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Conservation of mechanical energy

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data

    I have two blocks attached by a string (massless) over a pulley (frictionless, massless). Block 2 (which weighs less than block one) is released and the two blocks meet momentarily at the same height. I have the find the speed at which the meet at that moment. The blocks are separated by a distance of 1m.

    I have drawn a figure in paint (sorry for this pic): http://i36.tinypic.com/2076at1.png

    2. Relevant equations

    Emechf = Emechi
    Emech = Ug + K

    3. The attempt at a solution

    Emechf = Emechi
    <=> (m_1gh + m_2gh) + (1/2m_1v^2 + 1/2m_2v^2) = m_1gH (where H = 1m, and h is the height at which they meet)

    But then I have this h which is unkown and I can't figure out.

    Thanks a lot in advance.
  2. jcsd
  3. Oct 20, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    If the lower (and lighter) block starts out at height "h", the higher (heavier) block starts out at height "h + 1".

    At what height will they meet? Hint: One block lowers by the same distance that the other rises.
  4. Oct 20, 2008 #3
    Thanks a lot, I see it clearly now. I really appreciate you help.
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