# Q.> 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 ?My answer:mass per unit centimetre:4000/200=20gdisplacement=60cmtotal mass of 60cm chain=1200gwe know: W=F.sF=m.g=1200gm*10m/s=1.2kg*10m/ss=60cm=0.6mw=F.s=1.2*10*0.6=7.2Jso total work done is 7.2J

You have done it wrong.

As we learnt in

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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