Answer
The package has a speed of 3.41 m/s after sliding 2.80 meters.
Work Step by Step
(a) $W = -F_f~d$
$W = -mg~cos(\theta)~\mu_k~d$
$W = -(5.00~kg)(9.80~m/s^2)~cos(24.0^{\circ})(0.310)(2.80~m)$
$W = -38.9~J$
(b) $W = mg~sin(\theta)~d$
$W = (5.00~kg)(9.80~m/s^2)~sin(24.0^{\circ})(2.80~m)$
$W = 55.8~J$
(c) Since the normal force acts at an angle of $90^{\circ}$ to the direction of motion, the normal force does zero work.
(d) $W_{tot} = 55.8~J - 38.9~J = 16.9~J$
(e) $W_{tot} = K_2-K_1$
$W_{tot} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$
$v_2^2 = \frac{2~W_{tot}+mv_1^2}{m}$
$v_2 = \sqrt{\frac{(2)(16.9~J)+(5.00~kg)(2.20~m/s)^2}{5.00~kg}}$
$v_2 = 3.41~m/s$
The package has a speed of 3.41 m/s after sliding 2.80 meters.