Answer
(a) W 608 J
(b) W = -395 J
(c) W = 0
(d) W = -189 J
(e) W = 24 J
(f) v = 1.55 m/s
Work Step by Step
(a) $W = F~d = (160~N)(3.80~m)$
$W = 608~J$
(b) $W = -mg~sin(32.0^{\circ})~d$
$W = -(20.0~kg)(9.80~m/s^2)~sin(32.0^{\circ})(3.80~m)$
$W = -395~J$
(c) Since the normal force acts at an angle of $90^{\circ}$ to the direction of motion, the normal force does zero work.
(d) $W = -F_f~d$
$W = -mg~cos(\theta)~\mu_k~d$
$W = -(20.0~kg)(9.80~m/s^2)~cos(32.0^{\circ})(0.300)(3.80~m)$
$W = -189~J$
(e) $W_{tot} = 608~J -395~J-189~J$
$W_{tot} = 24~J$
(f) $\frac{1}{2}mv^2 = 24~J$
$v^2 = \frac{48~J}{m}$
$v = \sqrt{\frac{48~J}{20.0~kg}}$
$v = 1.55~m/s$