A 1.5 m long chain of mass 0.8 kg is kept on a horizontal table and a part of its length hangs from the edge of the table. When the length of the hanging part is one- third the total length of the chain, it starts sliding off the table. What will be the work done by friction when the whole length of the chain slides off the table ?

Explanation

When `1/3` of the chain is hanging, it starts sliding [Fig.1.29]. In this condition,
frictional force =weight of the hanging part of the chain
`or, mu xx2/3m//g=1/3 m//g`
`[mu`= coefficient of friction ,m =mass per unit length of the chain, and l= length of the chain]
`therefore mu =1/2=0.5`
When the whole length of the chain slides off the table, the effective frictional force on the chain=0 .

`therefore` Effective average frictional force on the chain
`=(2/3mu m//g+0)/(2)=1/3 mu m//g`.
The chain moves through a distance of `2/3` l against the effective friction. work done against friction is, therefore,
`W=1/3 mu m//g lxx2/3 l =2/9mumgl^2`
`=2/9xx0.5xx(0.8)/(1.5)xx9.8xx(1.5)^2 =1.3 J.`