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A uniform chain of length .L. and mass .m. is on a smooth horizontal table, with (1)/(n) th part of its length hanging from the edge of the table. Find the kinetic energy of the chain as it completely slips off the table.

Answer» <html><body><p></p>Solution :With respect to the top of the table, the initial potential energy of the chain, <br/>`U_(1)= PE` of the chain lying on the table + PE of the <a href="https://interviewquestions.tuteehub.com/tag/hanging-1015416" style="font-weight:bold;" target="_blank" title="Click to know more about HANGING">HANGING</a> <a href="https://interviewquestions.tuteehub.com/tag/part-596478" style="font-weight:bold;" target="_blank" title="Click to know more about PART">PART</a> of the chain<br/>`=L(1-(1)/(<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>))mg(0)+(m)/(n)g(-(L)/(2n))=-(mgL)/(2n^(2))`<br/>P.E of the chain, when it just slips off the table,<br/>`U_(2)=mg((-L)/(2))=-(mgL)/(2)`<br/>From law os conservation of energy<br/>`<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> K =- Delta U "" K_(f)-K_(i)=-(U_(f)-U_(i))`<br/>`because K_(i)=0 , K_(f)=-[-(mgL)/(2)-(-(mgL)/(2n^(2)))]`<br/>`K_(f)=(mgL)/(2)[1-(1)/(n^(2))]`<br/>If .V. is the velocity of the chain, then,<br/>`(1)/(2)mv^(2)=(mgL)/(2)[1-(1)/(n^(2))] <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> v=sqrt(gL[1-(1)/(n^(2))]`</body></html>


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