Answer
(a) W = 324 J
(b) W = -324 J
(c) W = 0
Work Step by Step
(a) Since the monitor moves at a constant speed, the force of friction must be equal in magnitude to the component of the weight directed down the ramp.
$W = F_f~d$
$W = mg~sin(\theta)~d$
$W = (10.0~kg)(9.80~m/s^2)~sin(36.9^{\circ})(5.50~m)$
$W = 324~J$
(b) We can find the work done by gravity.
$W = mg~d~cos(126.9^{\circ})$
$W = (10.0~kg)(9.80~m/s^2)(5.50~m)~cos(126.9^{\circ})$
$W = -324~J$
(c) Since the normal force acts at a $90^{\circ}$ angle to the direction of motion, the normal force does zero work.
