To solve the problem, we need to determine the coefficient of kinetic friction ($\mu_k$) between the block $M$ and the floor. Here’s the step-by-step process:

Given:

• Force $F = 140 \, \text{N}$
• Mass of the smaller block $m = 5 \, \text{kg}$
• Mass of the larger block $M = 20 \, \text{kg}$
• The system moves at a constant speed, implying zero acceleration.

Key Concepts:

Since the system is moving at a constant speed, the net force acting on the larger block $M$ must be zero. This implies that the pulling force $F$ is balanced by the kinetic friction force between $M$ and the floor.

Step 1: Analyze the forces on the larger block $M$

The frictional force $f$ acting on block $M$ is given by:

$f = \mu_k \cdot N$

where $N$ is the normal force. For block $M$, the normal force is its weight:

$N = M \cdot g = 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N}$

Step 2: Set up the force balance equation

Since the block moves at a constant speed:

$F = f$

Substituting the expressions for $f$:

$140 \, \text{N} = \mu_k \cdot 196 \, \text{N}$

Step 3: Solve for the coefficient of kinetic friction $\mu_k$

$\mu_k = \frac{140}{196} = 0.714$

Comparing this value with the provided options:

• A. $\mu_k = 0.32$
• B. $\mu_k = 0.41$
• C. $\mu_k = 0.57$
• D. $\mu_k = 0.69$
• E. $\mu_k = 0.75$

The closest option is:

Answer: D. $\mu_k = 0.69$

Would you like further details or explanations?

Related Questions:

1. How would the problem change if $m$ were not attached to a wall?
2. What would happen if the pulling force $F$ were increased?
3. How does friction affect the motion of objects in different scenarios?
4. What are the differences between static and kinetic friction?
5. How can friction be reduced in mechanical systems?

Tip:

Always double-check your calculations, especially when matching values against multiple-choice options; approximations can sometimes lead to slight variations in the final answer.