Question

Calculate the elastic potential energy when a 1000N/m spring is stretched by a 20N force

Answer 1

The Hooke's law says

$$F=kx$$

So, the deformation of a spring

$$x=\frac{F}{k}=\frac{20\:\rm N}{1000\:\rm N/m}=0.02\:\rm m$$

The elastic potential energy

$$E_p=\frac{kx^2}{2}=\frac{1000\:\rm N/m\times (0.02\:\rm m)^2}{2}=0.2\:\rm J$$