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A bucket of water of mass 10 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 10 minutes, during which time 4 kg of water drips out at a steady rate through a hole in the bottom. Find the work needed to raise the bucket to the platform. (Use ????=9.8m/s2.)

Answer :

Answer:

W = 2352 J

Explanation:

Given that:

  • mass of the bucket, M = 10 kg
  • velocity of pulling the bucket, v = 3[tex]m.min^{-1}[/tex]
  • height of the platform, h = 30 m
  • time taken, t = 10 min
  • rate of loss of water-mass, m = [tex]0.4 kg.min^{-1}[/tex]

Here, according to the given situation the bucket moves at the rate,

[tex]v=3 m.min^{-1}[/tex]

The mass varies with the time as,

[tex]M=(10-0.4t) kg[/tex]

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance

∆x =  3∆t meters

So, during this interval change in work done,

∆W = m.g∆x

For work calculation:

[tex]W=\int_{0}^{10} [(10-0.4t).g\times 3] dt[/tex]

[tex]W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}[/tex]

[tex]W= 2352 J[/tex]

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