Question

A spring becomes 15 cm long when the force of 40 N is used to pull the spring. The elastic potential energy stored in the spring is 2.2 J. What is the original length of the spring?

Answer 1

Given:

$l=0.15\:\rm m$

$F=40\:\rm N$

$E_p=2.2\:\rm J$

The lastic potential energy stored in the spring 

$$E_p=\frac{kx^2}{2}=\frac{Fx}{2}$$

So, the elongation of a spring

$$x=l-l_0=\frac{2E_p}{F}=\frac{2*2.2}{40}=0.11\:\rm m$$

Finally, the original length of the spring

$$l_0=l-x=0.15-0.11=0.04\:\rm m=4\:\rm cm$$