Solve qus no. 46

Dear Student ,
If the spring constant is k then the potential energy stored in the spring is ,
$$\begin{gathered} U=\frac{1}{2}k{x}^2 \\ Now when x=2 then potential energy=U=\frac{1}{2}k\times 4=2k \\ So when x=8 then the potential energy is , \\ U\text{'}=\frac{1}{2}k{x}^2=\frac{1}{2}k\times 64=32k=16\times 2k=16U . \end{gathered}$$
Regards