a chain of mass 0.80 kg and lengh l=1.5m rests on a rough surfaced table so that one of its ends hang over the edge...the chain starts sliding off the table all by itself provided the overhanging part equals n=1/3 of the chain lengh...what will be the total work performd by the friction forces acting on the chain by the moment it slides completely off the table..??

Let the total mass of the chain be 'm' and the total length be 'l'.

the length of the hanging part of the chain is = nl

where 'n' is the coefficient of friction and 'l' is the total length.

so, the mass of the hanging chain will be = (m/l).ln =  mn

and the weight of the hanging portion would be = mng

now, the mass of the chain of the table will be = m - mn = m(1-n)

the weight of the that portion will be  = mg.(1-n)

the force of normal reaction on the chaiin kept on the table will be = u.mg.(1-n)

now, in equilibrium -  weight of the hanging chain = normal reaction 

u.mg.(1-n) = mng

or

u = n / (1-n)

now, let us consider the length of the hangign chain be  = x

the length of chain on table = l -x

and

the mass of the chain on the table = (m/l).(l-x)

the weight of the chain on the table =(mg/l).(l-x)

now, the work done in moving the chain a small distance dx will be

dW = F.dx = - u[(mg/l).(l-x)].dx

we can integrate the above equation within length limits from nl to l. So,

dW = -_nl∫^lu[(mg/l).(l-x)].dx

now, by solving further, we get

W = (mgln/2). (n-1)

by putting in the values, the work done will be

W = -1.3 J