A box is sliding up an incline that makes an angle of 30.8° with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.105. The initial speed of the box at the bottom of the incline is 2.79 m/s. How far does the box travel along the incline before coming to rest?

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

Id k but good !