A chain of mass m and length  l rests on a rough surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the over hanging part equals  n=13 of the chain length. What will be the total work performed by the friction forces acting on the chain by the moment it slides completely off the table ?

at limiting equilibrium weight of the hanging part = frictional force on the table  i.e;mgl3= μmgl1-13= μmgl23   μ=0.5 if at any time x length is hanging then on the table i-x    and frictional force μmgll-xwork done by the frictional force till the entire length slips  -∫l3lμmgll-x dx=--∫l3lμmg dx-∫l3lμmglx dx=μmg2l3-μmg2ll2-l29 =μmg2l3 -4μmgl9=μmg2l9here μ=0.5 work done against friction =mgl9