Consider a situation as shown in the figure. The system is released from rest. When the block of mass `m` has falled a distance `L`, its speed becomes `(sqrt(gL))/(3)`. Find the friction coefficient `mu`.

Explanation

When the block of mass `m` has descended a distance `L`, distance traveled by the block of mass `2m` is 2L. IF the speed of `m` is `v`, speed of `2m` will be `2v`.
Work done against friction `=muxx2mgxx2L`
`=4mumgL`
Loss in `P.E=`gain in `K.E+`work donw against friction
`mgL=(1)/(2)mv^2+(1)/(2)xx2m(2v)^2+4mumgL`
`=(9)/(2)mv^2+4mumgL`
`=(9)/(2)mxx(gL)/(9)+4mumgL`
`(mgL)/(2)=4mumgLimpliesmu=(1)/(8)`