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Conservation of energy of a downhill skier

  1. Jul 14, 2008 #1
    1. The problem statement, all variables and given/known data

    A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates. The crest of the second hill is circular, with a radius of r = 47 m. Neglect friction and air resistance. What must be the height h of the first hill so that the skier just loses contact with the snow at the crest of the second hill?

    [​IMG]
    2. Relevant equations

    1/2mvf2+mgHf=1/2mvo2+mgho


    3. The attempt at a solution

    All the masses cancel out. I got the equation down to two unknowns, one is the initial height and the other is the final velocity. Now I'm stumped.

    This is the furthest I've gotten: 1/2vf2+g(47+h)=1/2vo2+g(47-h)
     
  2. jcsd
  3. Jul 14, 2008 #2

    Doc Al

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    Staff: Mentor

    You'll need more than conservation of energy. (Hints: Measure PE from the dashed line. What's the initial speed?)

    Figure out the speed that the skier must have at the top of the second hill to just lose contact. Hint: Newton's 2nd law.
     
  4. Jul 14, 2008 #3
    Ive been staring at this but i cant seem to squeeze a formula out of it..can you be a little more specifc..
     
  5. Jul 14, 2008 #4

    Doc Al

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    Another hint: What kind of motion must the skier execute when climbing the second hill? What's his acceleration?
     
  6. Jul 14, 2008 #5
    in going from the lowest point on to the top of the crest the skier is losing kinetic energy and gaining potential energy.

    Also in reference to your newton's second law hint, how does the energy-work theorm fit into this: Wnc=Ef-E0
     
  7. Jul 14, 2008 #6

    Doc Al

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    That's true, of course, but that hint about the "kind of motion" had to do with Newton's 2nd law, not energy conservation.
    It doesn't.

    What's Newton's 2nd law? Consider the shape of the second hill.
     
  8. Jul 14, 2008 #7
    Newtons second law is F=ma. and the second crest is a circle with a radius of 47m. Am i now trying to find the centripedal acceleration? I think I'm way off.
     
  9. Jul 14, 2008 #8

    Doc Al

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    Good.
    Yes!
    You are finally on track. :wink:
     
  10. Jul 14, 2008 #9


    Ok, how can I find centripedal acceleration if i do not have the velocity? Ac=v2/r
     
  11. Jul 14, 2008 #10

    Doc Al

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    Using Newton's 2nd law! Realize that your job is to find the speed required at the top of the second hill, then use that to figure out the height h of the starting point.
     
  12. Jul 14, 2008 #11
    Aha! I got it. Thank you doc al, you're help is sooooo greatly appreciated.
     
  13. Oct 23, 2011 #12
    Wait what would the Forces be in the F=ma then? the acceleration would be the centripetal acceleration and the m would cancel im guessing but what are the forces?
     
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