A uniform chain of mass m and length l is at the verge of sliding under the effect of gravity of the hanging part. Find the

a. coefficient of friction, between chain and table.
b. work done by friction and gravity till the chain leaves the table if the hanging part is pulled gently and released.
c. speed of the chain at the time of leaving the table in part (b) .
d. work done by the external force acting at the end A of the chain in slowly pulling the chain completely onto the table.

Explanation

a. Limiting friction acting on the upper part of chain will balance the weight of overhanging part.

`mu(m/3)g=(2m)/(3)gimpliesmu=2`
b. Let at any time, chain slides down by a distance x. Then
`f_l=mu[(l/2-x)m/l]g`
`W_f=-intf_ldx=-(mumg)/(l)underset0overset(l//3)int(l/3-x)dx`
`impliesW_f=-(mumgl)/(18)-(2mgl)/(18)=-(mgl)/(9)`
`impliesW_(gravity)=(2m)/(3)gl/3+m/3gl/6=(5mgl)/(18)`
c. `W_(gravity)+W_f=1/2mv^2`
`implies(5mgl)/(18)-(mgl)/(9)=1/2mv^2impliesv=sqrt((gl)/(3))`

d. `f_l=mu(1/3+x)m/lg`
`W_f=-intf_ldx`
`=-(mumg)/(l)underset0overset(2l//3)int(l/3+x)dx`
`=-(4mumgl)/(9)=-(8mgl)/(9)`
`W_(gravity)=-(2m)/(3)gl/3=-(2mgl)/(9)`
`W_(ext)+W_f+W_(gravity)=0`
`impliesW_(ext)=(8mgl)/(9)+(2mgl)/(9)=(10mgl)/(9)`