The system shown in figure is released is released from rest . Find acceleration of different string blocks and tenson in different strings.

Explanation

(i) pully `P` and `1kg` mass are attached with the same string. Therefore, if 1kg mass has an acceleration 'a' in upward direction , then pully P will have an acceleration 'a' downwards.
(ii) `2kg` and `3kg` blocks are attached with the same string passing over a moveable pully `P`. Therefore their relative acceleration, a, (relative to pully) will be same. Their net accelerations (relativeto ground) are as shown in figure.
(iii) Pully `P` is massless. Hence net force on this pulley should be zero. If `T` is the tenson in the string connecting `2kg` and `3kg` mass, then tenson in the upper string will be `2T`.
Now writing the equation ,`F_("net") =ma` for three blocks, we have.
`1kg` block

`2T-10=1xxa` ...(i)
`2kg` block
`T-20=(a_(r)-a)` ...(ii)
`3kg` block
`30-T=3(a_(r)+a)` ...(iii)
Solving Eqs. `(i),(ii)` and `(iii)` we get,
`T = 8.28N`, `a = 6.55m//s^(2)` and `a_(r) = 0.7m//s^(2)`.
Now, accelration of `3kg` block is `(a+a_(r))` or `7.25m//s^(2)` downwards and acceleration of `2kg` is `(a_(r) - a)`or `-5.85m//s^(2)` upwards. Since ,this comes out to be negative, hence acceleration of `2kg` block is `5.85m//s^(2)` downwards.