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 $F(x)=-kx+a{x}^3$. Here, $k$ and a are positive constants. For $x\ge 0$, the functional form of the potential energy $U(x)$ of the particle is

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

$F=-\frac{dU}{dx}$
$\therefore dU=-F.dx$
or $U(x)=-\underset{0}{\overset{x}{\int }}(-kx+a{x}^3)dx$
$U(x)=\frac{k{x}^2}{2}-\frac{a{x}^4}{4}$
$U(x)=0atx=0$, and $x=\sqrt{\frac{2k}{a}}$
$U(x)$=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
$x=0$.