Work and gravitational potential energy involving an optical illusion

AI Thread Summary
The discussion centers on calculating the work done by a climber pushing a 300 N wheelbarrow up an optical illusion staircase with a ramp at a 15° angle. The ramp consists of four sections with varying lengths, and the climber's effort is questioned due to the illusion of returning to the starting point. The participant attempted to calculate the work using the formula for work done against gravity but is uncertain about the accuracy of their approach. The key point is that despite the physical effort, the climber does no net work because they end up at the same elevation. Clarification on the correct formula and approach is sought to resolve the confusion.
MikWillis
Messages
2
Reaction score
0

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
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top