Question

A uniform chain of mass M and length L is lying on a frictionless table in such a way that its 1/3 part is

hanging vertically down. The work done in pulling the chain up the table is​

Answer 1

Answer:

Solution :

Method 1:The hanging part of the chain which is to be pulled can be considered as a point mass situated at the centre of the hanging part. The equivalent diagram is drawn. The work dine in bringing the mass up will be equal to the change in potential energy of the mass.

W=change in potential energy

=M3×g×L6=MgL18

Method 2: The per uni length of the chain =ML. let us consider a finitesimally small length of the chain dx at a distance x from the bottom. To move dx to the top, a force equal to the weight of chain x will have to be applied upwards.

Weight of chain of lengthx=(MLx)g

Small amount of work done in moving dx to the top

dW=F→.dx−→=Fdx=(MLx)gdx

The total amount of work done in moving the one third length of the hanging chain on the table will be

W=∫L30MLxgdx=MLg

∫L30xdx=MLg[x22]L30=MgL18