Question

A particle is projected to the right with initial kinetic energy T0but subject to a force
F(x) = −kx + kx3/A2, where k and A are positive constants. Find (a) the potential
energy function V (x) for this force; (b) the kinetic energy, and (c) the total energy of
the particle as a function of its position.

Answer 1

a) I believe this is right...
\[-\nabla U= \textbf F\]
simplifies to
\[ \frac{\partial U }{\partial x} \hat { \textbf x}=-\textbf F\]
The potential energy is only in the x direction, and only along the direction of F, so we can change the partials to differentials and drop the x hat

\[ \mathrm{d} U(x) = -F(x) \mathrm{d}x \]
U is zero at x=0, then we'll rename U to V(x), so we can say
\[ V(x) = -\int_0^x F \mathrm{d}x=\int_0^x \Big( kx - \frac{kx^3}{A^2} \Big) \mathrm{d}x\]