A uniform chain of length L and mass M overhangs a horizontal table with its two third part n the table. The friction coefficient between the table and the chain is `mu`. Find the work done by the friction during the period the chain slips off the table.

Explanation

Let at a certain instant length c of the string is hanging vertically as shown in figure .

Let N be the normal reaction applied by the table on portion of string lyng on the table . Normal reaction will be equal to the weight of string lying on the ytable .
`N = m/L (L-x) g `
Instanteneous force of friction applied by teh table can be written as follows :
`F = mu m/L (L-x)g`
Let string further slip off by a distance dx , then for this duration we can assume friction to be constant and work done can be written as follows :
`dW = -Fdx = - mu m/L (L-x) g dx `
Negative sign is due to thef fact that force and displacement are opposite to each other . Total work done by the friction can be calculated by integrating above relation nder suitable limits .
` rArr " " W = mu(mg)/L int_(L//3)^(L) (L-x) dx`
`rArr " " W = mu (mg)/L [ Lx - (x^(2))/2] _(L//3)^(L)`
`rArr " " W = - mu (Mg)/L [{L^(2) - (L^(2))/2 } { (L^(2))/3 - (L^(2))/(18)}] `
` rArr " " W = -mu (mg)/L [{(L^(2))/2}] - {(5L^(2))/18}] `
`rArr " " W = - (2mumgL)/9 `