Answering the two questions in reverse order:
-- No. I don't need to know how the speed of the person changed before I can answer the question. I can answer it now.
-- The NET work done by the gravitational force is zero.
-- As the person and his girl-friend go up the first half of the wheel, the motor does positive work and gravity does negative work.
-- After they pass the peak at the top and come down the second half of the wheel, the motor does negative work and gravity does positive work, even though the couple may be interested in other things during that time.
-- The total work done by gravity in one complete revolution is zero.
-- The total work done by the motor in one complete revolution is only what it takes to pay back the energy robbed by friction and air resistance.
The work done by the gravitational force is zero.
Work done by a conservative force is path independent. Which means it only depends on the initial and final position of the body. The gravitational force is a conservational force and the gravitational potential energy depends only upon the height of the body.
Let the lowest point of the body is at some height h, then the initial gravitational potential energy of the person is:
PE(initial) = mgh
The final position of the person is also at a height h, thus, the final gravitational potential energy :
PE(final) = mgh
According to the work-energy theorem:
work done = - change in potential energy
work done = -(mgh - mgh) = 0
Thus, the work done is zero in the given case.
Learn more about work done:
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