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−kx + [itex]\frac{kx3}{A2}[/itex], where k and A are positive constants. It is projected

from x = 0 to the positive x direction with initial velocity v0 =

A[itex]\sqrt{\frac{k}{2m}}[/itex]. Find:

(a) the potential energy V (x),

(b) the kinetic energy T(x),

(c) the turning points of the motion.

So I don't know if I am on the right track, I feel I am missing something. For V(x) (potential energy), i got:

[itex]\frac{1}{2}[/itex] kx[itex]^{2}[/itex] - [itex]\frac{1}{4}[/itex] [itex]\frac{kx^{4}}{A^{2}}[/itex]

Then I found T(x) (kinetic energy) to be:

T[itex]_{o}[/itex] - [itex]\frac{1}{2}[/itex]kx[itex]^{2}[/itex] - [itex]\frac{1}{4}[/itex] [itex]\frac{kx^{4}}{A^{2}}[/itex]

I know I am given an initial V[itex]_{o}[/itex], but where would I plug that in to find V(x) or T(x)?

Any help is very much appreciated.

Regards