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

(a) The maximum compression of the spring is 0.60 meters. (b) The maximum value of $v_0$ is 1.50 m/s.
(a) The potential energy in the spring will be equal to the block's initial kinetic energy. $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$ $x^2 = \frac{mv^2}{k}$ $x = \sqrt{\frac{mv^2}{k}}$ $x = \sqrt{\frac{(5.00~kg)(6.00~m/s)^2}{500~N/m}}$ $x = 0.60~m$ The maximum compression of the spring is 0.60 meters. (b) The potential energy in the spring will be equal to the block's initial kinetic energy. $\frac{1}{2}mv_0^2 = \frac{1}{2}kx^2$ $v_0^2 = \frac{kx^2}{m}$ $v_0 = \sqrt{\frac{kx^2}{m}}$ $v_0 = \sqrt{\frac{(500~N/m)(0.150~m)^2}{5.00~kg}}$ $v_0 = 1.50~m/s$ The maximum value of $v_0$ is 1.50 m/s.