A 12.0-kg package in a mail-sorting room slides 2.00 m down a chute that is inclined at 53.0∘ below the horizontal. The coefficient of kinetic friction between the package and the chute’s surface is 0.40. Calculate the work done on the package by .

(a) friction,
(b) gravity, and
(c) the normal force.
(d) What is the net work done on the package?

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j