Consider the arrangement shown in the figure. Find the acceleration of the block
(a) A monkey moves upward with respect to the rope with acceleration `a_(0)`
(b) A monkey moves downward with respect to the rope with acceleration `a_(0)`
Assume the surface smooth, pulley and string light.

Explanation

Let `a` be the acceleration of the block w.r.t. to the ground`=` acceleration of the rope w.r.t. the ground
(a) When the monkey moves up,
`a_("monkey"//"ground")=a_("monkey"//"rope")-a_("rope"//"ground")=a_(0)-a`
Monkey`:`

`T=mg=m(a_(0)-a)` (i)

`T=Ma` (ii)
Solving (i) and (ii), we get
`Ma-ma=ma_(0)-ma`
`a=(m(g+a_(0)))/(M+m)`
(b) When the monkey moves down.
`a_("monkey"//"ground")=a_("monkey"//"rope")+a_("rope"//"ground")=a_(0)+a`
Money

`mg-T=m(a_(0)+a)` (I)
Block`:`

`T=Ma` (ii)
Solving (i) and (ii), we have
`mg-Ma=ma_(0)+ma`
`a=(m(g-a_(0)))/(M+m)`