#### 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.