Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$F=-\frac{dU}{dx}\Rightarrow dU=-F.dx\Rightarrow U=-{\int }_0^x(-kx+a{x}^3)dx\Rightarrow U=\frac{k{x}^2}{2}-\frac{a{x}^4}{4}$
$\therefore \text{We get U=0 at x = 0 and x =}\sqrt{\frac{2k}{a}}\text{Also we get U = negative for x}$ > $\sqrt{\frac{2k}{a}}$

From the given function we can see that F = 0 at x = 0 i.e. slope of U-x graph is zero at x = 0.