Answer
(a) W = 374 J
(b) W = -333 J
(c) W = 0
(d) The net work is 41 J
(e) W = 352 J
Work Step by Step
(a) $W = F~d = (72.0~N)(5.20~m)$
$W = 374~J$
(b) $W = mg~d~cos(\theta)$
$W = (128.0~N)(5.20~m)~cos(120.0^{\circ})$
$W = -333~J$
(c) The normal force does zero work because it makes an angle of $90^{\circ}$ with the direction of motion.
(d) The net work done on the carton is $374~J-333~J$, which is $41~J$.
(e) $W = F~d~cos(\theta)$
$W = (72.0~N)(5.20~m)~cos(20.0^{\circ})$
$W = 352~J$