a description to length of string method..
Take accl^n of A and of pulley 2 = a , that of pulley 1 = a// and that of B = a/ ...
The strings are named as shown in MY diagram (Right hand side figure)
Let B represent the position of the block B. Let it is displaced to B/ .
The position of 1st pulley is C. It goes to C / .
The 2nd pulley is in A along with the block A. It goes to A / .
E and F are the fixed ends.
The length of the string will remain constant.
So for first string.....
BD + DC + CE = B/ D + DC/ + C / E
or BB/ + B/ D + DC + CE = B/ D + DC + CC/ + C/C + CE ... (from figure)
or BB/ = 2CC/
Thus displacement of block B = 2 x (displacement of pulley 1)
Diff.. twice with respect ot time..
we get accl^n of block B (a/ ) = 2 x accl^n of pulley 1 (a// )
or a/ = 2 a//__________(1)
Now for second string....
CA + AF = C/ A/ + A/ F
or CC/ + C/ A + AF = C/ A + AA/ + A/ A + AF
or CC/ = 2 AA/..
Thus displacement of pulley 1 = 2 x (displacement of pulley 2)
= 2 x (displacement of block A)
Diff. twice with respect to time..
we get accl^n of pulley 1 (a// ) = 2 x accl^n of block A (a)
or a// = 2a ________(2)
equation (1) and (2) implies
a/ = 2 (2a)
or a/ = 4a.
i,e, accl^n of block B = 4 x (accl^n of block A) ___________(3)
Now let tension in string joining B with E be T
Since pulley 1 and 2 are massless, so force on them is zero ...
i,e tension in string joining C and F is = T + T = 2T
and tension in string joining the block A to the 2nd pulley = 2T + 2T = 4T
So writing force equations of A and B... respectively we get...
5g - 4T = 5a ________ (4)
and T - k
(10g) = 10 a/
or T - k
(10g) = 10 (4a) ........... [by (3) ]
or T - 0.1(100) = 40a
or T = 40a + 10 _______ (5)
Put value of T from equ^n (5) in equ^n (4).... to get ---->>>
5g - 4( 40a + 10) = 5a
or 50 - 160a - 40 = 5a
or 165a = 10
or a = accl^n of block A = 10/165 = 0.06 m/s^2