Question

The potential energy of a conservative system is given by $U=A{x}^2-Bx$, where $x$ represents the position of the particle. $A$ and $B$ are positive constants. Then

• A. Force acting on the system will be $(B-2Ax)$
• B. At equilibrium, potential energy will be $\frac{-B^2}{4A}$
• C. system is in stable equilibrium.
• D. None

Answer 1

Answer: (A), (B), (C)

The correct options are
A Force acting on the system will be $(B-2Ax)$
B At equilibrium, potential energy will be $\frac{-B^2}{4A}$
C system is in stable equilibrium.

Given $U=A{x}^2-Bx$
$F=-\frac{dU}{dx}=-(2Ax-B)=B-2Ax$
At equilibrium,
$F=0=B-2Ax$
$x=\frac{B}{2A},$
$\frac{d^{2}U}{d{x}^2}=2A=+ve$, hence stable equilibrium.
$\text{Potential energy at equilibrium}U=A\times \frac{B^2}{4A^2}-B\times \frac{B}{2A}=\frac{-B^2}{4A}$