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

Let 'dx' be the length of an element at distance x from the table.
Mass of the element, 'dm' \[= \left( \frac{\text{ m }}{\text{ l }} \right) \text{ dx }\] 

Work done to putting back this mass element on the table is \[\text{ dW }= \left( \frac{\text{ m}}{\text{l}} \right) \times \text{ x } \times \text{ g } \times \text{ dx } \]

So, total work done to put \[\frac{1}{3}\]  part back on the table

\[W = \int_0^{1/3} \left( \frac{\text{m}}{\text{l}} \right) \text{ gx dx }\]

\[ \Rightarrow \text{W} = \left( \frac{\text{m}}{\text{l}} \right) \text{g} \left[ \frac{x^2}{2} \right]^{1/3} \]

\[ = \frac{\text{mgl}}{18 \text{l}} = \frac{\text{mgl}}{18}\]

The work to be done by a person to put the hanging part back on the table is \[\frac{\text{ mgl } }{18}\] .