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

(a) The rock's kinetic energy when it leaves the spring will be equal to the energy $U_s$ stored in the spring initially. We can find the speed of the rock. $KE = U_s$ $\frac{1}{2}mv^2=\frac{1}{2}kx^2$ $v^2=\frac{kx^2}{m}$ $v=\sqrt{\frac{kx^2}{m}}$ $v=\sqrt{\frac{(1000~N/m)(0.30~m)^2}{0.400~kg}}$ $v = 15~m/s$ The rock leaves the spring with a speed of 15 m/s. (b) The potential energy of the rock at maximum height will be equal to the rock's kinetic energy when it leaves the spring. We can find the maximum height $h$ reached by the rock. $PE = KE$ $mgh = \frac{1}{2}mv^2$ $h = \frac{v^2}{2g}$ $h = \frac{(15~m/s)^2}{(2)(9.80~m/s^2)}$ $h = 11.5~m$ Since the rock's maximum height is less than the 15 meters required to reach the fruit, the contestants will go hungry.