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

(a) The energy stored in the spring will be equal to the ball's kinetic energy when it leaves the barrel. $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$ $v^2 = \frac{kx^2}{m}$ $v = x~\sqrt{\frac{k}{m}}$ $v = (0.0600~m)~\sqrt{\frac{400~N/m}{0.0300~kg}}$ $v = 6.93~m/s$ (b) The kinetic energy in the case with friction will be equal to the kinetic energy in the frictionless case reduced by the negative work done by friction. $K_2 = K_1 - F~d$ $\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2 - F~d$ $v^2 = \frac{mv_1^2-2F~d}{m}$ $v = \sqrt{\frac{mv_1^2-2F~d}{m}}$ $v = \sqrt{\frac{(0.0300~kg)(6.93~m/s)^2-(2)(6.00~N)(0.0600~m)}{(0.0300~kg)}}$ $v = 4.90~m/s$ (c) The ball has the greatest speed at the point when the force of the spring is equal to the resisting force. After this point, the resisting force is greater than the force of the spring and the ball starts to decelerate. $kx = F$ $x = \frac{F}{k} = \frac{6.00~N}{400~N/m}$ $x = 0.015~m$ The ball has the greatest speed when x = 0.015 meters. We can find the speed of the ball at this point. $K_2+U_2 = K_1+U_1+W$ $\frac{1}{2}mv^2+\frac{1}{2}k(0.015~m)^2 = 0+\frac{1}{2}k(0.0600~m)^2 -(6.00~N)(0.045~m)$ $v^2 = \frac{k((0.0600~m)^2-(0.015~m)^2)-(2)(0.27~J)}{m}$ $v = \sqrt{\frac{(400~N/m)((0.0600~m)^2-(0.015~m)^2)-0.54~J}{0.0300~kg}}$ $v = 5.20~m/s$