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Solving for potential energy after time t.

  1. Dec 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A stone of mass 0.2kg is dropped from the top of a building 80m high. After t seconds it has fallen x meters and has a velocity of v. What is the potential energy of the stone t seconds later

    2. Relevant equations
    The answer is 2(80 - x) = 160 - 10t^2

    Using suvat: s = ut + (1/2)at^2

    3. The attempt at a solution
    I started with PE = mgh
    PE = 0.2 * 10 * h
    PE = 2h

    Then I figured h = (1/2)*10*t^2
    or h = 5t^2

    So now I have

    PE = 10t^2

    Not sure if I'm headed in the right direction or not.
     
  2. jcsd
  3. Dec 2, 2015 #2

    cnh1995

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    That gives PE=0 at the instant of dropping. Where should PE be 0 and where should it be maximum?
     
  4. Dec 2, 2015 #3

    cnh1995

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    It is coming downwards.
    What is s?
    What is h?
    Are they both same in this problem?
     
  5. Dec 2, 2015 #4
    I got it, or at least I understand the answer now. 80 - (80 - x) gives the remaining distance with regards to x, and this equal to the same measurement given (1/2)(g)t^2. Set them equal and shuffle around a bit. Thanks
     
  6. Dec 2, 2015 #5

    cnh1995

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    Right.
    x=80-h.
     
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