Question

A rod of mass M hinged at O is kept in equilibrium with a spring of stiffness k as shown in figure. The potential energy stored in the spring is

• A. $\frac{{\left(mg\right)}^2}{4k}$
• B. $\frac{{\left(mg\right)}^2}{2k}$
• C. $\frac{{\left(mg\right)}^2}{8k}$
• D. $\frac{{\left(mg\right)}^2}{k}$

Answer 1

The correct option is C $\frac{{\left(mg\right)}^2}{8k}$

As the rod is in equilibrium torque about the hing must be 0.
torque due to gravity is ${\tau }_g=mg\frac{l}{2}$( as mg force acts on the COM)
if the spring exerts a force F then ${\tau }_s=Fl$
${\tau }_g={\tau }_s⟹F=\frac{mg}{2}$
we Know that force exerted by spring when it stretches by length x is $kx$
$\therefore kx=\frac{mg}{2}⟹x=\frac{mg}{2k}$
PE stored in the spring is $PE=\frac{1}{2}k{x}^2=\frac{1}{2}k(\frac{mg}{2k}{)}^2=\frac{(mg{)}^2}{8k}$