The coefficient of static friction is 0.60 between the two blocks in figure. The coefficient of kinetic friction between the lower block and the floor is 0.20. Force F⃗ causes both blocks to cross a distance of 5.0 m, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

To solve this problem, we need to analyze the forces acting on the system and determine the minimum time required for the motion, given that the top block must not slide on the lower block. Here's a step-by-step breakdown:

### Step 1: Understanding the forces involved

- **Static friction** between the two blocks (top and lower block) resists the relative motion between them. The maximum static friction force, \( f_s \), is given by:
  \[
  f_s = \mu_s m_1 g
  \]
  where \( \mu_s = 0.60 \) is the coefficient of static friction, \( m_1 \) is the mass of the top block, and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

- **Kinetic friction** between the lower block and the floor resists the motion of the system. The kinetic friction force, \( f_k \), is given by:
  \[
  f_k = \mu_k (m_1 + m_2) g
  \]
  where \( \mu_k = 0.20 \) is the coefficient of kinetic friction, and \( m_2 \) is the mass of the lower block.

### Step 2: Forces acting on the system

The force \( F \) must accelerate both blocks together across a distance of 5.0 m without causing the top block to slide on the lower block. This means the acceleration of the system, \( a \), must be such that the static friction force is not exceeded.

1. The total mass of the system is \( m_1 + m_2 \).
2. The net force required to accelerate the system is:
   \[
   F - f_k = (m_1 + m_2) a
   \]
   where \( F \) is the applied force and \( f_k \) is the kinetic friction.

3. For the top block not to slide, the static friction must provide enough force to cause the same acceleration \( a \):
   \[
   f_s \geq m_1 a
   \]
   Substituting \( f_s = \mu_s m_1 g \), we get:
   \[
   \mu_s m_1 g \geq m_1 a
   \]
   This gives the maximum acceleration \( a \):
   \[
   a \leq \mu_s g
   \]
   Substituting \( \mu_s = 0.60 \) and \( g = 9.8 \, \text{m/s}^2 \), we find:
   \[
   a \leq 0.60 \times 9.8 = 5.88 \, \text{m/s}^2
   \]

### Step 3: Kinematic equation to find minimum time

The blocks start from rest and move a distance of 5.0 m. The kinematic equation that relates distance \( d \), acceleration \( a \), and time \( t \) is:
\[
d = \frac{1}{2} a t^2
\]
Solving for time \( t \):
\[
t = \sqrt{\frac{2d}{a}}
\]
Substitute \( d = 5.0 \, \text{m} \) and \( a = 5.88 \, \text{m/s}^2 \):
\[
t = \sqrt{\frac{2 \times 5.0}{5.88}} = \sqrt{1.7007} \approx 1.30 \, \text{seconds}
\]

### Conclusion:

The least amount of time in which the motion can be completed without the top block sliding on the lower block is approximately **1.30 seconds**.

---

Would you like more details on any part of the solution? Here are some follow-up questions to expand your understanding:

1. What happens if the coefficient of static friction is lower than 0.60?
2. How would the answer change if the blocks start with an initial velocity instead of from rest?
3. What effect would a higher coefficient of kinetic friction have on the minimum time?
4. How can we calculate the force \( F \) needed to achieve this motion?
5. What if the distance traveled were different, say 10 meters instead of 5 meters?

**Tip:** Always ensure the static friction is enough to prevent slipping, especially when dealing with multiple bodies in contact during motion.

Solve a friction problem using Newton's laws of motion. Learn how to calculate the minimum time for two blocks to travel 5 meters without sliding, using coefficients of static and kinetic friction. This physics problem includes key concepts such as friction forces, kinematic equations, and Newton’s second law. Suitable for Grades 11-12.

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