A uniform chain of length L and mass M overhangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is μ . The work done by friction during the period, the chain slips of the table is

Step-by-step solution

The linear mass density is M L . The small work done for the slippage on the small distance d l is given by d W = - μ M L g l d l Now, total work done W = ∫ 0 2 L 3 - μ M g L l   d l =     - μ M g L l 2 2 0 2 L 3 = - μ M g 2 L 4 L 2 9 - 0 ⇒       W = - 2 9 μ M g L