Consider the situation as shown in the figure. The pulley has moment of inertia `I` and radius `r`. The table is rough and the coefficient of kinetic friction is `mu_(k)`. The system is released from the rest. Calculated speed of block of mass `3m` as a function of the distance `h` it has descended. Assume no slipping.

Explanation

Loss in the gravitational potential of block of mass `3m`
`=` gain in the kinetic energy (block `+` pulley) `+` work done against friction
`3mgh=(1)/(2)xx3mv^(2)+(1)/(2)xxmv^(2)+(1)/(2)Iomega^(2)+mu_(k)mgh`
`mgh(3-mu_(k))=2mv^(2)+(1)/(2)I((v)/(r))^(2)=v^(2)(2m+(I)/(2r^(2)))`
`=2mv^(2)(1+(I)/(4mr^(2)))`
`v=sqrt((gh(3-mu_(k)))/(2(1+(I)/(4mr^(2)))))`