Question on work,energy and power

[matsci0000]

Que- A uniform chain of length L and mass m is held on a smooth table with one-fourth of its length hanging over the edge. If g is the acceleration due to gravity, find the work required to pull the hanging part on the table.


CHECK MY SOLUTION---------AND POINT OUT MY MISTAKE.

Weight of the hanging part of the chain=mg/4

Therefore, work done=(mg/4)(L/4)
=mgL/16...Answer


I am getting this answer. But the correct answer is mgL/32.
Is there any role of center of mass in this question?

 

[Matterwave]

This is because the mg/4 isn't all concentrated at L/4 distance from the table. It requires less work because some of that mass is distributed closer to L/4 distance to the table. Intuitively, you can see that the total work should be half of what you computed since then, the "average" displacement of the chain (taking into account the mass distribution) is only L/8 away from the table. Mathematically, you need to compute this differently.

 

[Doc Al]

Therefore, work done=(mg/4)(L/4)

Hint: Are all parts of the hanging piece raised by the same distance?

Is there any role of center of mass in this question?

Sure. Using the center of mass will allow you to get the answer quicker.

Moderator's Note: Multiple threads merged.

 

[matsci0000]

What is the significance of center of mass in this question?

 

[Mindscrape]

You can use the center of mass as an average for all the work done. Less work is done at the top of the chain than the bottom of the chain. It will all even out if you just look at the center of the chain, that is the center of the hanging part of the chain.