To find the magnitude of the horizontal force $F$ needed to drag block $B$ to the left at a constant speed, let's analyze the forces on both blocks $A$ and $B$. Here's the step-by-step solution:

Step 1: List Given Data

• Weight of block $A$: $W_A = 1.40 \, \text{N}$
• Weight of block $B$: $W_B = 4.20 \, \text{N}$
• Coefficient of kinetic friction $\mu_k = 0.30$
• The system moves at a constant speed, meaning acceleration $a = 0$.

Step 2: Determine Normal Forces

Since the weights of the blocks act vertically, the normal forces for each block are given by their weights:

• Normal force on $A$: $N_A = W_A = 1.40 \, \text{N}$
• Normal force on $B$: $N_B = W_B = 4.20 \, \text{N}$

Step 3: Calculate Frictional Forces

The kinetic frictional force $f_k$ for each block is given by: $f_k = \mu_k \cdot N$

• Friction force on $A$: $f_{kA} = \mu_k \cdot N_A = 0.30 \cdot 1.40 = 0.42 \, \text{N}$
• Friction force on $B$: $f_{kB} = \mu_k \cdot N_B = 0.30 \cdot 4.20 = 1.26 \, \text{N}$

Step 4: Analyze Forces on Block $A$

Block $A$ is connected to block $B$ via a light flexible cord that goes over a frictionless pulley. Since block $B$ is moving at a constant speed, block $A$ will also move at a constant speed to the left. Therefore, the forces on block $A$ in the horizontal direction must balance out.

Let $T$ be the tension in the cord connecting $A$ and $B$. The forces on block $A$ in the horizontal direction are:

1. Tension $T$ pulling $A$ to the left.
2. Frictional force $f_{kA} = 0.42 \, \text{N}$ opposing the motion to the right.

Since block $A$ moves at constant speed: $T = f_{kA} = 0.42 \, \text{N}$

Step 5: Analyze Forces on Block $B$

For block $B$, the forces in the horizontal direction are:

1. The applied force $F$ pulling $B$ to the left.
2. The frictional force $f_{kB} = 1.26 \, \text{N}$ opposing the motion to the right.
3. The tension $T = 0.42 \, \text{N}$ pulling block $B$ to the right (because it’s attached to $A$).

Since block $B$ is also moving at a constant speed: $F = f_{kB} + T$ Substitute the values: $F = 1.26 + 0.42 = 1.68 \, \text{N}$

Final Answer

The magnitude of the horizontal force $F$ necessary to drag block $B$ to the left at a constant speed is: $\boxed{1.68 \, \text{N}}$

Would you like further details or have any questions?

Additional Questions

1. How would the force $F$ change if the coefficient of kinetic friction increased?
2. What would happen to the tension $T$ if block $A$ was heavier?
3. How would the result differ if the pulley had friction?
4. What is the effect of changing the direction of $F$ on the motion of blocks $A$ and $B$?
5. If $F$ were applied to block $A$ instead, how would the system's behavior change?

Tip

In problems involving friction, identifying all surfaces in contact and calculating individual frictional forces separately helps in balancing the equations efficiently.