Work Required to Pull Chain onto Table (I.I.T 1985)

[prateek_34gem]

Homework Statement

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)

Homework Equations

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.

 

[LowlyPion]

Homework Statement

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)

Homework Equations

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.

$$Work = \Delta PE =M*g*h = \frac{m}{3}*g*\frac{L}{2*3} = \frac{m*g*L}{18}$$

Your weight is MG/3 as you noted but the height that you raise the center of mass is 1/2 the length, which makes it L/6, not L/3.

 

[prateek_34gem]

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