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Potential energy of a spring and transverse motion

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Picture6.png

    L = Stretched distance
    L0 = Unstretched distance

    2. Relevant equations
    W = 1/2kL2


    3. The attempt at a solution
    In attempting to help a friend with this problem, having taken this same physics class under a different professor, i was completely stumped by this problem. I know that Fx = kL, but i can't make any headway on figuring out the part of the problem that involves the transverse of the spring that leads to a distance pulled of xm with the same applied force that stretches the spring the same distance. Is there something easy that i'm missing?
     
  2. jcsd
  3. Feb 13, 2012 #2
    For a) you need to parameterise the amount it has stretched in fig b as a function of x. Geometrically think of take the stretched spring in b, and rotating it back to vertical. Then consider the relation between the extrat height L(x) and the distance it was moved x. After this I presume you are okay?
     
  4. Feb 13, 2012 #3
    So basically, because the work done in each of the two scenarios is the same?

    Fx*xm = 1/2kL^2
    xm = (kL^2)/(2Fx)
     
  5. Feb 13, 2012 #4
    You do not know that the work done is the same, you just follow the prescribed method to determine it. I'm getting a bit confused by what you mean by

    What do you think is the answer to question a), Ie L(x)=? once you know this, then you can consider the force needed to extend the spring by L(x), and that is the answer to question b).
     
  6. Feb 13, 2012 #5
    By rotating b back vertically, the distance it was pulled is xm

    L(x) = xm
     
  7. Feb 13, 2012 #6
    Exactly, so now you can find the magnitude of the force easily enough, the rest follows from this.
     
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