## University Physics with Modern Physics (14th Edition)

We can find the force constant $k$ of the spring. $kx = mg$ $k = \frac{mg}{x} = \frac{(3.15~kg)(9.80~m/s^2)}{0.0140~m}$ $k = 2205~N/m$ We can find the distance the spring stretches when it stores 10.0 J of potential energy. $\frac{1}{2}kx^2 = 10.0 ~J$ $x^2 = \frac{20.0~J}{2205~N/m}$ $x = \sqrt{\frac{20.0~J}{2205~N/m}}$ $x = 0.0952~m$ The total length of the spring would be 12.00 cm + 9.52 cm, which is 21.52 cm.