Rectilinear motion of a particle

[Johnson]

A particle of mass m moves without friction subject to a force F(x) =
−kx + $\frac{kx3}{A2}$, where k and A are positive constants. It is projected
from x = 0 to the positive x direction with initial velocity v0 =
A$\sqrt{\frac{k}{2m}}$. 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:
$\frac{1}{2}$ kx$^{2}$ - $\frac{1}{4}$ $\frac{kx^{4}}{A^{2}}$

Then I found T(x) (kinetic energy) to be:
T$_{o}$ - $\frac{1}{2}$kx$^{2}$ - $\frac{1}{4}$ $\frac{kx^{4}}{A^{2}}$

I know I am given an initial V$_{o}$, but where would I plug that into find V(x) or T(x)?

Any help is very much appreciated.

Regards

 

[Johnson]

Anyone? Do I need to include other attempts?

 

[danielakkerma]

Hi,
It seems to me you've done very well so far.
You did correctly work out, that under a central force expressible as:
$\vec{F} = -\vec{\nabla}{V}$
So that in your case,
$V(x) = -\displaystyle \int_0^x{Fdx}$
Here, under the constraints on the given question, it is okay to take the potential energy equal to some base/initial value of your choice(at x = 0), say even zero.
But recall that m*x''t = F(x), so it may be best to solve a differential equation for x(t), in order to find out the kinetic energy, which will be, in this case K = mx'(t)^2/2;
Try that,
Daniel