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Question on forces and energy

  1. Dec 9, 2009 #1
    I just watched this http://www.youtube.com/watch?v=eVl5zs6Lqy0" video...

    If W = KE ---> Fd = [mv^2] / 2

    If in a spring, force is equal to a constant (k) times how far its been stretched (d), then subbing kd for F in the first equation...

    Kd^2 = [mv^2]/2
    Rearranging, E = kd^2

    But in my textbook, it says that E = 1/2 [kd^2]
    Where did the 1/2 come from? Even in the video, W = Fd, so from the graph, you times F by d, getting a SQUARE, not a triangle, but he says you times F by d, and divide by 2 since your looking at the area (triangle) underneath...

    Any help?
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Dec 9, 2009 #2
    [tex] U = \frac{1}2kx^2[/tex]

    is the potential energy of the spring. You're talking about the kinetic energy.
    Last edited by a moderator: Apr 24, 2017
  4. Dec 9, 2009 #3

    If on a F vs x graph, you want to find work...

    F = k x
    W = F d

    Then shouldn't work = k xd = kx^2?
  5. Dec 9, 2009 #4
    Yes....but the net work is only equal to the change in kinetic energy. What your book is saying (spring's potential energy) has nothing to do with what you're saying. For a conservative force (ie. only a position-dependent force) like Hooke's law, the force will equal the negative gradient of the potential. So now we have:

    [tex] F = -kx[/tex]

    Take integral and get:

    [tex]U = \frac{1}2kx^2[/tex]
    Last edited: Dec 9, 2009
  6. Dec 9, 2009 #5
    https://www.physicsforums.com/latex_images/24/2483850-2.png [Broken]
    whats that big s line thing for lol
    Last edited by a moderator: May 4, 2017
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