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Potential Energy of a Spring

  1. Feb 23, 2006 #1
    [​IMG]
    (ignore values in image)

    A 2.4 kg block is placed against a spring on a frictionless 17o incline. The spring, whose spring constant is 59.2 N/cm , is compressed 35.3 cm and then released.

    What is the elastic potential energy of the compressed spring in J?


    According to my calculations and my book, the potential energy of a spring is .5kx2. That didn't work, so I converted cm to m in k and x. That didn't work either. The only difference between this problem and a similar one in the book is that this spring is at an angle. Could someone give me a hint as to what the trick with this problem is?
     
  2. jcsd
  3. Feb 23, 2006 #2

    Hootenanny

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    What formula are you using?
     
  4. Feb 23, 2006 #3

    Hootenanny

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    Work done is force multiplied by distance, in the case tension multiplied by extension/compression. As tension is given by:
    [tex] T = \frac{kx}{l} [/tex]
    Then the work done or elastic potential is given by:
    [tex] \int_{0}^{e} \frac{kx}{l} \Rightarrow E_p = \frac{ke^2}{2l} [/tex]
    Where e is extension and l is origional length.

    I imagine you will also need to factor in gravity.
     
  5. Feb 23, 2006 #4
    I don't know the original length, unless there's a way to determine that from the spring constant and compression.

    I also can't think of how I can factor in gravity. I had no trouble with a problem where I calculated the height that a ball was shot by a vertical spring. I just calculated the work done by the spring and found the velocity from W=K2-K1. I didn't need to do anything with gravity until the ball had left the spring. Why wouldn't that be the case here aswell?
     
  6. Feb 23, 2006 #5

    Hootenanny

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    You can work out the origonal length of the spring because you know the force applied to it, the weight of the block (gravity). Don't forget to resolve.
     
  7. Feb 23, 2006 #6
    Sigh, I got it. I was dividing the 59.2 N/cm by 100 instead of multiplying. Sorry about that.

    As an aside, how do you make those equation images? Are you using excel, or is there something more convenient?
     
  8. Feb 23, 2006 #7
    The elastic spring potential energy does not take any gravitational potential energy into account by definition. So, as you said, [tex]S=(1/2)kx^2[/tex].
    Thus:
    [tex]k= \frac{59.2N}{1cm} * \frac{100cm}{1m} = 5920N/m [/tex]
    [tex]S=(1/2)(5920N/m)(0.353m)^2=368.8J [/tex].

    If that isn't the answer your book gives, your book is incorrect.

    -Dan
     
  9. Feb 23, 2006 #8

    Hootenanny

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    What is the answer you book gave? You can type them directly using a code called Latex. There are a few tutorials on these forums about using it. Click on equation to display the code used.
     
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