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Completely lost

  1. Oct 19, 2006 #1
    A 19.0 kg child on a 1.00 m long swing is released from rest when the ropes of the the swing make an angle of 32.0° with the vertical.

    (a) Neglecting friction, find the child's speed at the lowest position.

    L - Lcostheta = h

    1 -1cos32 = .15

    mgh = 1/2mv^2

    (19)(9.8)(.15) = 1/2(19)v^2
    27.93 = 9.5v^2
    v = 1.7 m/s

    (b) If the speed of the child at the lowest position is 1.50 m/s, what is the mechanical energy lost due to friction? J

    I do not understand this question at all but i think its like this:

    KE = 1/2(19)(1.5)^2 then i would get mgh and subtract that by KE?

    KE = 21.38
    27.93 - 21.38 = 6.55 J?

    please help me out i seriously dont understand this question.
  2. jcsd
  3. Oct 19, 2006 #2
    okay i gave that answer a try and its incorrect

    edit: Nevermind it seem I did do it correctly but the online homework was picky with the answer and wanted a whole number.
    Last edited: Oct 19, 2006
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