Question

The mass of a bucket full of water is 15kg it is being pulled up from 15m deep well. Due to hole in the bucket 6kg water flows out of the bucket . The work done in drawing the bucket out of the well is

plz explain your ans in detail

Answer 1

Let's first find out the rate (say R) at which mass of bucket ( filled with water ) changes as it is lifted up ( due to leakage ) with respect to height ( to which it has been pulled ) :
Total Change in Mass = 6Kg
Total height it being pulled up = 15m
then ;
R = 6/15 Kg/m .
Assume it be constant throughout the process !!
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Further ;
Let's suppose the bucket is lifted by a very small height (ds ) then ;
mass of bucket at that instant =
15 - [(6/15)×ds]
Then work done is given by :
(using W = mgh )
dW = { 15 - [(6/15)×ds] } × g × s .
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On integrating the above equation we get :
W = [15×g×s ] - [(6/15)×(s²/2)×g]
Now s ranges from 0 to 15 m & taking g=10m/s² . using it in above equation we get ;
W =( 15×10×15) - {(6/15)×[(15)²/2] × 10}
On solving we get The Work done in the whole process which is
W = 1800 J
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hope it helps!

Answer 2

mass varies from deep of well to 15m height .
let ¢ is mass per unit height .
so,
dm =¢dx
=6/15dx

M =m -dm

now,

workdone = -change in potential energy
= F.x
= ( m -dm)g .x
= mgx -dm.gx
= 15 × 10 × 15 -6/15 gx.dx
=2250 -6/15 x10x 225/2
=2250 -3 x 15x 10
= 2250 -450
= 1800 Joule