Question

A chain is held on a frictionless table with (1/n)th of its length hanging over the edge. If the chain has a length L and ,mass M , how much work is required to pull the hanging part back on the table?

• A. $\frac{MgL}{8n^2}$
• B. $\frac{MgL}{4n^2}$
• C. $\frac{MgL}{2n^2}$
• D. $\frac{MgL}{n^2}$

Answer 1

The correct option is C $\frac{MgL}{2n^2}$

The correct option is C.

Given,

$Length=l$

$Mass=m$

Let h is the height of the table $(h=0)$

Then,

The final gravitational potential energy

The initial potential energy of $(\frac{n-1}{n}{)}^{th}$ of the chain which is already on the chain is also zero.

So, We have to find out the gravitational potential energy of $(\frac{1}{n}{)}^{th}$, Which is hanging.

The mass of the small portion $dm$ of the chain having arbitrary length $dh$ will be:

$dm=\frac{m}{L}dh$

Here,

L is the length of the chain, m is the mass of the chain, his the length of the chain and $\frac{m}{L}$

is unit mass.

So, work is required.

$W=u\left(x\right)-U{x}_0$

$=-{\int }_{x_0}^x{f}_x\left(x\right)dx$

$=-{\int }_{\frac{L}{n}}^0\left(\frac{m}{L}dh\right)gh$

$=-{\int }_{\frac{L}{n}}^0\frac{mg}{L}hdh$

$=-\frac{mg}{L}{\int }_{\frac{L}{n}}^{0}hdh$

$=-\left(\frac{mg}{L}\right)-\frac{L^2}{2n^2}$

$=\frac{mgL}{2n^2}$