Answer:
4.1. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. In other words, when one object exerts a force on another object, the second object exerts an equal force in the opposite direction on the first object.
4.2. Here's a labeled free-body diagram for Block A:
```
T (tension in the string)
↑
│
│
│
│
│
F (applied force)
──→ (direction of motion)
```
In this diagram, "T" represents the tension in the string, and "F" represents the applied force at an angle of 30° to the horizontal. The arrow indicates the direction of motion.
4.3. To find the frictional force acting on block A as it accelerates, we can use Newton's Second Law:
\[F_{\text{net, A}} = m_A \cdot a\]
Where:
- \(F_{\text{net, A}}\) is the net force acting on block A.
- \(m_A\) is the mass of block A (given as 15 kg).
- \(a\) is the acceleration (given as 2.08 m/s²).
Rearranging the equation to solve for \(F_{\text{net, A}}\):
\[F_{\text{net, A}} = 15 kg \cdot 2.08 m/s² = 31.2 N\]
Now, we need to consider the frictional force, which opposes the motion and acts in the direction opposite to the applied force. So, the frictional force is 31.2 N in the opposite direction of motion, making it:
Frictional force on block A = -31.2 N
However, since you want it in magnitude, it's 31.2 N.
4.4. To calculate the mass of block B, we can use the fact that block A and block B are connected by a string, so they experience the same acceleration. Therefore, we can use the following equation:
\[F_{\text{net, B}} = m_B \cdot a\]
Where:
- \(F_{\text{net, B}}\) is the net force acting on block B, which is the tension in the string.
- \(m_B\) is the mass of block B (unknown).
- \(a\) is the acceleration (given as 2.08 m/s²).
We already calculated that the tension in the string is 31.2 N. Plugging in the values:
\[31.2 N = m_B \cdot 2.08 m/s²\]
Now, solving for \(m_B\):
\[m_B = \frac{31.2 N}{2.08 m/s²} \approx 15 kg\]
So, the mass of block B is approximately 15 kg.