EXIT QUESTION: One end of a light inextensible string is attached to a tool box of mass 2.5 kg which is lying on a horizontal table. The string passes over a smooth pulley and is tied at the other end to a bag of mass 1.4 kg. a Draw a diagram showing the forces acting on the tool box. b If the tool box is just on the point of sliding, find a value for μ, the coefficient of friction.
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Alinear relationship would look like a straight line on the graph. it could have a positive negative or 0 slope. so it could be a line potentially going in any direction but it must be perfectly straight.

10 is the . . .

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