Answer :
Answer:
Therefore, the acceleration of the system is 2.5 m/s and the tension force in the rope is 10 N.
Explanation:
To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Additionally, we can use the fact that the tension force in the rope is the same throughout the system.
Let's denote the acceleration of both blocks as \( a \) and the tension force in the rope as \( T \).
For the first block (4 kg):
\[ F_{\text{net}} = m_1 \cdot a \]
\[ T - F_{\text{friction}} = m_1 \cdot a \]
\[ T = m_1 \cdot a \] (since friction is ignored)
For the second block (6 kg):
\[ F_{\text{net}} = m_2 \cdot a \]
\[ T = m_2 \cdot a \]
Since the tension in the rope is the same for both blocks, we can set the equations for tension equal to each other:
\[ m_1 \cdot a = m_2 \cdot a \]
Plugging in the given values:
\[ 4 \cdot a = 6 \cdot a \]
\[ 4a = 6a \]
\[ 4 = 6 \]
This equation is not possible; it implies that the acceleration is infinite, which is not realistic. The reason for this contradiction is that we neglected to include the external force acting on the system (the force of 25 N).
Let's correct this and include the external force in our calculations.
The net force acting on the system is the external force minus the force due to the friction.
\[ F_{\text{net}} = F_{\text{ext}} - F_{\text{friction}} \]
Since friction is ignored, \( F_{\text{friction}} = 0 \), so
\[ F_{\text{net}} = F_{\text{ext}} \]
Now, we can use Newton's second law to find the acceleration:
\[ F_{\text{net}} = m_{\text{total}} \cdot a \]
\[ F_{\text{ext}} = (m_1 + m_2) \cdot a \]
\[ 25 = (4 + 6) \cdot a \]
\[ 25 = 10a \]
\[ a = \frac{25}{10} \]
\[ a = 2.5 \, \text{m/s}^2 \]
Now that we have the acceleration, we can find the tension force in the rope using one of the equations we derived earlier:
\[ T = m_1 \cdot a \]
\[ T = 4 \cdot 2.5 \]
\[ T = 10 \, \text{N} \]
Therefore, the acceleration of the system is \( 2.5 \, \text{m/s}^2 \) and the tension force in the rope is \( 10 \, \text{N} \).
