Physics - NLM Question with Solution | TestHub

The work done in stretching the spring, $W=\frac{1}{2}K{x}^2$ $K=$ force constant of spring $x=$ extension in length of the spring $$\begin{array}{l}{W}_1=\frac{1}{2}K{x}_1^2.\\ W_2=\frac{1}{2}K{x}_2^2\\ x_1=1\\ x_2=1+1=2\\ \frac{W_1}{W_2}=\frac{\frac{1}{2}K(1{)}^2}{\frac{1}{2}K(2{)}^2}=\frac{1}{4}\\ W_2=4W_1\\ W_1=10\mathrm{J}\\ W_2=40\mathrm{J}\end{array}$$ $\therefore$ More work required $=40\mathrm{J}-10\mathrm{J}=30\mathrm{J}$