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.

Answer

B

Explanation

Let x length of chain is one the table at a particular instant.
So, work done by friction force o a small element dx
`dw_1=muRx=(m/ldx gx)`
`[when dm=m/ldx]`
total work done by friction
`w_1=int_(2L/3)^0 uM/l gx dx`
`:. W_1=mu m/lg[x^2/2]_0^(21L/3)`
`mu m/lg[(4l^2)/18]`
`=(2mumg.l)/9`