We need to determine the least amount of time in which block A can be pulled across 5.0 meters without sliding on block B. Here's how we approach the problem step by step:

Step 1: Understand the forces

Block A (4 kg) is on top of block B (3 kg), and they both experience friction:

• Static friction between A and B prevents A from sliding on B.
• Kinetic friction between block B and the floor slows down the system as it moves.

The goal is to find the maximum acceleration that prevents block A from sliding on block B.

Step 2: Maximum static friction

The static friction force between block A and block B is what holds A in place relative to B. The maximum possible static friction force is given by: $f_{\text{max}} = \mu_s \cdot N_{\text{AB}}$ where:

• $\mu_s = 0.60$ (coefficient of static friction between A and B),
• $N_{\text{AB}}$ is the normal force between A and B, which is equal to the weight of block A: $N_{\text{AB}} = m_A \cdot g = 4 \cdot 9.8 = 39.2 \, \text{N}.$ Thus: $f_{\text{max}} = 0.60 \cdot 39.2 = 23.52 \, \text{N}.$ This is the maximum force that friction can provide to prevent block A from sliding on block B.

Step 3: Maximum acceleration of the system

To prevent block A from sliding, the maximum possible acceleration $a_{\text{max}}$ of the entire system must be such that the static friction force can counteract the relative motion between the two blocks. The maximum acceleration is found by dividing the maximum static friction by the mass of block A: $a_{\text{max}} = \frac{f_{\text{max}}}{m_A} = \frac{23.52}{4} = 5.88 \, \text{m/s}^2.$

Step 4: Total forces acting on the system

Next, we calculate the total force needed to accelerate both blocks together. The total mass of the system is: $m_{\text{total}} = m_A + m_B = 4 + 3 = 7 \, \text{kg}.$

The friction between block B and the floor opposes the motion. The kinetic friction force between B and the floor is: [ f_k = \mu_k \cdot N_{\text{B}} = 0.20 \cdot m_B \cdot g = 0.20 \cdot 3 \cdot 9.8 = 5.88 \