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To solve the problem, we need to determine the coefficient of kinetic friction \(\mu_k\) between the block of mass \(M\) and the floor.

### Given:
- Mass of block \(M = 20 \, \text{kg}\)
- Mass of block \(m = 5 \, \text{kg}\)
- Force \(F = 140 \, \text{N}\)
- The system moves at a constant speed, meaning the acceleration \(a = 0\).

### Key Points:
- Since the system is moving at a constant speed, the net force acting on block \(M\) must be zero.
- The frictional force \(f_k\) between block \(M\) and the floor must exactly balance the applied force \(F\).

### Forces acting on block \(M\):
1. **Horizontal Direction**:  
   - Applied force \(F = 140 \, \text{N}\) to the left.
   - Frictional force \(f_k\) opposing the motion.

   Since the block moves at a constant speed, the frictional force must be equal to the applied force:
   \[
   f_k = F = 140 \, \text{N}
   \]

2. **Vertical Direction**:
   - The normal force \(N\) acting on block \(M\) is the sum of the gravitational forces of block \(M\) and block \(m\) (since \(m\) is placed on top of \(M\) and is attached to the wall).
   \[
   N = (M + m) \cdot g
   \]
   where \(g\) is the acceleration due to gravity (\(g = 9.8 \, \text{m/s}^2\)).

   Plugging in the values:
   \[
   N = (20 \, \text{kg} + 5 \, \text{kg}) \times 9.8 \, \text{m/s}^2 = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 245 \, \text{N}
   \]

### Determining the Coefficient of Kinetic Friction \(\mu_k\):
- The frictional force is given by:
   \[
   f_k = \mu_k \times N
   \]
   Substituting the values:
   \[
   140 \, \text{N} = \mu_k \times 245 \, \text{N}
   \]

- Solving for \(\mu_k\):
   \[
   \mu_k = \frac{140 \, \text{N}}{245 \, \text{N}} \approx 0.571
   \]

### Answer:
The coefficient of kinetic friction \(\mu_k\) between block \(M\) and the floor is approximately **0.571**.

---

Do you want further details or have any questions? Here are some related questions to consider:

1. What would happen if the applied force \(F\) were increased?
2. How would the friction coefficient change if block \(m\) were removed?
3. What role does the string attached to block \(m\) play in this scenario?
4. How would the problem change if there were friction between block \(m\) and \(M\)?
5. How can the concept of normal force be applied in inclined planes?

**Tip:** When dealing with frictional forces, always ensure the normal force is correctly calculated, as it directly affects the frictional force.

A block of mass M is pulled toward the left with a force of F = 140 N and moves at a constant speed. There is no friction at the surface between the two masses, but there is friction between M and the floor. The block m does not move since it is attached to a wall by a string. Use m = 5 kg, and M = 20 kg. What is the coefficient of kinetic friction between M and the floor?

Determine the coefficient of kinetic friction between a block M of mass 20 kg and the floor when it is pulled with a force of 140 N at constant speed. The block m of 5 kg is attached to a wall. This problem covers frictional forces, Newton's laws of motion, and normal force calculation. Suitable for Grades 9-12.

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