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Energy, Power, Work related question

  1. Jan 3, 2009 #1
    This isn't the actual problem given in the assignment. I've twisted it a bit because I want to see were I specifically go wrong in my calculations.
    1. A pitcher rotates a 0.25kg ball around a vertcal circular path of radius 0.6m before releasing it. The pitcher exerts a 50N force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 20m/s. If the ball is released at the bottom of the circle, what is the speed upon release?



    2. KE = 1/2mv^2
    PE = mgh
    Energy at top = Energy at bottom.
    1/2mvi^2+mgh = 1/2mvf^2+mgh.


    3.vf = sqrt (1/2mvi^2+mgh-mgh)/m
     
  2. jcsd
  3. Jan 3, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi alevis! Welcome to PF! :smile:
    What about the work done by the pitcher? :wink:

    (and you're missing a 2)
     
  4. Jan 5, 2009 #3
    thanks alot
     
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