A spring of spring constant  $5\times {10}^3$ N/m  &  stretched initially by 5 cm from the unstretched position.  Then work done required to stretch it further by another 5cm is

$$\begin{gathered} work done to stretch spring is W=\frac{1}{2}K({x_2}^2-{x_1}^2) \\ given spring cons\tan t is K=5 x{10}^{3}N/m; x_2=10cm=10x{10}^{-2}m;x_1=5cm=5x{10}^{-2}m \\ W=\frac{1}{2}\times 5\times {10}^3({10}^2-5^2)\times {10}^{-4} \\ W=\frac{5\times {10}^3}{2}\times 0.0075 \\ W = 18.75 joule \end{gathered}$$