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Potential/Kinetic Energy Again.

  1. Oct 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A skier, of mass 50 kg, pushes off the top of a hill with an initial speed of 5 m/s. Neglecting friction, how fast in m/s will he be moving after dropping 17 m in elevation? (Neglect friction, use g=10m/s2)


    2. Relevant equations
    GPE = MGH
    KE =(1/2)mv^2


    3. The attempt at a solution
    KE = (1/2)mv^2
    KE = (1/2)(50)(25)
    KE = 625

    GPE = mgh
    625 = (50)(10)h
    625 = 500h
    1.25 = h

    If he's only 1.25 meteres up, I can't drop him seventeen meters, so I know I'm doing something wrong. Please help.

    Also. I'm sorry to post so often, but this homework is killing me.
     
  2. jcsd
  3. Oct 8, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Unfortunately you've taken things a little backwards. The skier starts at 17m.

    That gives you m*g*h of potential energy turned into KE at the bottom.
    That yields 50*10*17 that goes into kinetic energy. 8500N-m
    You also start out with KE = to 1/2m*v2 = 625N-m

    So at the bottom he has a Total of 9125N-m

    That's all KE which again is 1/2m*v2
    So... 9125 = 1/2m*v2 = 1/2*(50)*v2
    v2 = 9125*2/50 = 365
    v= 19.1m/s
     
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