Question

A spring has a spring constant of 20 N/m. How much potential energy is stored in the spring as it stretched 0.20 m

Answer 1

The elastic potential energy (Ep) is given by $Ep = \frac{1}{2}*k*x^2$

Data: 
Ep = ? (Joule)
k = 20 N/m
x (displacement) = 0.20 m

Solving:
$Ep = \frac{1}{2}*k*x^2$
$Ep = \frac{1}{2}*20*0.20^2$
$Ep = \frac{20*0.04}{2}$
$Ep = \frac{0.8}{2}$
$\boxed{\boxed{Ep = 0.4\:J}}\end{array}}\qquad\quad\checkmark$