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A man places a chain (of mass ‘m’ and length ‘l’) on a table slowly. Initially the lower end of the chain just touches the table. The man drops the chain when half of the chain is in vertical position. Then work done by the man in this process is :

Option: 1

-mg\; \frac{l}{2}   


Option: 2

-\frac{m\: g\, l}{4}


Option: 3

-\frac{3\: m\: g\: l}{8}


Option: 4

-\frac{m\: g\: l}{8}


Answers (1)

best_answer

 

Potential Energy stored when particle displaced against gravity -

U= -\int fdx= -\int \left ( mg \right )dx\: cos180^{0} 

- wherein

m=mass \: of \: body

g= acceleration \: due \: to\: gravity

dx= small\: displacement

 

 

The work done by man is negative of magnitude of decrease in potential energy of chain

\Delta U=mg\frac{L}{2}-\frac{m}{2}g\frac{L}{4}=3m\: g\frac{L}{8}

\therefore \; \; W=-\frac{3\: m\: g\: l}{8}

Posted by

Ritika Harsh

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