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Q. The ratio of spring constants of two springs is $2: 3$. What is the ratio of their potential energy, if they are stretched by the same force?

Work, Energy and Power

Solution:

The spring forces are
$\because F=k_{1} x_{1}$ and $F=k_{2} x_{2}$
$\therefore k_{1} x_{1}=k_{2} x_{2}$
$\Rightarrow \frac{k_{1}}{k_{2}}=\frac{x_{2}}{x_{1}}$
$\Rightarrow \frac{( PE )_{1}}{( PE )_{2}}=\frac{k_{1} x_{1}^{2}}{k_{2} x_{2}^{2}}$
$=\frac{k_{1}}{k_{2}} \times\left(\frac{k_{2}}{k_{1}}\right)^{2}=\frac{k_{2}}{k_{1}}=\frac{3}{2}$
$\left[\because \frac{k_{1}}{k_{2}}=\frac{2}{3}\right]$