Amount of work

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lamaclass

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I wanted to know if I did both parts of this problem correctly:

A) A bucket weighs 70 lbs when filled with water and is lifted at a constant rate, by a mechanical wrench, from the bottom of a well that is 60 feet deep. If the chain that is being used to lift the bucket is .55 lbs/ft, find the amount of work required to lift the full bucket from the bottom of the well all the way to the top.

F = W * D

Delta F = W * L

= .55 * delta y

D: = 60-y

= IN, limits 0 and 60, (.55) (60-y) dy + 70(60)

B) I forgot to tell you that the bucket has a hole at the bottom. It weighs 35 lbs when it reaches the top becuase of the water leaking at a constant rate. Compute the real work required to lift the full bucket from the bottom to the top of the well. Assume the bucket is lifted at a constant rate as well.


IN, limits 0 and 60, (.55) (60-y) dy + 35(60)

Do I have the correct set up for both parts?
 
No, this doesn't look correct.

First of all the equation for work is W = F*d and is thus

\(\displaystyle W = \int^b_a F(x)\,dx\)

Work is a force over a given distance. In order to set this up, you have to ask yourself, "what is changing in this problem?" It's the mass. As the bucket is winched up, the mass of the chain/bucket system is reduced. So how can you relate the mass of the chain/bucket system to any given height? ie. What is F(x)?

In part A, the water is a constant, so there is no need to relate it to the height. The chain weight, however, is variable. It changes as the height changes. ie. 0.55(60 - x), this will give you the weight of the chain at any height. Make sense?

Let me know if this helps
 
It does make sense, however I'm a little confused because I think I did show the wieght of the chain. And it looked like the weight of the bucket was a constant variable so I thought you might have to take the regular formula of it and also apply it to the weight of the water as well.
 
Yes (if I understand you), in part A the bucket and water is a constant, so you add it to the weight of the chain and integrate the whole thing, since the constant weight is still displaced a given distance. so...

Int [70 + 0.55(60 - x)] dx

so, how about part b? How do you relate the changing weight of the bucket/water system to the height now? (and then you add that to the chain as well)

Oh, I see what you did now, I was confused for a bit there. It looks correct.
 
Could you use the orginal function of the change in the chain and then apply the weight of 35 lbs and multiply it into 60?
 
No, I'm not sure I understand what you mean. OK, you have the function for the chain and that stays the same for part b. You have to add to that the function for the change in weight of the bucket. Think of it like this...

When x = 0, weight of the bucket = 70

When x = 60, weight of the bucket = 35

Can you come up with a simple formula to explain this at any value of x between 0 and 60?
 
Part b

\(\displaystyle W = \int^{60}_0 [70 - \frac{35}{60}x + 0.55(60 - x)]\,dx\)[/spoiler:2oxn7hf7]
 
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