Three blocks of masses \( m_1 = 2\text{ kg} \), \( m_2 = 3\text{ kg} \) and \( m_3 = 5\text{ kg} \) are placed on a horizontal frictionless surface and a force of \( 30\text{ N} \) pulls the system as shown below. The value of tension \( T \) will be
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In connected body systems, the tension in any string is always equal to the sum of all masses behind it multiplied by the common acceleration of the system. This allows you to find any tension instantly without drawing complete free-body diagrams.
Step 1: Understanding the Question:
Three connected blocks are being pulled on a frictionless surface by a horizontal force of \( 30\text{ N} \). Since the specific string for tension \( T \) is not explicitly labeled in the diagram, we will calculate the tension in both connecting strings. Step 2: Key Formula or Approach:
1. Common Acceleration of the System (\( a \)): Since the blocks are connected, they move together with a common acceleration:
\[ a = \frac{F_{\text{net}}}{M_{\text{total}}} = \frac{F}{m_1 + m_2 + m_3} \]
2. Tension in a string: Apply Newton's second law (\( F_{\text{net}} = m a \)) on individual blocks or subsystems of blocks. Step 3: Detailed Explanation:
First, find the common acceleration of the system:
- Total mass: \( M_{\text{total}} = 2\text{ kg} + 3\text{ kg} + 5\text{ kg} = 10\text{ kg} \)
- Pulling force: \( F = 30\text{ N} \)
- Acceleration:
\[ a = \frac{30\text{ N}}{10\text{ kg}} = 3\text{ m/s}^2 \]
Now we analyze the two strings:
- Case 1: Tension \( T_1 \) in the string between \( m_2 \) (\( 3\text{ kg} \)) and \( m_3 \) (\( 5\text{ kg} \)):
This string is responsible for accelerating both \( m_1 \) and \( m_2 \). Using the subsystem of \( (m_1 + m_2) \):
\[ T_1 = (m_1 + m_2) a = (2 + 3) \times 3 = 5 \times 3 = 15\text{ N} \]
- Case 2: Tension \( T_2 \) in the string between \( m_1 \) (\( 2\text{ kg} \)) and \( m_2 \) (\( 3\text{ kg} \)):
This string is only responsible for accelerating the block \( m_1 \). Using the block \( m_1 \):
\[ T_2 = m_1 a = 2 \times 3 = 6\text{ N} \]
Since both \( 15\text{ N} \) and \( 6\text{ N} \) are in the options, they are both correct depending on which string is chosen. Step 4: Final Answer:
The tension is \( 15\text{ N} \) (option A) for the string between the \( 3\text{ kg} \) and \( 5\text{ kg} \) masses, or \( 6\text{ N} \) (option C) for the string between the \( 2\text{ kg} \) and \( 3\text{ kg} \) masses.
