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Homework Help: Energy and Work

  1. Nov 3, 2004 #1
    A ball is attached to a horizontal cord of length L whose other end is fixed.

    a. If the ball is released what will be its speed at the lowest point of its path?

    (1/2)mv^2 = mgr, or v^2 = 2gr, v = sqrt(2gr)???

    b. A peg is located a distance h directly below the point of attachment of the cord. If h = 0.80 L, what will be the speed of the ball when it reaches the top of its circular path about the peg?



    + = path of circular motion
    O = ball

    Any help to start is appreciated.
  2. jcsd
  3. Nov 3, 2004 #2
    The first one is correct.(r=L).
    When the ball reaches its top circular path, it has both kinetic energy and potential energy.
    Initial energy,E(initial) = mgL
    Total energy when the ball reaches its top circular path,E(final) = [tex]\frac{1}{2}mv^2 + mg*2(L-h).[/tex]
    Law of conservation of energy : E(initial) = E(final)
    Last edited: Nov 4, 2004
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