1. Forces acting on the block

A block sliding down a rough incline experiences:

• Weight: $Mg$ vertically downward
• Normal reaction: $N$ perpendicular to the plane
• Friction: $f$ along the plane, opposite the motion

Since the block slides down with constant velocity, its acceleration is zero.

2. Resolve forces along the incline

Component of weight along the incline: $Mg\sin\theta$

Since acceleration is zero along the plane, $Mg\sin\theta = f$

3. Resolve forces perpendicular to the incline

Perpendicular component of weight: $Mg\cos\theta$

So normal reaction is $N = Mg\cos\theta$

4. Find the contact force

The contact force is the resultant of normal reaction and friction, because both are contact forces.

So its magnitude is $R = \sqrt{N^2 + f^2}$

Substitute: $R = \sqrt{(Mg\cos\theta)^2 + (Mg\sin\theta)^2}$

$R = Mg\sqrt{\cos^2\theta + \sin^2\theta}$

$R = Mg$

5. Correct option

$\boxed{A: Mg}$

6. Comparison with stored answer

Stored correct answer: $A$

Our derived answer is also $A$, so they agree.