Question

A uniform chain of masa, m and length, 1 is held on a frictionless table such that one third of its length hangs over the edge Calculate the work done to pull the hanging part of the chain back on the table?​

Answer 1

Answer:

Length of hanging chaing is l/3 , and mass of hanging chain is M=m/3  .

The center of mass of hanging chain is at h=2l/3=6l from the table top.

So work done ,W=Mgh=3×6mgl=18mgl

Answer 2

Dear student,

It is given that a uniform chain of mass m and length l is held on a frictionless table such that one third of its length hangs over the edge .

We need to find how much work will be done in pulling up the hanging part of the chain on the table. We can find this with the help of relation between work done by conservative forces and change in potential for initial reference kept at zero i.e.

                Wc.f. = u

or,              Wg = u

           

We have;   suspended length = l/3

                  suspended mass = m/3

centre of mass of suspended mass= m/6

Putting the given values in the formula, we get:

               u = - [ - Wg ]

                  = Wg

                  = mgh

                  = m/3 × g × l/6

               u = mgl/ 18

Hence, the work done to pull the hanging part of the chain back on the table is mgl/18.