Question

A person pulls a bucket of water from a well of depth $h$ . If the mass of uniform rope is $m$ and that of the bucket full of water is $M$ , the work done by the person is:
(A) $\left( {M + \dfrac{m}{3}} \right)gh$
(B) $\dfrac{1}{2}\left( {M + m} \right)gh$
(C) $\left( {M + \dfrac{m}{2}} \right)gh$
(D) $\left( {\dfrac{M}{2} + m} \right)gh$

Answer 1

First of all, we will find the work done to raise the bucket from the well and work done to pull the rope. The summation of these two will give the net work done. Whole mass of the rope is concentrated in the centre of the rope.

Complete step by step answer:
In the given question, we are supplied with the following data:
A person is pulling a bucket from a well whose depth is $h$ .
The mass of a uniform rope is $m$ .
The mass of the bucket which is full of water is $M$ .
We are asked to find the work done by the person in polling the bucket from the well.
To begin with, when the person is pulling the bucket from the well, he is doing work against the potential energy stored by the bucket along with the rope. The rope extends from the top of the well to the bottom of the well whose height is given as $h$ . For this reason, the length of the rope is also $h$ . Although the length of rope is $h$ , the centre of mass is located at the point halfway of the rope i.e. the centre of mass of the rope is located at a point whose length is $\dfrac{h}{2}$ .
The work done by the person is purely against the potential energy of the bucket full of water and the rope.
For the first case:
The work done in lifting the bucket of the given mass to the given height is written as:
${E_1} = Mgh$ …… (1)
Where,
${E_1}$ indicates the work needs to be done in lifting the bucket from the well.
$M$ indicates the mass of the bucket full of water.
$g$ indicates the acceleration due to gravity.
$h$ indicates height of the well.
Again, while pulling the rope, the person has to bring the centre of mass of the rope to the ground level. So, the work done pulling the rope is equivalent to the change in potential energy of the rope, which can be written as:
${E_2} = mg \times \dfrac{h}{2}$ …… (2)
Where,
${E_2}$ indicates the work done in pulling the rope.
$m$ indicates the pass of the rope.
$g$ indicates the acceleration due to gravity.
$\dfrac{h}{2}$ indicates the effective height or the distance of the centre of mass from the person’s grip.
Now, the net work done $\left( E \right)$ is given by the sum of the work done on the bucket and the rope.
$E = {E_1} + {E_2} \\\ \Rightarrow E = Mgh + mg\dfrac{h}{2} \\\ \therefore E = \left( {M + \dfrac{m}{2}} \right)gh \\\$
Hence, the work done by the person in polling the bucket from the well is $\left( {M + \dfrac{m}{2}} \right)gh$ .

The correct option is C.

Note: While solving this problem, many students seem to make a mistake in taking the height of the rope. They usually take as $h$ which is wrong. Rather it will be $\dfrac{h}{2}$ as the centre of mass of the rope is located at half way. This can cause serious issues while calculating the correct result.