In the following arrangement the system is initially at rest . The 5 kg block is now released . Assuming the pulleys and strings to me massless and smooth . We have to relation between acceleration of all blocks .

I tried as

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**Revised Answer**

Your attempt is correct. Your method works because the tensions in the ropes are internal forces, and the sum of work done by internal forces is zero. Work done **by** one part of the system is work done **on** another part of the system. "The system" here consists of the 3 masses and the moving pulley.

This gives you the equation of constraint :

$x_A+x_B+2x_C=0$.

(The same relation could be obtained by considering what happens to C if A and B move up by $x_A$ and $x_B$ respectively.)

From the above we get the relation between accelerations :

$a_A+a_B+2a_C=0$.

To get more equations, so that we can solve for the accelerations, we have to consider external forces - eg the pull of gravity.

An important observation (which I overlooked) is that blocks B and C are initially at rest on a floor and cannot move downwards.

The weight of A+B > C, so C will accelerate upwards and the "free" pulley on the LHS will accelerate downwards. Since the system starts from rest the direction of acceleration is also the direction of velocity - ie motion.

So either A or B (or both) will move downwards. Since the weight of B > A then B would move down - if it were not prevented by the floor. So it must be that A moves down while B rests on the floor $(a_B=0)$. Applying the restraint equation this leaves $a_A=-2a_C$.

Apply $F=ma$ to blocks A and C - remember that +ve acceleration is upwards :

$T-5g=5a_A=-10a_C$

$2T-8g=8a_C$.

Eliminating $T$ we get

$2g=28a_C$

$a_C=\frac{1}{14} g$

$a_A=-2a_C=-\frac17 g$.

This immediately gives $a_B=0$ and $a_A=2a_C$. After which, the solution is a lot simpler. I shall revise my answer later.

...

Then what could be the mistake in your answer