The 1.5kg cylinder C travels along the path described by r sin m  0.6   . If arm OA is rotating counterclockwise with an angular velocity of   3 / rad s, determine the force exerted by the smooth slot in arm OA on the cylinder at the instant 60o   . The spring has a stiffness of 100 / N m and is unstretched when 30o   . The cylinder is in contact with only one edge of the slotted arm. Neglect the size of the cylinder. Motion occurs in the vertical plane (which means to account for gravity)

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

The answer is  the potential energy of the bob at position b is 1.47 joules.if the maximum height of the bob is 0.45 meters when it swings back,1.323   joules of energy was transformed to thermal energy.hope this

B? i think so but hope that's right and answers your question

Answer4 million;