# Work Energy Question

1. Sep 18, 2008

### prateek_34gem

1. The problem statement, all variables and given/known data

Uniform chain of length 'L' and mass of 'M' is lying on smooth table and one third of it's length is hanging vertically down over the edge of the table. If 'g' is acceleration due to gravity , then work required to pull the hanging part onto the table is :

A)MgL
B)MgL/3
C)MgL/9
D)MgL/18
(I.I.T :- 1985)

3. The attempt at a solution

mass of hanging part will be m/3. so it will be exerting mg/3 force downward.
so this amount of force is required to pull it up.when the chain will be up its displacement along table will be l/3 because now hanging part is on table.
so

F=Mg3
S=L/3
Work = F*S= MgL/9

But it is the wrong answer.

Last edited: Sep 18, 2008
2. Sep 18, 2008

### art.t

Try lifting the center of mass of the hanging chain up on to the table and using the formula (work done against gravity = m*g*h).

The center of mass is halfway down the part that's hanging off, which was L/3. So the center of mass is L/6 below the table top. The mass of the hanging part is 1/3 of the total mass...Go for it.

3. Sep 20, 2008

### prateek_34gem

ya i got it . thnx a ton