# A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is accelelation due to gravity, the work done to pull the hanging part on to the table is?

Ramesh V
70 Points
12 years ago

unit length of rope is M/L

given L/3 part of rope is hanging

we have to solve using integration methods, lets say 'x' part of rope is consided with an element 'dx' which is pulled up with limits from 0 to L/3

Potential energy = mgh

here dE= integral of ( Mx/L*g*dx)

E = integral of ( M/L*g*x.dx) limits from 0 to L/3

P.E = MgL/18

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regards

Ramesh

Abhishek Gupta
44 Points
6 years ago
Work Done = mass of chain hanged * g * distance of chain from centre of mass of hanged part of chain
So, work done = M/3 * g * L/3*2
(This length is taken L/6 NOT L/3 becoz distance is taken from centre of mass)
So, WORK DONE = MgL/18  (ANSWER)
Sonam Tyagi
16 Points
4 years ago
It's is given that one third part of chain is hanging therefore two third part of chain is on the table .
Now, L/3 part is hanging and the mass of the hanging chain is M/3.
So, Work done = mass of hanging chain * g * and distance of chain from centre of mass of hanging chain
M/3*g*L/6
MgL/18
SV Channel
24 Points
3 years ago
The length of hanging part=L/3
I.e.n=3
We have,W=mgl/2n^2
W=mgl/2×3×3
W=mgl/18
This is the answer for your question
Vaibhav