Have come across that acceleration of center of mass is the net force divided by the total mass,
my query is if a block of mass m is placed on an incline of mass M and the incline is dragged to the right by a horizontal force F, all surfaces being frictionless,
this case is bit complicated since the block is sliding down the incline and the incline itself is moving right by the force F,
will acceleration of center of mass be given by the same same expression,
$a = F/(M + m)$
You are correct and to justify it, consider the system M+m. The only external force in the x direction is F, therefore the acceleration of its center of mass is : $ a_{cm} = \frac{F}{M+m} $
In classical mechanics, Newton's Second Principle of Dynamics states that the time derivative of the momentum of a system is equal to the resultant of the external forces,
$ \dfrac{d \mathbf{Q}}{dt} = \mathbf{R}^e $.
It's quite easy to prove that the momentum van be written as the product of the total mass of the system and the velocity of its centre of mass, $\mathbf{Q} = m^{tot} \mathbf{v}_G$. If the system is close (i.e. doesn't exchange mass with the environment, so that its mass is constant in classical mechanics), it's possible to write the Second Principle of Dynamics as
$ m^{tot} \dfrac{d \mathbf{v}_G}{dt} = \mathbf{R}^e \qquad$ i.e. $\qquad m^{tot} \mathbf{a}_G = \mathbf{R}^e $.
Before going on, please remember that force, acceleration (and velocity and position) are vector quantities and the Second Principle of Dynamics is a vector equation. In a 2D problem, like the one depicted in the image, you need to write down two components, here let's call them $x$ in the horizontal direction, and $y$ the vertical direction
$x: \quad m^{tot} a_{G,x} = F$
$y: \quad m^{tot} a_{G,y} = N_y - m^{tot} g$,
being $m^{tot} = m + M$ the total mass of the system, $F$ the horizontal external force applied, $N_y$ the normal reaction acting to the lower surface of the triangle due to the presence of the horizontal frictionless surface where the system is sliding, and $g$ the gravity field.
From the first equation, you can readily get the horizontal component of the acceleration of the centre of mass
$ a_{G,x} = \dfrac{F}{m^{tot}} = \dfrac{F}{m + M}$,
while in the y-equation you have two unknowns, $N_y$ and $a_y$, so that you need another independent equation to solve the problem.
In general the reaction $N_y$ and the weight of the system do not cancel out.
TODO: solve the problem for $a_{G, y}$
