The displacement from equilibrium of an oscillating weight suspended by a spring and subject to the damping effect of friction is given by y(t) = 2e−t cos 4t, where y is the displacement (in centimeters) and t is the time (in seconds). find the displacement when t = 0, t = 1 4 , and t = 1 2 . (round your answers to two decimal places.)

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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F  you double the  force, you double the acceleration, but  if  you double the  mass, you velocity. all bodies accelerate at the  same  rate regardless of size or  mass.