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 30N pulls the system as shown below. The value of tension in the string between $m_2$ and $m_3$ will be
To find the tension in the string between blocks \(m_2\) and \(m_3\), we need to analyze the forces acting on the blocks and utilize Newton's second law of motion.
First, calculate the total mass of the system:
\(m_{\text{total}} = m_1 + m_2 + m_3 = 2\,\text{kg} + 3\,\text{kg} + 5\,\text{kg} = 10\,\text{kg}\)
Now, calculate the acceleration of the system using the force applied:
\(F = 30\,\text{N}\)
Using Newton's second law \(F = m \cdot a\),
\(a = \frac{F}{m_{\text{total}}} = \frac{30\,\text{N}}{10\,\text{kg}} = 3\,\text{m/s}^2\)
To find the tension \(T\) in the string between \(m_2\) and \(m_3\), consider only the mass \(m_3\). The only force causing its acceleration is the tension in the string:
Using \(T = m_3 \cdot a\),
\(T = 5\,\text{kg} \cdot 3\,\text{m/s}^2 = 15\,\text{N}\)
Thus, the tension in the string between \(m_2\) and \(m_3\) is \(15\,\text{N}\).
The correct answer is: \(15\,\text{N}\)
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