A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 200 N/m. The spring-block system is initially stretched a distance “d” beyond the spring's equilibrium position and is released from rest. As the block moves to the right, its speed is 50.0 cm/s when the spring-block system is 4.00 cm from the equilibrium position.
Find d, the initial displacement of the spring-block system (in cm).

Domain eukarya; kingdom plantae should be the answer.

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