https://file.aiquickdraw.com/mathbot/file/1726181238176zyzkjqqk.jpg

Let's solve the problem step by step:

### A. Free-Body Diagrams of Boxes A and B

**Free-Body Diagram for Box A:**
- Force \( F = 10.0 \, \text{N} \) acting to the right.
- Normal force (\( N_A \)) acting upward.
- Gravitational force (\( m_A g = 3.00 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)) acting downward.
- Force of interaction (\( F_{\text{interaction}} \)) from Box B acting to the left.

**Free-Body Diagram for Box B:**
- Force of interaction (\( F_{\text{interaction}} \)) from Box A acting to the right.
- Normal force (\( N_B \)) acting upward.
- Gravitational force (\( m_B g = 1.50 \, \text{kg} \times 9.8 \, \text{m/s}^2 \)) acting downward.

### B. Equations of Motion for Boxes A and B

Using Newton's second law, \( F = ma \):

1. **For Box A:**
   \[
   F - F_{\text{interaction}} = m_A a
   \]
   where \( F = 10.0 \, \text{N} \), \( m_A = 3.00 \, \text{kg} \).

2. **For Box B:**
   \[
   F_{\text{interaction}} = m_B a
   \]
   where \( m_B = 1.50 \, \text{kg} \).

### C. Acceleration of Boxes A and B

Both boxes accelerate together since they are in contact. To find the acceleration:

1. Substitute the equation of Box B into the equation of Box A:
   \[
   10.0 \, \text{N} - F_{\text{interaction}} = 3.00 \, \text{kg} \times a
   \]
   \[
   F_{\text{interaction}} = 1.50 \, \text{kg} \times a
   \]

2. Combine the equations:
   \[
   10.0 \, \text{N} = 3.00 \, \text{kg} \times a + 1.50 \, \text{kg} \times a
   \]
   \[
   10.0 \, \text{N} = (3.00 + 1.50) \, \text{kg} \times a
   \]
   \[
   10.0 \, \text{N} = 4.50 \, \text{kg} \times a
   \]
   \[
   a = \frac{10.0 \, \text{N}}{4.50 \, \text{kg}}
   \]
   \[
   a \approx 2.22 \, \text{m/s}^2
   \]

### D. Magnitude of the Force of Interaction Between Box A and B

Using the acceleration found in part C:

\[
F_{\text{interaction}} = m_B \times a = 1.50 \, \text{kg} \times 2.22 \, \text{m/s}^2
\]
\[
F_{\text{interaction}} \approx 3.33 \, \text{N}
\]

### Summary of Answers
- **Free-Body Diagrams:** Include forces acting on each box as described.
- **Equations of Motion:** Derived for each box.
- **Acceleration of Boxes A and B:** Approximately \( 2.22 \, \text{m/s}^2 \).
- **Force of Interaction Between Box A and B:** Approximately \( 3.33 \, \text{N} \).

Would you like more details on any part, or have any questions?

Here are 5 related questions for further exploration:
1. How would the acceleration change if there was friction involved?
2. What happens to the interaction force if Box B's mass is doubled?
3. How would the system behave if the force was applied to Box B instead of Box A?
4. What are the implications if the surface was inclined instead of horizontal?
5. How does the interaction force affect the net force on each box?

**Tip:** Always double-check unit consistency when calculating physical quantities.

Boxes A and B are in contact on a horizontal frictionless surface. Box A has mass m_A = 3.00 kg while box B has mass m_B = 1.50 kg. A horizontal force with strength F = 10.0 N acts on box A. Solve for the free-body diagrams, equations of motion, acceleration, and the force of interaction between the boxes.

Step-by-step solution to a physics problem involving boxes A and B on a frictionless surface. Learn how to calculate the free-body diagrams, equations of motion, acceleration, and interaction force using Newton's Second Law. Suitable for high school physics students.

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