# A man pulls a bucket of water from a well of depth H. if the mass of uniform rope and bucket full of water are m and M respectively. The value of work done by man isA.(M+m/2).g.hB.(M+m)ghC.(m+M)/2.G.HD. none

utkarsh0110 and cherry.berry has given absolutely correct answer. So, you can refer to their answers.

Here in the answer work done is computed using the fact that work done is equal to the change in potential energy. The work done in lifting the mass M of the bucket to the height H is MgH. In the case of the rope, first center of mass of the rope is being calculated which is at the depth H/2 from the top of the well as mass is uniformly distributed along its whole length. Second, the change in potential energy has been computed which is from the depth of the center of the mass of the rope i.e. mgH/2. Therefore, work done in lifting the whole system is

@utkarsh0110 and cherry.berry. Good answer. You both get thumb up. Keep posting!!

• 30

It should probably be a...since d rope is uniform, we wud consider the movement of centre of mass of the rope...now given that d rope is uniform n d depth is H, the centre of mass wud b at 2H...so//

total work done = change in potential energy (this is the work energy theorum)

hence,

W = MgH + mg(H/2)

This is because the water n bucket r at the depth of d well n r brought up whereas d COM of d rope is at H/2 and is brought up..

with a little algebra..

W = (M+m/2)gH

Note : u shud b sure wid d symbols..in first n second option u hv used small h and in the ques n 3rd option capital H...it causes blunders at time..

• 21

Work = force x distance = mass x g x height

for rope w1 = m'.g.h' = mgH/2 (centre of mass of rope = h/2 from both ends)

for bucket w2 = m''.g.h'' = MgH

total W = w1 + w2 = gH(M+m/2) = (M+m/2).g.H

• 86
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