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Homework Help: Lenard Jones potential

  1. Oct 22, 2007 #1
    I'm really stuck on b.) of this question.

    http://www.physics.brocku.ca/Courses/2P20/problems/prob_lenardjones.pdf [Broken]

    I think I can use T = E - U, and solve for v from within T and integrate for x...

    but what do I do with E?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 22, 2007 #2
    can i ignore E since its a constant? I got something like,
    [tex] x = /frac{-6r^6A + 12B}{2r^13} t^2 [\tex]
  4. Oct 22, 2007 #3
    The minimum is the point at which dV/dr=0
  5. Oct 22, 2007 #4
    I'm getting the ridiculous value of :

    r = [tex]\sqrt[6]{\frac{-12B}{6A}}[/tex]

    by deriving the two quotients, allowing A' and B' to cancel in the process (constants) cancelling out the high denominators as much as possible... etc..
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