If an object on a spring became four times as massive, how would that affect the period of its oscillation on the spring?

The formula for the time period oscillation of a spring is:
T = 2π√(m/k) where m is the mass of the object attached to the spring and k is the spring constant of the spring.

Multiplying m by 4 we have:
2π√(4m/k)
= 2π x 2√(m/k)
or 2 x (2π√(m/k)) or 2T.

Therefore, when object is four times as massive, then period will double.

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