)...
     
  5. Nov 24, 2007 #4

    Doc Al

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    They gave you the coefficient of kinetic friction. Use that to find the friction force. (Look it up if you need to.)
     
  6. Nov 24, 2007 #5
    Ok, so I just realized that they gave the the coefficient of kinetic friction. Fk = coeff. x normal force. From what I understand, the normal force cancels out because there is no accelleration in the y direction...I just can't connect how this information helps me with the mass part of it.
     
  7. Nov 24, 2007 #6

    Doc Al

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    Good.
    Use that fact to figure out the normal force, and then the friction force.
     
  8. Nov 24, 2007 #7
    So normal force = 0, is this correct? If so, then that makes my friction force = 0. I still cant connect that with the mass part of this. The only thing I can think of is that the friction force balances the weight force and so my mg will equal zero too....??
     
  9. Nov 24, 2007 #8

    Doc Al

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    Not at all. Consider all forces acting on the skier in the y-direction (perpendicular to the slope). As you stated earlier, the acceleration in that direction is zero, thus the net force must be zero. Use that fact to figure out the normal force.
     
  10. Nov 24, 2007 #9
    Ok, so Fn = mgcos(25) So I still have m in there. This is my overall equation:

    Square root: 2(mgsin25 - coeff. k(mgcos25)) (cos theta) (s)/ m

    I don't get it! Do my masses cancel somehow??

    I can see how the masses on top might, but what happens to the denominator? (My algebra is obviously lacking...)
     
    Last edited: Nov 24, 2007
  11. Nov 24, 2007 #10

    Doc Al

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    Excellent.


    OK. Looks like you're solving for the speed at the point where the skier goes over the cliff. No problem. (What's theta?)

    Well, yes. Look at what you wrote.

    It's the denominator that does the canceling in the numerator! ab/a = b

    FYI: You don't have to do this in two steps. You can compare initial energy to final energy and use that to solve for the final speed in one step.
     
    Last edited: Nov 24, 2007
  12. Nov 24, 2007 #11
    Theta is 0 degrees because the force and the net force and the displacement go in the same direction.

    I can use: (Sum of forces) x s = 1/2 mv^2 - 1/2 mv (initial) ^2 and that final velocity is the final after she flies off of the cliff? So I use the second displacement here and the final velocity from the first part for initial...is this correct?

    And is my sum of forces the same as above, or do I need a new force diagram, in which the only forces acting after she flies off of the cliff is weight?
     
    Last edited: Nov 24, 2007
  13. Nov 24, 2007 #12

    Doc Al

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    Good.
    Right. This is what I thought you did.
    Right. Hint: You'll save time by not bothering to solve for the final velocity from the first part, but just stick with the final KE (which is what you really need).
    The forces are different--there's no normal force, only weight.
     
  14. Nov 25, 2007 #13
    Thank You So Much Doc Al!!!
     
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