The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is:

Let extension produced in a spring be x initially. In stretched condition spring will have potential energy

$\mathrm{U} = \frac{1}{2}{\mathrm{kx}}^2$

Where k is spring constant or force constant

$\therefore \frac{{\mathrm{U}}_1}{{\mathrm{U}}_{2 }} = \frac{{{\mathrm{x}}^2}_1}{{{\mathrm{x}}^2}_2}$-----------------(i)

Given, ${\mathrm{U}}_1 = \mathrm{U}, {\mathrm{x}}_1 = 2 \mathrm{cm}, {\mathrm{x}}_2 = 8 \mathrm{cm}$

Putting these values in Eq.(i), we have

$\frac{\mathrm{U}}{{\mathrm{U}}_2} = \frac{{\left(2\right)}^2}{{\left(8\right)}^2} = \frac{4}{64} = \frac{1}{16}$

$\therefore {\mathrm{U}}_2 = 16 \mathrm{U}$