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Question

The chain of length L is released from rest on a smooth fixed incline with x = 0 as shown in the figure. Determine velocity v of the chain when a half of the length has fallen (Take L=425m) in m/s. (Neglect edge effect of inclined). (Answer upto two digit after decimal places)

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Solution

We know that total energy is conserved
εi=εfεi=initial potential energy+initial kinetic energyεf=final potential energy+final kineti+energymgL2sinθ+0=m1g(Lx)2sinθm2gx2+12mv2Now if the mass is considered as uniformly distributed thenm1=mL(Lx)m2=mLxmgL2sinθ+0=mL(Lx)g(Lx)2sinθmLxgx2+12mv2gLsinθ=gL(L2+x22Lx)sinθgx2L+v2As it is given half of the length is left sox=L2v=58gl
Taking g=10 m/s2
Putting value of L=425

v=1 m/s

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