Q. 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 $F(x) = - kx + ax^{3}$. 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)
Explanation
(d) $F = \frac{- dU}{dx} \Rightarrow dU = - F\ dx$
$\Rightarrow U = - \int_{0}^{x}{\left( - Kx + ax^{3} \right)dx = \frac{kx^{2}}{2}} - \frac{ax^{4}}{4}$
$\therefore$ We get $U = 0$ at $x = 0$ and $x = \sqrt{2k/a}$
And also $U =$ negative for $x > \sqrt{2k/a}$
So $F = 0$ at $x = 0$
$i.e.$ slope of $U - x$ graph is zero at $x = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.