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How come I can't use m(v^2/r)=kx for this problem?

  1. May 8, 2012 #1
    Relaxed spring is sitting on a horizontal surface. A block is attached at one of its ends is kicked with a horizontal velocity, v1, given to it. The block will move and stretch. Find the distance, x, the spring will stretch. X is in meters.




    Energy 1/2 mv1^2 = 1/2 mv2^2 + 1/2 k x^2
    Angular momentum mv1l0 = mv2(l0+x)




    The problem was solved using kinetic energy and conservation of angular momentum formulas.

    My question is, why doesn't m(v^2/r)=kx apply for this problem? Is it because m(v^2/r) is to only be used in uniform circular motion and the object in the problem moves in the shape of an ellipse?
     
  2. jcsd
  3. May 8, 2012 #2

    cepheid

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    I have no idea what path this thing moves on, but if the string stretches, then it can't be a circle.
     
  4. May 8, 2012 #3
    understood, but does m(v^2/r) only apply to a circle?
     
  5. May 8, 2012 #4
    Yes.
     
  6. May 8, 2012 #5
    thank you!
     
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