In the scaled diagram, two forces, F1 and F2 , act on a 4.0-kilogram block at point P. Force F1 has a magnitude of 12 Newtons, and is directed toward the right. Force F2 has a magnitude of 5 Newtons and is directed toward the left.

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

Ibelieve the answer is d. densityhope my answer has you : )