Question

A uniform chain of mass m and length l overhangs a table with its two third part on the table. Find the work to be done by a person to put the hanging part back on the table.

Answer 1

The work to be done by a person to put the hanging part back on the table is :

• Let us take a small element of length dx at a distance x from edge of the table.

• Let λ = mass per unit length.

λ = m/l

• mass of element dx = (m/l)×x

• Let work done to put dx element on be dW, then

• dW = ∫ (m/l).x.g.dx from 0 to l/3

• So, total work done to put 1/3 length of chain back on table is

• W = (m/l)×g×(l^2/18)

= mgl/18

Answer 2

A uniform chain of mass m and length l overhangs a table with its two third part on the table. The work to be done by a person to put the hanging part back on the table is $W_{f}=\frac{2 \mu M g L}{9}$

Explanation:

Step 1:

Given in the question

In this question since one third of the hanging part mass is m/3 and this weight is going to be in the middle of the hanging component mass. L/3 hangs part length Let, x length of chain is on the table at a particular instant.

Step 2:

So, work done by frictional force on a small element ‘dx’

$\mathrm{d} \mathrm{W}_{\mathrm{t}}=\mu R x=\mu\left(\frac{M}{L} d x\right) g x$  

where

$\mathrm{d} \mathrm{x}=\frac{M}{L} d x$

Total work done by friction  

$W_{f}=\int_{\frac{z L}{3}}^{0} \mu\left(\frac{M}{L} d x\right) g x$

$W_{f}=\mu\left(\frac{M}{L} g\right)\left[\frac{x^{2}}{2}\right]_{\frac{2 L}{3}}^{0}$

$\begin{aligned}&W_{f}=\mu\left(\frac{M}{L} g\right) \frac{4 L^{2}}{18}\\&W_{f}=\frac{2 \mu M g L}{9}\end{aligned}$