Respuesta :

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

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

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