**1. Introduction**

In this problem, we have an arrangement of blocks where a smaller block of mass m is placed on top of a larger block of mass M. The blocks are on a horizontal surface and are connected by a string passing over a pulley. The pulley is attached to a weight of mass W hanging vertically. We are asked to find the acceleration of the larger block M along the negative x-axis.

**2. Free Body Diagram**

To analyze the problem, we start by drawing a free body diagram for each block.

For the smaller block of mass m:
- There are two forces acting on it: the weight mg and the tension T in the string.
- The normal reaction N1 is perpendicular to the surface and cancels out the vertical component of the weight.

For the larger block of mass M:
- There are three forces acting on it: the weight Mg, the tension T in the string, and the normal reaction N2.
- The normal reaction N2 cancels out the vertical component of the weight, while the horizontal component of the weight provides the net force for acceleration.

**3. Equations of Motion**

Applying Newton's second law to each block, we can write the following equations of motion:

For the smaller block:
- ΣFy = N1 - mg = 0
- N1 = mg

For the larger block:
- ΣFx = T - Mg = Ma
- T = Ma + Mg

**4. Tension in the String**

To find the tension T in the string, we can substitute the value of N1 from the equation for the smaller block into the equation for the larger block:

- T = Ma + Mg
- T = m(a + g)

**5. Acceleration of the Larger Block**

We can now express the acceleration a in terms of the given quantities:

- T = m(a + g)
- a = (T/m) - g

Substituting the given value of T/m as N cos(theta)/M, we get:

- a = (N cos(theta)/M) - g

Finally, rearranging the equation, we obtain:

- a = N cos(theta)/M - g

Therefore, the acceleration of the larger block M along the negative x-axis is equal to N cos(theta)/M.

Where is The figure