A uniform chain of mass m and length l overhangs a table with two-thirds of its length resting on the table. Determine the work to be done by a person to put the hanging portion back onto the table.

Step-by-step solution

Mass of the hanging portion of the chain is m' = m 3 . The center of mass of the hanging portion is at a depth h = l/3 2 = l 6 below the edge of the table. Work done to pull the hanging portion back onto the table is equal to the change in potential energy of the hanging portion. W = m'gh = ( m 3 )g ( l 6 ) = mgl 18