A 2 kg block is attached to a spring (k = 10 N cm!) at point A as shown in the figure. The spring is compressed 50 cm
by the block When released the block travels over the
frictionless track until it is launched into the air at point B
Determine the velocity at point B if height, h, is 45 m

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