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

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

We know $F=-\frac{dU}{dx}$
$\Rightarrow dU=-F\cdot dx$
$\Rightarrow U=-\underset{0}{\overset{x}{\int }}\left(-kx+a{x}^3\right)dx$
$\Rightarrow U=\frac{k{x}^2}{2}-\frac{a{x}^4}{4}$
$\therefore$ We get $U=0$ at $x=0$ and $x=\sqrt{\frac{2k}{a}}$
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 $x=0$.