A small block, with a mass of 350 g, starts from rest at the top of an incline. It slides without friction down the incline, around the inside
of the loop, and continues on the level track to the right. The height
of the incline is 110 cm, and the radius of the loop is 15 cm.
Adam?
To: Every
Type mess
a. Find the initial potential energy of the block.
b. Find the velocity of the block at the bottom of the loop.
c. Find the velocity of the block at the top of the loop.
d. What is the normal force on the block at the bottom of the
loop?
e. What is the normal force on the block at the top of the loop?

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