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anonymous

  • 5 years ago

a bucket of water of mass 20kg is pulled at a constant velocity up to a platform 40 meters above the ground. this takes 10 minutes , during which time 5kg of water drips out at a steady rate through a hole in the bottom . find the work required to raise the bucket to the platform.

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  1. anonymous
    • 5 years ago
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    from the information above, we know the average mass of the bucket of water may be m=20-5/2=17.5kg. as the bucket of water is pulled at a "constant velocity" , the mechanical energy is in conservation. in other words, the work required to raise the bucket to the platform transformed into the potential energy of the bucket of water. thus the answer should be W=mgh=17.5*9.8*40=6860J, irrelevant to the time.

  2. anonymous
    • 5 years ago
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    thank u . u got the right answer. but do you know how i can do this problem using calculus

  3. anonymous
    • 5 years ago
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    i'm sorry, network problem just now.

  4. anonymous
    • 5 years ago
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    its fine take ur time . ty for helping me

  5. anonymous
    • 5 years ago
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    hey if u want you want u can use http://www.twiddla.com/515340 to work out the problem

  6. anonymous
    • 5 years ago
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    the rate of the dripping wate, according to the problem, is 5kg/10min=5kg/600s=1/120 kg/s(convenient to use ISU here). you can consider a function on the mass of the bucket of water to the time as m(t)=20-t/120.

  7. anonymous
    • 5 years ago
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    from t=0 to 600, m(t)dx=(20-t/120)dx=20dx-(t/120)dx=20t-t^2/240=10500kg*s. W=mgh=m(t)dx*gh/t=10500*9.8*40/600=6860J

  8. anonymous
    • 5 years ago
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    i'm terribly sorry for wasting you so much time... just mixed up with the Newton-Leibniz Formula.

  9. anonymous
    • 5 years ago
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    its fine no worries

  10. anonymous
    • 5 years ago
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    isnt the function suppose to b in terms of force, not mass

  11. anonymous
    • 5 years ago
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    it is the same... actually..

  12. anonymous
    • 5 years ago
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    hey do u think u can help me with another problem

  13. anonymous
    • 5 years ago
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    id love to, you can ask that question on the left column.

  14. anonymous
    • 5 years ago
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    k ty

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