Question

A uniform chain of L length and M mass, two-third part of the chain is on the frictionless table and one-third part is vertically suspended, work done to pull the whole chain on the table is

Answer 1

Answer:

work done = force * displacement

or w=fs

Explanation:

as work done(w)= fs

so, we have to find force (displacement=0)

force = mass * acceleration due to gravity

or F = ma

Answer 2

Dear student,

It is given that a  uniform chain of L length and M mass, two-third part of the chain is on the frictionless table and one-third part is vertically suspended. We need to find the work done to pull the whole chain on the table.

We know by the concept of gravitational potential or work done by conservative forces for a reference which is at rest that;

                     u = - [Wg]

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 = - [ - mgh]

                 u = + mgh

                 u = m/6 × g × l/3

                 u = mgl / 18          

Hence, the work done to pull the whole chain on the table is mgl/ 18.