anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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.
anonymous
  • anonymous
thank u . u got the right answer. but do you know how i can do this problem using calculus
anonymous
  • anonymous
i'm sorry, network problem just now.

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anonymous
  • anonymous
its fine take ur time . ty for helping me
anonymous
  • anonymous
hey if u want you want u can use http://www.twiddla.com/515340 to work out the problem
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
i'm terribly sorry for wasting you so much time... just mixed up with the Newton-Leibniz Formula.
anonymous
  • anonymous
its fine no worries
anonymous
  • anonymous
isnt the function suppose to b in terms of force, not mass
anonymous
  • anonymous
it is the same... actually..
anonymous
  • anonymous
hey do u think u can help me with another problem
anonymous
  • anonymous
id love to, you can ask that question on the left column.
anonymous
  • anonymous
k ty

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