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What is the skier's initial velocity?

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    A 50.0kg skier starts at A with an unknown initial velocity. The skier then goes over a hill (point B) which is 4.00m above point A. At point B (top of the hill), the skier experiences a Force normal 5/6 of his/her weight. The hill has a radius of 14.6m. What is the skier's initial velocity?
    (Problem continues but only need help for this part)

    2. Relevant equations

    K=1/2mv2
    G=mgh
    Fc=(mv2)/r

    3. The attempt at a solution

    So since the skier's "apparent weight" is 5/6 of his/her real weight, then force normal is = 5mg/6.
    Since only gravity is holding the skier in a circle, then force centripetal = 5/6 force gravity.
    ([STRIKE]m[/STRIKE]v2)/r = (5/6)([STRIKE]m[/STRIKE]g)
    v=sqrt(5/6)(9.81)(14.6)
    v=10.92m/s
    This is the speed of the skier at the top of the hill

    Then we set the conservation of energy equation
    Kinitial = Khill + G
    K = 1/2mv2 + mgh = (0.5)(50.0)(10.922) + (50.0)(9.81)(4.00)
    K = 2983.875J + 1962J = 4945.875J
    1/2mv2 = 4945.875J
    v= sqrt [(4945.875J)(2)/(50.0kg)]
    v=14.07m/s


    It seems logical to me but for some reason it is not correct : /
    Help appreciated thanks :)
     
  2. jcsd
  3. Oct 1, 2011 #2

    ehild

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    Draw the free body diagram and think over how is the centripetal force related to gravity and normal force.

    ehild
     
  4. Oct 1, 2011 #3
    Hi Virtuozzo and hi PF! I'm losing my virginity here :blushing:

    You've got the right idea for the second (conservation of energy) part, only your velocity from the first part is wrong. For the first part, note that the skier isn't moving in the vertical direction, so this is a statics problem: The sum of all forces (in the y-direction) equals zero, and you've got three distinct forces at play as ehild mentioned, so your equation should have three terms. Hint: the top-of-the-hill velocity should be less than half of the value that you got.

    Also, it's good practice to wait until the end of the problem to plug in your numbers to ensure accuracy. Good luck!
     
    Last edited: Oct 1, 2011
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