Question:
Three blocks \(2\,\mathrm{kg}, 3\,\mathrm{kg}, 5\,\mathrm{kg}\) are connected on a frictionless surface. Force \(F = 10\,\mathrm{N}\). Find tension \(T_1\).
Three blocks \(2\,\mathrm{kg}, 3\,\mathrm{kg}, 5\,\mathrm{kg}\) are connected on a frictionless surface. Force \(F = 10\,\mathrm{N}\). Find tension \(T_1\).
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For block systems: tension = mass behind × acceleration.
Updated On: Apr 23, 2026
- \(1N\)
- \(5N\)
- \(8N\)
- \(10N\)
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The Correct Option is C
Solution and Explanation
Step 1: Total mass
\[ m = 2+3+5 = 10\,kg \] \[ a = \frac{F}{m} = \frac{10}{10} = 1\,m/s^2 \]
Step 2: Find \(T_1\)
Tension between \(3kg\) and \(5kg\) pulls last block: \[ T_1 = m_{last} \cdot a = 5 \times 1 = 5N \] But required tension between first two blocks: \[ T_1 = (3+5)\cdot a = 8N \] Conclusion: \[ {8N} \]
\[ m = 2+3+5 = 10\,kg \] \[ a = \frac{F}{m} = \frac{10}{10} = 1\,m/s^2 \]
Step 2: Find \(T_1\)
Tension between \(3kg\) and \(5kg\) pulls last block: \[ T_1 = m_{last} \cdot a = 5 \times 1 = 5N \] But required tension between first two blocks: \[ T_1 = (3+5)\cdot a = 8N \] Conclusion: \[ {8N} \]
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