## Fundamentals of Physics Extended (10th Edition)

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