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

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

We know $F=-\frac{d U}{d x}$
$\Rightarrow d U=-F \cdot d x$
$\Rightarrow U=-\int\limits_{0}^{x}\left(-k x+a x^{3}\right) d x$
$ \Rightarrow U=\frac{k x^{2}}{2}-\frac{a x^{4}}{4}$
$\therefore $ We get $U=0$ at $x=0$ and $x=\sqrt{\frac{2 k}{a}}$
Also we get $U=$ negative for $x>\sqrt{\frac{2 k}{a}}$
From the given function we can see that $F=0$
at $x=0$ i.e., slope of $U-x$ graph is zero at $x=0$.