Check please, Work integral

  • Thread starter Sheneron
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  • #1
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Homework Statement


A bucket that weighs 80lbs when filled with water is lifted from the bottom of a well that is 75ft deep. The bucket has a hole in it and weighs only 40lb when it reaches the top of the well. The rope weighs 0.65 lb/ft. Find the work required to life the bucket from the bottom to top of the well.

The Attempt at a Solution



Here is what I did:
Set the bottom a y=0 and top at y=75, and got the following:

[tex]W_{bucket}= 80 - \frac{40}{75}y[/tex]

[tex]W_{rope}= 48.75 - 0.65y[/tex]
So:
[tex]W_{total} = 128.75 - \frac{71}{60}y [/tex]

[tex]\int_{0}^{75} (128.75 - \frac{71}{60}y)75 dy[/tex]

Would that be the right integral set up?
 

Answers and Replies

  • #2
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Everything appears fine up until the integral. What is 75 doing in the integrand?
 
  • #3
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Does it make a difference? It can come out, but it all integrates the same...
 
  • #4
Dick
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Mark44 is right, what is 75 doing in the integrand?
 
  • #5
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are you saying it shouldn't be there at all? I multiplied the weight by the height it had to move, that was my reason
 
  • #6
Dick
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are you saying it shouldn't be there at all? I multiplied the weight by the height it had to move, that was my reason
You multiply the weight by the height if the weight is constant. Then you don't need to integrate. If the weight isn't constant then you need to integrate over the height. You don't do BOTH.
 
  • #7
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Yeah ok that makes sense. So it would be:

[tex]\int_{0}^{75} (128.75 - \frac{71}{60}y)dy[/tex]
 
  • #8
Dick
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Yes.
 
  • #9
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Thanks
 

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