A meter stick is balanced at its center (i. e. at the 50 cm mark). At the 22 cm mark is placed a 158 gram mass. Where should a 259 gram mass be placed, so that the stick remains balanced. Write your answer as the cm mark on the meter stick (round to one place of decimal).

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

The force is push or pull

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