Find the acceleration of the block of mass `m` in the arrangement shown in the following figures. The masses of the pulleys and the threads, as well as the friction are assumed to be negligible.

Explanation

(a)
If pulley `A` goes down by `x_(0)` (also the block of mass `4m`), the block of mass `m` moves up by `2kg`
If the acceleration of `4m` is `a`, the acceleration of `m` will be `2a`

`4mg-2T=4mg` (i)

`T-mg=mxx2a` (ii)
`(i)+2(ii)rArr2mg=4ma+4ma`
`rArra=(g)/(4)`
Acceleration of `m:2a=(g)/(2)`
(b)

`2mg=T-2mxx2a` (i)
`2T-mgsin30^(@)=mxxa` (ii)
`a=(7g)/(18)`
Acceleration of `m:(7g)/(10)`
Observe carefully. the block moving together with the movable pulley hasd acceleration `a`, another block `2a`.