1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Check please, Work integral

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    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.

    3. 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?
     
  2. jcsd
  3. Oct 13, 2008 #2

    Mark44

    Staff: Mentor

    Everything appears fine up until the integral. What is 75 doing in the integrand?
     
  4. Oct 13, 2008 #3
    Does it make a difference? It can come out, but it all integrates the same...
     
  5. Oct 13, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Mark44 is right, what is 75 doing in the integrand?
     
  6. Oct 13, 2008 #5
    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
     
  7. Oct 13, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  8. Oct 13, 2008 #7
    Yeah ok that makes sense. So it would be:

    [tex]\int_{0}^{75} (128.75 - \frac{71}{60}y)dy[/tex]
     
  9. Oct 13, 2008 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes.
     
  10. Oct 13, 2008 #9
    Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Check please, Work integral
  1. Integral please check (Replies: 6)

  2. Check my work please? (Replies: 4)

  3. Please check my work (Replies: 2)

Loading...