Question:
A uniform chain of mass in and length I 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.
Solution:
$W=(m / l) d x g(x)$
Total work done is :
$\mathrm{W}=1 / 3 \int_{0}(\mathrm{~m} / \mathrm{l}) \mathrm{dx} \mathrm{g}(\mathrm{x})$
$\mathrm{W}=\mathrm{mgl} / 18$