Physics - W.E.P. Question with Solution | TestHub

For equilibrium, the force $F=-\frac{dU}{dx}$ must be zero. Given $U(x)=A{x}^{-2}-B{x}^{-1}$, we have $\frac{dU}{dx}=-2A{x}^{-3}+B{x}^{-2}$. Setting $\frac{dU}{dx}=0$, we get $-2A{x}^{-3}+B{x}^{-2}=0\Rightarrow B{x}^{-2}=2A{x}^{-3}\Rightarrow B=\frac{2A}{x}\Rightarrow x=\frac{2A}{B}$. To check for stable equilibrium, we evaluate $\frac{d^{2}U}{d{x}^2}$. $\frac{d^{2}U}{d{x}^2}=6A{x}^{-4}-2B{x}^{-3}$. At $x=\frac{2A}{B}$, $\frac{d^{2}U}{d{x}^2}=6A{\left(\frac{B}{2A}\right)}^4-2B{\left(\frac{B}{2A}\right)}^3=6A\frac{B^4}{16A^4}-2B\frac{B^3}{8A^3}=\frac{3B^4}{8A^3}-\frac{B^4}{4A^3}=\frac{B^4}{8A^3}$. Since $A$ and $B$ are positive, $\frac{d^{2}U}{d{x}^2}>0$, indicating stable equilibrium.