Answer
(a) The potential energy in the spring is 52.0 J.
(b) The potential energy in the spring is 3.25 J.
Work Step by Step
(a) We can find the force constant $k$ of the spring.
$kx = 520~N$
$k = \frac{520~N}{x} = \frac{520~N}{0.200~m}$
$k = 2600~N/m$
We can find the potential energy in the spring when it is stretched 0.200 m.
$E = \frac{1}{2}kx^2$
$E = \frac{1}{2}(2600~N/m)(0.200~m)^2$
$E = 52.0~J$
The potential energy in the spring is 52.0 J.
(b) We can find the potential energy in the spring when it is compressed 0.0500 m
$E = \frac{1}{2}kx^2$
$E = \frac{1}{2}(2600~N/m)(0.0500~m)^2$
$E = 3.25~J$
The potential energy in the spring is 3.25 J.
