At t = 0 a block with mass m = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the equilibrium position of the spring. the block is attached to a massless spring of spring constant k = 61.2 n/m and slides on a frictionless surface. at what time will the block next pass x = 0, the place where the spring is unstretched? t1 =

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

Answerapplying a direct current to a non magnetic material in order to create a electromagnetic field; plz check first good luck;