A 70kg teacher (including the parachute) is skydiving! As the parachute opens,
the system experiences a 1000N drag.
How many forces are acting on the system?
2
What is the magnitude of the net force on the system?
N
What is the magnitude of the system's acceleration?
m/s2
Describe the system's speed:
decreasing

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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

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