Answer
62.7 J
Work Step by Step
In Problem 19a, we found that the spring constant was k = 784 N/m. To find the elastic potential energy of the compressed spring, we use
$U = \frac{1}{2} kx^{2}$
The spring is compressed 10.0 cm and then an additional 30.0 cm, so x = 10.0 cm + 30.0 cm = 40.0 cm = 0.40 m. Plugging in these values:
$U = \frac{1}{2} kx^{2}$
$ = \frac{1}{2} (784 N/m)(0.40 m)^{2}$
= 62.7 J
