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Question: A chain of mass \(m\) and length \(L\)is over hanging from the edge of a smooth horizontal table suc...

A chain of mass mm and length LLis over hanging from the edge of a smooth horizontal table such that 34th{\dfrac{3}{4}^{th}} of its length is lying on the table. The work done in pulling the chain completely on to the table is:
A) mgL16\dfrac{{mgL}}{{16}}
B) mgL32\dfrac{{mgL}}{{32}}
C) 3mgL32\dfrac{{3mgL}}{{32}}
D) mgL8\dfrac{{mgL}}{8}

Explanation

Solution

In this problem, 14th{\dfrac{1}{4}^{th}} of the chain is hanging from the table. Work will be done by gravity in pulling the chain downwards. The work done will be given as the mass of the hanging part of the chain multiplied by the centre of mass of the hanging part multiplied by the acceleration due to gravity.

Complete step by step solution:
We are given that 14th{\dfrac{1}{4}^{th}} part of the chain is hanging. The mass of the chain is mm and its length is LL .
Therefore, the mass of the hanging part will be, m4\dfrac{m}{4}
And the length of the hanging part will be, L4\dfrac{L}{4}

Work will be done in pulling the chain completely on to the table. This work done will be equal to the change in the potential energy of the hanging part. The only force acting on the hanging part is the force due to gravity. This gravitational force will act at the centre of mass of the hanging part.

Thus, we have work done WW ,
W=PEW = \vartriangle PE

The change in potential energy is given as mghmg\vartriangle h where h\vartriangle h is the change in the height of the body.
In our case, the hanging part has mass m0=m4{m_0} = \dfrac{m}{4} and the change in height is the change in the height of the centre of mass, L0=L8{L_0} = \dfrac{L}{8} as centre of mass of the hanging part will be at the centre of the hanging part.

W=m0×l0×g \Rightarrow W = {m_0} \times {l_0} \times g
Here, gg is the acceleration due to gravity.

Substituting the values, we get
W=m4×l8×g\Rightarrow W = \dfrac{m}{4} \times \dfrac{l}{8} \times g
W=mgl32\Rightarrow W = \dfrac{{mgl}}{{32}}
The work done in pulling the chain completely on to the table is mgl32\dfrac{{mgl}}{{32}} .

Thus, option B is the correct option.

Note: The work is done against gravity. As the chain does not gain any velocity thus, the work done will only change the potential energy of the hanging part. The center of mass of the hanging part lies at the center of it. The chain is of uniform mass and uniform length. There is no friction acting on the surface.