A group of particles of total mass 48 kg has a total kinetic energy of 320 J. The kinetic energy relative to the center of mass is 80 J. What is the speed of the center of mass?

K=.5mv^2 + relative Ke
320=.5(48)v^2+80
V=3.16

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