A spring of spring constant 5×${10}^3$ N/m is stretched initially by 5 cm from the upstretched position. The work required to further stretch the spring by another 5cm is:

$\Rightarrow \mathrm{W}={\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}$

$\Rightarrow \mathrm{work} \mathrm{done}= \mathrm{difference} \mathrm{in} \mathrm{PE} \mathrm{of} \mathrm{spring}$

$\mathrm{W}=\frac{1}{2}\mathrm{K}({\mathrm{x}}_{\mathrm{f}}^2-{\mathrm{x}}_{\mathrm{i}}^2)$

$\mathrm{x}=\mathrm{distance}\mathrm{from}\mathrm{unstretched}\mathrm{position}$

${\mathrm{x}}_{\mathrm{f}}^{ }=10\mathrm{cm}=0.1\mathrm{m}; {\mathrm{x}}_{\mathrm{i}}^{ }=0.05\mathrm{m}; \mathrm{K}=5\times {10}^3\mathrm{N}/\mathrm{m}$

$\mathrm{W}=\frac{1}{2}\times 5\times {10}^3[0{.1}^2-0{.05}^2]$

$\mathrm{Work} \mathrm{required} \mathrm{to} \mathrm{further} \mathrm{stretch} \mathrm{the} \mathrm{spring}=18.75 \mathrm{J}.$