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Conservation Of Mechanical Energy

  1. Apr 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.30 , its speed is 1.00 .
    If the total mass of the two blocks is 18.0 , what is the mass of the more massive block?
    Take free fall acceleration to be 9.80 .
    I set the heavier block as m1.

    2. Relevant equations
    m1gh1+0.5m1v1^2=m2gh2+0.5m2v2^2

    3. The attempt at a solution
    They both start at height 0 and velocity 0 and so the initial PE and KE is going to be 0, and so the initial Mechanical energy is also 0 (I'm pretty sure this is wrong but don't know how to fix it). Then m1 drops 1.30m so thats h1 and m2 goes up -1.30m=h2. v should be equal between both, m1=1, m2=-1.

    m1(gh1+0.5v^2)=m2(gh2+0.5v^2)
    m1/m2=(gh2+0.5v^2)/(gh1+0.5v^2)
    =(9.8*-1.30+1/2*-1^2)/(9.8*1.30+1/2*1^2)
    Therefore
    m1=-m2

    So its wrong ahah I was wondering if someone could help me?
    Thanks in advance!:shy:
     
  2. jcsd
  3. Apr 10, 2008 #2

    tiny-tim

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    Hi tizzful! :smile:

    erm … it's not "-1^2" … :redface:

    No wonder they came out minus each other! :rolleyes:
     
  4. Apr 10, 2008 #3
    actually the -1 gets squared and so it becomes one.. Its negative because the height is negative because down is positive and up is negative.. But from what you're saying why isn't it -1? It's also in the opposite direction...
     
  5. Apr 10, 2008 #4

    tiny-tim

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    Noooooo … the kinetic energy mv^2/2 is always positive!

    It depends only on speed, not direction!

    You have too much imagination! :smile:
     
  6. Apr 10, 2008 #5
    ahahah thank you! But I know KE is always positive because if velocity is negative it gets squared making it positive.. And thats what happened in this case.. But i still can't figure out the answer.. I think there's something wrong with me saying initial ME = 0...
     
  7. Apr 10, 2008 #6

    tiny-tim

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    ah … I see now … your basic equation is wrong …

    KE gained is (m1 + m2)v^2/2

    PE gained is (m1 - m2)gh. :smile:
     
  8. Apr 10, 2008 #7
    When you have gotten the answer using that method (which is probably the easiest), you can also try doing it by finding the acceleration on the big mass, and then use s=(at^2)/2. You will end up with the exactly same equation. :)
     
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