A uniform chain of length 2 m is kept on a table such that a length of  60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg . What is the work (in Joule) done in pulling the entire chain on the table ?

Answers (1)

Definition of work done by variable force -

W=\int \vec{F}\cdot \vec{ds}

- wherein

\vec{F} is variable force and \vec{ds} is small displacement

Consider a small part dx at a depth x from table. 

Work done in lifting this small portion is dw = dm gx

Total work done =\int dw=\int_{0}^{h}(\frac{m}{l}dx)gx

    =\frac{mg}{l}\int_{0}^{h}x dx=\frac{4\times 10}{2}\times \frac{(0.6)^{2}}{2}

=\frac{mg}{l}\int_{0}^{h}x dx=\frac{4\times 10}{2}\times \frac{(0.6)^{2}}{2}=3.6J

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