A stationary light, smooth pulley can rotate without friction about a fixed horizontal axis. A light rope passes over the pulley. One end of the rope supports a ladder with man and the other end supports a counterweight of mass `M`. Mass of the man is `m`. initially, the centre of mass of the counterweight is at a height `h` from that of man as shown in Fig.
If the man starts to climb up the ladder slowly, calculate work done by him to reach his centre of mass in level with that of the counterweight.

Answer

`W=(Mmgh)/(M-m)`

Explanation

Mass of ladder `=M-m`, because net mass on both sides of string is same. Let in the given process, block `M` rises by height `y`, then ladder will come down by distance `y` and when man comes in the level of `M`, then height raised by man w.r.t ground will be `h+y`.
Now cenre of mass (man`+`ladder) system will be raised by `y` , then

`y=(m(h+y)+(M-m)(-y))/(2(M-m))`
Now work done by man
`W=`total change in potential energy of system
`implies W=Mgy+mg(h+y)-(M-m)gy`
Put the values of `y` and solve to get `W=(Mmgh)/(M-m)`