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

(a) $\frac{1}{2}kx^2 = E$ $x^2 = \frac{2E}{k}$ $x = \sqrt{\frac{2E}{k}} = \sqrt{\frac{(2)(1.20~J)}{800~N/m}}$ $x = 0.0548~m = 5.48~cm$ The spring should be compressed a distance of 5.48 cm. (b) The energy stored in the spring at maximum compression will be equal to the initial potential energy of the book. Let $y$ be the distance that the spring is compressed at the point of maximum compression. Then the change in height of the book is also $y$. $U_s = PE$ $\frac{1}{2}ky^2 = mgy$ $y = \frac{2mg}{k}$ $y = \frac{(2)(1.60~kg)(9.80~m/s^2)}{800~N/m}$ $x = 0.0392~m = 3.92~cm$ The spring will be compressed a maximum distance of 3.92 cm.