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amistre64

Find the amount of work done by lifting 150 ft of wire that weighs 4 lbs/foot.

  • 2 years ago
  • 2 years ago

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  1. elecengineer
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    :|

    • 2 years ago
  2. satellite73
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    i will watch. good morning sensei

    • 2 years ago
  3. amistre64
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    Gmornin :)

    • 2 years ago
  4. amistre64
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    im sure we are starting at one end and winding the whole thing up, at least that is what I pictured

    • 2 years ago
  5. satellite73
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    honestly if i pretended to help i would be making things up. never had a physics class in my life.

    • 2 years ago
  6. amistre64
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    (150-x) would be the length of each "portion" which then weighs 4 lb/ft maybe: \[\int_{0}^{150}(150-x)4\ dx\]

    • 2 years ago
  7. elecengineer
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    its very easy

    • 2 years ago
  8. amistre64
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    im sure it is, but im wondering if I did the integral correctly ...

    • 2 years ago
  9. satellite73
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    i will learn something...

    • 2 years ago
  10. amistre64
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    i got another one about fluid density pressure after this :)

    • 2 years ago
  11. satellite73
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    physics class?

    • 2 years ago
  12. abtrehearn
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    The total weight of the cable is 600 lb. Find the vertical distance the center of mass moves in ft., then multiply that distance by 600, and you have the ansewr in ft. lb. The hard part is finding the vertical distance. The problem, as stated, does not provide enough information on that score.

    • 2 years ago
  13. amistre64
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    calculus1

    • 2 years ago
  14. elecengineer
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    It assumes it moves vertically, just enough to leave the ground.

    • 2 years ago
  15. abtrehearn
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    Then the vertical distance is 75 ft.

    • 2 years ago
  16. abtrehearn
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    So W = (600lb)(75ft) = 45,000 ft. lb.

    • 2 years ago
  17. amistre64
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    so the 75 comes from 150/2 right?

    • 2 years ago
  18. amistre64
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    how would we set that up as an integral?

    • 2 years ago
  19. abtrehearn
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    I picture one end being lifted up until the other end is just about to leave the ground. Is that the rright scenario?

    • 2 years ago
  20. abtrehearn
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    Half the length. Yes, Amistre.

    • 2 years ago
  21. amistre64
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    i dunno if thats the right scenario; i believe the question was asking, my interpretation is this, you have 150 feet of cable suspended by a winch; how much work is required to roll it all up onto the winch?

    • 2 years ago
  22. amistre64
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    weight of cable being 4 lb/ft

    • 2 years ago
  23. abtrehearn
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    If you want to set it up as an integral, you can start with\[\int\limits_{0}^{150}F(x) dx,\]where F(x) is the force needed to lift x feet of cable off the ground.

    • 2 years ago
  24. abtrehearn
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    If the weight is 4 lb/ft, then x ft weighs (4x) lb, so 4x is the integrand.

    • 2 years ago
  25. abtrehearn
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    The integral evaluatex to 2x^2, where x = 150. Works out to 2* 22500 = 45,000 ft. lb.

    • 2 years ago
  26. amistre64
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    so if anything, I would have been off by makeing the force = length*weight? [150ft - (x ft off the ground) ] * 4 lb/ft? \[\int_{0}^{150}(150-x)4\ dx\] \[\int_{0}^{150}(500-4x)\ dx\implies500x-2x^2|^{150}\]

    • 2 years ago
  27. abtrehearn
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    :\[W = \int\limits_{0}^{150}4x dx = 45,000.\]The length is x ft, and the weight pre foot is 4 lb/ft.

    • 2 years ago
  28. amistre64
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    thanx :)

    • 2 years ago
  29. abtrehearn
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    Remember the definition of work?

    • 2 years ago
  30. abtrehearn
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    :^)

    • 2 years ago
  31. amistre64
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    Work = Force x Distance yes :)

    • 2 years ago
  32. abtrehearn
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    That's the idea ; ^)

    • 2 years ago
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