To solve this problem, we can use the concept of friction and Newton's second law of motion.

1. Understanding the problem:
A heavy block of mass M is placed on a conveyor belt moving with a speed v. The coefficient of friction between the block and the belt is μ. We need to find the distance the block will slide on the belt before coming to rest.

2. Analyzing the forces acting on the block:
- The weight of the block acts vertically downwards with a force of mg (where g is the acceleration due to gravity).
- The normal force acts perpendicular to the surface of the belt and is equal to the weight of the block, which is mg.
- The frictional force acts horizontally opposite to the direction of motion and is given by f = μN.

3. Applying Newton's second law of motion:
The net force acting on the block is the difference between the applied force (friction) and the opposing force (weight). Therefore, we can write:
ma = f - mg

4. Solving for acceleration:
Substituting the values of f and N, we get:
ma = μmg - mg
a = μg - g
a = g(μ - 1)

5. Calculating the time taken to stop:
The block will come to rest when its final velocity becomes zero. Using the equation of motion v = u + at, where u is the initial velocity (which is the speed of the conveyor belt), we can write:
0 = v + a(t)
t = -v/a
t = -v/(g(μ - 1))

6. Calculating the distance traveled:
The distance traveled by the block can be calculated using the equation of motion s = ut + (1/2)at^2. Since the initial velocity u is equal to v, we have:
s = vt + (1/2)at^2
s = v(-v/(g(μ - 1))) + (1/2)(g(μ - 1))^2(-v/(g(μ - 1)))^2
s = -v^2/(g(μ - 1)) - v^2/(2g(μ - 1))
s = -2v^2/(2g(μ - 1)) - v^2/(2g(μ - 1))
s = -3v^2/(2g(μ - 1))
s = v^2/(2g(1 - μ))

7. Simplifying the expression:
Since the coefficient of friction (μ) is less than 1, the denominator (1 - μ) will be positive. Therefore, we can simplify the expression as follows:
s = v^2/(2g(1 - μ))
s = v^2/(2g - 2gμ)
s = v^2/(2g(1 - μ))
s = v^2/2gμ

8. Comparing the answer choices:
The correct answer is option 'D', which matches the derived expression for the distance traveled by the block: v^2/(2gμ).