Work and gravitational potential energy involving an optical illusion

[MikWillis]

Homework Statement

You see an optical illusion of an ever-upward spiral staircase. The climber trudges up and up and never gets anywhere, going in circles instead. Suppose the staircase is provided with a narrow ramp, allowing the tired stair-climber to push a wheelbarrow up the stairs. The loaded wheelbarrow weighs 300.0 N, and the ramp makes an angle of 15.0° with the horizontal, all along its length. The ramp consists of four straight sections, with slant lengths 12.0 m, 8.0 m, 20.0 m, and 20.0 m. How much work does the climber do on the wheelbarrow when he pushes it up the ramp from the red marker, all the way around the loop, and (supposedly) back to the red marker again? An ordinary inclined-plane computation will give an accurate value for the work. (In the illusory illustration, the fact that he ends up where he started means that, impossibly, he does NO work.)

Homework Equations

Work done against gravity: W=(delta)PE

Work done by gravity: W=-(delta)PE

W=F*vertical displacement

The Attempt at a Solution

I found the work done on all four sections and tried adding them together. It didn't work.

(12*sin15*300)+(8*sin15*300)+(20*sin15*300)+(20*sin15*300)=4658.7428=4660

 

[member 553083]

JI'm not sure if this is the right formula or if I should be using the work done against gravity formula. Any help would be appreciated.