Answer to Question #344170 in Physics for tom

Question

How much energy is in the elastic potential energy store of a spring with a spring constant of 4 N/m if it is stretched from 1 m to 1.1 m in length?

Answer 1

$$W=\frac{kx_2^2}{2}-\frac{kx_1^2}{2}\\ =\frac{4}{2}*(1.1^2-1.0^2)=0.42\:\rm J$$