A uniform chain is just at rest over a rough horizontal table with its `(1)/(eta)th` part of length hanging verticaliy find the co-efficient of static friction between the chain and the table

Explanation

With respect to the top of the table, the initial potential energy of the chain, `U_i=PE` of the chain lying on the table+PE of the over hanging part of the chain

`=(2m)/(3)gxx0+(m//3)g(-l//6)=-(mgl)/(18)`
PE of chain at the instant of slip,
`U_f=0+mg(-l//2)=-(mgl)/(2)`
Since only gravity is acting on the chain, we have
`K=-(U_f-U_i)=[(-mgl)/(2)-((-mgl)/(18))]=4/9mgl`
Since `K_i=0`, `K_f=4/9mgl`.