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

We can find the spring constant of the rubber band. Note that the rubber band has a force of 60 N when it is stretched 0.30 meters. $kx = F$ $k = \frac{F}{x}$ $k = \frac{60~N}{0.30~m}$ $k = 200~N/m$ We can find the energy stored in the rubber band when it is stretched 30 cm. $U_s = \frac{1}{2}kx^2$ $U_s = \frac{1}{2}(200~N/m)(0.30~m)^2$ $U_s = 9.0~J$ When the rock leaves the rubber band, the kinetic energy will be equal to the energy stored in the rubber band initially. We can find the speed of the rock. $KE = U_s$ $\frac{1}{2}mv^2 = 9.0~J$ $v^2 = \frac{(2)(9.0~J)}{m}$ $v = \sqrt{\frac{(2)(9.0~J)}{m}}$ $v = \sqrt{\frac{(2)(9.0~J)}{0.050~kg}}$ $v = 19.0~m/s$ The speed of the rock is 19.0 m/s