## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) Let $y$ be the compression distance of the spring. Then the block also falls a distance of $y$ from its initial height. The energy stored in the compressed spring will be equal to the block's initial potential energy. $U_s = PE$ $\frac{1}{2}ky^2 = mgy$ $y = \frac{2mg}{k}$ $y = \frac{(2)(5.0~kg)(9.80~m/s^2)}{490~N/m}$ $y = 0.20~m$ The spring compresses a distance of 0.20 meters. (b) The force with which the spring pushes up on the block will be equal to the block's weight. Let $y$ be the compression distance. $ky = mg$ $y = \frac{mg}{k}$ $y = \frac{(5.0~kg)(9.80~m/s^2)}{490~N/m}$ $y = 0.10~m$ The spring compresses a distance of 0.10 meters. (c) The two answer are different because in part (b) the force of the hand pulling up on the block does negative work on the block which removes some energy from the system.