Two springs have their force constants ${\text{k}}_{\text{1}}{\text{\hspace{0.17em}and\hspace{0.17em}k}}_{\text{2}}$and they are stretched to the same extension. If ${\text{k}}_{\text{2}}>{\text{k}}_{\text{1}}$work done is

Work done stretching spring stored in the form of elastic potential energy

$\text{U=}\frac{1}{2}\text{K}{x}^2$

$\text{W}=\text{U}=\frac{1}{2}\text{K}{x}^2$

$\text{W}\propto \text{K}$

When x is same, here we get work done is more in spring ${\text{k}}_{\text{2}}$ because

${\text{k}}_{\text{2}}>{\text{k}}_{\text{1}}$