Physics Question on work, energy and power

Question

A particle, which is constrained to move along x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F (x) $=-kx+ax^3$ . Here, k and a are positive constants. For $x \ge 0$ , the functional form of the potential energy U (x) of the particle is

Answer 1

Solution

$F=-\frac{dU}{dx}$
$\therefore \hspace10mm dU=-F.dx \, or \, U(x)=-\int_0^x(-kx+ax^3)dx$
$\hspace10mm U(x)=\frac{kx^2}{2}-\frac{ax^4}{4}$
$\hspace10mm U(x)=0 \, at \, x=0 \, and \, x=\sqrt{\frac{2k}{a}}$
$\hspace10mm U(x)=negative \, for \, x=\sqrt{\frac{2k}{a}}$
From the given function, we can see that
F = O at x = 0 i.e. slope of U-x graph is zero at
x = 0. Therefore, the most appropriate option is (d).