Inclined plane with massless, frictionless winch

AI Thread Summary
A 100 kg box is being pulled up a 45° inclined plane by a massless, frictionless winch with a constant angular velocity of 3.14 radians/sec. The coefficient of friction between the box and the plane is 0.25, and the winch completes 10 revolutions. The work done on the box is calculated using the formula W = mgh + uNd, where mgh represents gravitational potential energy gained and uNd accounts for frictional work. The correct answer for the work done is believed to be 13620 J, but the calculations need to accurately reflect the forces at play. Understanding the opposing forces and their impact on the distance moved is crucial for solving this problem correctly.
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A 100 kg box is pulled up an inclined plane by a massless, frictionless winch to which it is attached by a massless rope. The plane makes an angle of 45° with respect to the horizontal. The coefficient of friction between the box and the plane is μ = 0.25 (with μs = μk). The winch has a radius of 0.25 m and is turned at a constant angular velocity of ω = 3.14 radians/sec.

How much work is done on the 100 kg box by the rope when the winch makes 10 complete revolutions?

(a) 4380 J
(b) 5235 J
(c) 9262 J
(d) 12531 J
(e) 13620 J


So I did d = [2(.25)(3.34)]10 since we need 10 rev. And then...
W = mgh + uNd
W = mg(d sin 45) + u(mg cos 45)d
but I'm not getting the right answer, I think I'm suppose to get 13620J, but how do I get that?



 
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I'm not so sure that W = mgh + uNd
Isn't mgh the gravitational potential that is gained by doing the work?

I would sum up the forces that are opposing the pull of the rope (parallel to the plane) and multiply it by the distance it moves
 
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