# Lenard Jones potential

1. Oct 22, 2007

### Oblio

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. Oct 22, 2007

### Oblio

can i ignore E since its a constant? I got something like,
$$x = /frac{-6r^6A + 12B}{2r^13} t^2 [\tex] 3. Oct 22, 2007 ### christianjb The minimum is the point at which dV/dr=0 4. Oct 22, 2007 ### Oblio I'm getting the ridiculous value of : r = [tex]\sqrt[6]{\frac{-12B}{6A}}$$

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..