Answer
(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.
Work Step by Step
(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.
