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Elastic Potential Energy problem

  1. Jan 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A spring has a force constant of 30000 N/m. How far must it be stretched for it's potential energy to be 47 J? Answer in units of m.

    2. Relevant equations
    Us= 1/2k(∆x)^2

    3. The attempt at a solution
    47= 1/2 (30000)(x)^2
    -14953= x^2
    √14953= 122.28
  2. jcsd
  3. Jan 26, 2009 #2

    Doc Al

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    Staff: Mentor

    So far, so good.
    Oops... you subtracted instead of divided. Redo this step.
  4. Jan 26, 2009 #3


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    Homework Helper

    You subtracted when you should have divided.
  5. Jan 26, 2009 #4
    So you are saying to do 47/15000?
    that would equal .0031... then would i do the square root of it or leave it at that?

    ohhh nvm i did that on one of my attempts but i forgot to put .0 in front of .0557

    =] Thanks a lot. I apperciate it!
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