Calculating Work on a Bucket Hanging in a Well

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To calculate the work done on a 6.75 kg bucket being raised 4.00 m in a well, the force of tension must equal the gravitational force when the bucket is moved at a constant speed. The work done by the person pulling the bucket is calculated as 3.60 J, while gravity does -0.900 J of work. The total work on the bucket is therefore 2.70 J. This indicates that the energy expended in raising the bucket is significantly less than the gravitational force acting on it. The discrepancy in expected answers highlights the importance of understanding the forces involved in the system.
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Homework Statement


An old oaken bucket of mass 6.75 kg hangs in a well at the end of a rope. The rope passes over a frictionless pulley at the top of the well, and you pull horizontally on the end of the rope to raise the bucket slowly a distance of 4.00 m. (a) How much work do you do on the bucket in pulling it up? (b) How much work does gravity do on the bucket? (c) What is the total work on the bucket?

m=6.75 kg, s=4.00 m


Homework Equations


F=ma, W=Fs


The Attempt at a Solution


I know that this should be easy, right? It's just one-dimensional movement... but how do I know what the force of tension is? I drew a free body diagram for the bucket - the only forces are FG and FT, but I only know that FT is greater than FG (if it's accelerating upward). I don't know how to actually find the force of tension acting on the bucket.
 
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Is the bucket accelerating if it's being "raised slowly"? I think that you can presume a constant speed over the 4.00 m.
 
So, if FT=mg then, wouldn't the answer to part (a) be (6.75 kg)(9.8 m/s2)(4.00 m) = 264.6 J? But that would make the total work equal zero, with gravity doing -264.6 J. The back of the book says the answers are (a): 3.60 J, (b): -.900 J, (c): 2.70 J
 
Well, it's a mystery. 3.60 J is enough energy to raise a 6.75 kg bucket about 5.4 cm.
 

Thread 'Errors and significant figures'

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...
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Thread 'A cylinder connected to a hanging mass'

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...
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