## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

When the book is placed on the spring, the spring is compressed by 6 cm. The force that the spring pushes is equal to the weight of the book. Therefore; $kx = mg$ $k = \frac{mg}{x}$ $k = \frac{(2.2~kg)(9.80~m/s^2)}{0.060~m}$ $k = 360~N/m$ The spring constant is 360 N/m.