Pulley problem, wrong decomposition of forces The mass of the red block is $M$. The rope is inextensible and its mass isn't relevant. The mass of the sphere is $M\sqrt{2}$. The angle between the rope and the horizontal is $45°$. 

I'm looking for the acceleration of the body but I'm doing something wrong. I want to understand my mistake.
Well, let's say that $\vec{F_m}$ is the force of the rope (the weight of the sphere). We can decompose it as $\vec{F_{mx}}$ and $\vec{F_{my}}$. 
$\vec{F_{p}}$ is the weight of the block. 
We arrive that $|\vec{F_{my}}|=|\vec{F_{p}}|$ and this is 100% true, in fact the test asks me to demonstrate it. So there is no friction. Then the only force remained is $\vec{F_{mx}}$. So we can calculate the acceleration in this way:
$a = |\vec{F_{mx}}|/M = Mgcos(45)\sqrt{2}/M=g$
However this isn't the solution, so there is a force which I'm not considerating. The solution in fact is $4.03 m/s^2$
EDIT: The question isn't a duplicate: we're talking about acceleration.
 A: The mistake is that $F_m$ is set equal to $M\sqrt{2}g$. 
This would be the case if the ball was stationary; if it was hanging there without falling, without accelerating. But it is in fact accelerating.
When you let go, the ball and box start to move. Gravity accelerates the ball downwards. They may not be moving in the very first initial moment, but they still accelerate. Otherwise they would never gain any speed.
Now, go back to where you found the value of $F_m$. I assume you found that by setting up $M\sqrt{2}g-F_m=0$ with Newton's 1st law on the ball. And then you correctly assumed that this $F_m$ pulls all the way along the string to the box. Since we now realized that Newton's 1st law isn't valid here, when the ball isn't stationary, we must instead use Newton's 2nd law and include the missed acceleration term:
$$M\sqrt{2}g-F_m=M\sqrt{2}a_{ball}$$
A: Try to avoid answering questions by looking at the results you are trying to obtain.
Start from the very beginning, without any prior knowledge of the acceleration and tension. All you know is the mass of the objects and the inclination of the rope. This eliminates the factuality of the $\vec F_{p}+\vec F_{my}=0$ relation in the question. This relation might show up in our answer but its not stated true in the question.
Since we are trying to find the acceleration of the block horizontally, begin with finding the acceleration of the rope itself by resolving the weight of the block along the rope to get the acceleration of the block along the rope and setting up equations of motion for both the objects. After calculating, you are one (or two) manipulation away from your answer.
A: Consider the tension present in the rope to be T.
So, for the sphere,
$$M\sqrt2g - T = M\sqrt2a_1$$
Similarly, for the block,
$$\frac{T}{\sqrt2} = Ma_2$$
[Here, $a_1$ and $a_2$ are the acceleration of sphere and block respectively.]
Now, assuming that the rope remains taut, the components of $a_1$ and $a_2$ along the rope wouybe equal.
Thus,
$$a_1=\frac{a_2}{\sqrt2}$$
On solving, you should probably reach to the correct answer.
What mistake you were making is 
$$\vec{|F_p|} = \frac{T}{\sqrt2}$$
However, this won’t be the case.
The correct equation I think would be
$$\vec{|F_p|} = \frac{T}{\sqrt2} + N$$
Where N is the force exerted by the ground in contact on the block
