Two springs P and Q having stiffness ${\mathrm{K}}_{1 }\mathrm{and} {\mathrm{K}}_2\left(<{\mathrm{K}}_{1 }\right)$ respectively are stretched equally. If equal forces are applied on two springs, then

Work done =W= $\frac{1}{2}k{x}^2=\frac{F^2}{2K}$

$\mathrm{W}\propto \frac{1}{\mathrm{K}}$

Spring Q has lesser force constant than P. So Q will develop less restoring force than P. As a result , Q will suffer more extension. Since force is same in both the cases, more work will be done on Q.