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

(a) $v = 1.7~m/s$ (b) If we round off, the package barely makes it onto the truck with a final kinetic energy of zero.
(a) The sum of the kinetic energy and the potential energy in the truck will be equal to the initial energy stored in the spring. $KE+PE=U_s$ $KE = U_s-PE$ $\frac{1}{2}mv^2 = \frac{1}{2}kx^2-mgh$ $v^2 = \frac{kx^2-2mgh}{m}$ $v = \sqrt{\frac{kx^2-2mgh}{m}}$ $v = \sqrt{\frac{(500~N/m)(0.30~m)^2-(2)(2.0~kg)(9.80~m/s^2)(1.0~m)}{2.0~kg}}$ $v = 1.7~m/s$ (b) We can find the work that the sticky spot does on the package. $W_f = -mg~\mu_k~d$ $W_f = -(2.0~kg)(9.80~m/s^2)(0.30)(0.50~m)$ $W_f = -2.94~J$ Let's assume that the final kinetic energy is zero. We can find the maximum possible height of the package. $PE = U_s+W_f$ $mgh = \frac{1}{2}kx^2+W_f$ $h = \frac{\frac{1}{2}kx^2+W_f}{mg}$ $h = \frac{\frac{1}{2}(500~N/m)(0.30~m)^2-2.94~J}{(2.0~kg)(9.80~m/s^2)}$ $h = 0.998~m \approx 1.0~m$ If we round off, the package barely makes it onto the truck with a final kinetic energy of zero.