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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><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :With respect to the top of the table, the initial potential energy of the chain <br/> `U_(1)`= PF of the chain lying on the table + PE of the hanging part of the chain <br/> `= L (1- (1)/(n)) mg (0)(m)/(n)<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> (-(L)/(2n))= - (<a href="https://interviewquestions.tuteehub.com/tag/mgl-2176708" style="font-weight:bold;" target="_blank" title="Click to know more about MGL">MGL</a>)/(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 <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> of conservation of energy <br/> `Delta K= - Delta <a href="https://interviewquestions.tuteehub.com/tag/u-1435036" style="font-weight:bold;" target="_blank" title="Click to know more about U">U</a> K_(r)- K_(i) = - (U_(f)- U_(i))` <br/> `because K_(i)= 0, K_(f)= - [-(mgL)/(2)- (-(mgL)/(2n^(2)))]` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_B_C06_SLV_023_S01.png" width="80%"/> <br/> If .V. is the velocity of the chain, then, `(1)/(2) mv^(2)= (mgL)/(2) [1-(1)/(n^(2))]` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_B_C06_SLV_023_S02.png" width="80%"/></body></html>


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