Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Given:
$F\left(x\right)=-kx+a{x}^3$

As we know that,

$F=-\frac{dU}{dx}$

$\therefore dU=-Fdx$

taking integration both side

$U\left(x\right)=-\underset{0}{\overset{x}{\int }}(-kx+a{x}^3)dx$

$U\left(x\right)=\frac{k{x}^2}{2}-\frac{a{x}^4}{4}$

So, $U\left(x\right)=0$ at,

$\frac{k{x}^2}{2}-\frac{a{x}^4}{4}=0\Rightarrow x^2(\frac{k}{2}-\frac{a{x}^2}{4})=0$

$\therefore x=0$ and $x=\pm \sqrt{\frac{2k}{a}}$

So, we can see that $U\left(x\right)$ will be negative for $x>\sqrt{\frac{2k}{a}}$

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

Therefore, the most appropriate option is $\left(D\right)$.