A particle with mass $m$ moves under influence of a force $F=-cx^3$, with $c$ a constant. What is the potential energy function $V(x)$?

And if it starts to move from rest from position $x=-a$, what is the velocity at $x=0$?


closed as off-topic by David Z Apr 23 '15 at 12:01

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  • $\begingroup$ I think if the potential is only a function of position, then you can say that F = -$\nabla V(x)$. So if you integrate F with respect to x, I think you should be able to find V(x). $\endgroup$ – honeste_vivere Apr 23 '15 at 11:52
  • $\begingroup$ so $V(x)=-\frac{c}{4}x^4$? but what is then the velocity at $x=0$? $\endgroup$ – JimmyP Apr 23 '15 at 11:54
  • $\begingroup$ I think you have a sign error, but once you have potential, you can just conserve energy to get the answer for which you are looking. $\endgroup$ – honeste_vivere Apr 23 '15 at 11:58
  • $\begingroup$ you can use conservation of energy; at the start is it at rest but has potential energy $V(-a)$. At the end it has converted some (or all, depending on your integration constant) of this to kinetic energy, from which you can find the velocity $\endgroup$ – danimal Apr 23 '15 at 11:58
  • $\begingroup$ alternatively you can use Newton's second law and do some integration... $\endgroup$ – danimal Apr 23 '15 at 11:58