In the system shown in figure `m_(B)=4kg` , and `m_(A)=2kg` . The pulleys are massless and friction is absent everywhere. The acceleration of block `A` is.

Options

A. `10//3m//s^(2)`
B. `20//3m//s^(2)`
C. `2m//s^(2)`
D. `4m//s^(2)`

Answer

A

Explanation

For movable pulley
`a_(B)=(O+a_(A))/(2)`
`a_(A)=2a_(B)` (i)
Free body diagrams of `A` and `B` :
Equation of motion of block `A` and `B`
For `A` : `T-(m_(A)g)/(2)=m_(A)a_(A)=m_(A)(2a_(B))`
For `B` : `m_(B)g-m_(A)g=[4m_(A)+m_(B)]a_(B)`
`a_(B)=([m_(B)-m_(A)]g)/([4m_(A)+m_(B)])=([4-2]xx10)/([4xx2+4])`
`=(20)/(12)=(5)/(3)m//s^(2)`
Hence `a_(A)=2a_(B)=(10)/(3)m//s^(2)` .