#### Answer

(a) The spring should be compressed a distance of 5.48 cm.
(b) The spring will be compressed a maximum distance of 3.92 cm.

#### Work Step by Step

(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.