Question

A chain is place on a frictionless table such that it's 1/5part is hanging down over the edge of the table if the length of the chain be l and mass be m then how much work will be done in pulling up the hanging part of the chain onthe table

Answer 1

Mass of part of the chain hanging over = L/5 * m/L
= m/5.

Change in the height of L/5 part = L/10.

Work done = m g h
= m g L/50 units

Answer 2

Dear student,

It is given that a  chain is place on a frictionless table such that it's 1/5th part is hanging down over the edge of the table if the length of the chain be l and mass be m.

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/5

                  suspended mass = m/5

centre of mass of suspended mass= m/10

Putting the given values in the formula, we get:

               u = - [ - Wg ]

                  = Wg

                  = mgh

                  = m/10 × g × l/5

              u =  mgl/50

Hence, the mgl/50 work will be done in pulling up the hanging part of the chain on the table.