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Three blocks of masses $m_{1}, m_{2}$ and $m_{3}$ are connected by massless strings as shown on a frictionless table. They are pulled with a force $T_{3} = 40 N$ If $m_{1} = 10 kg, m_{2} = 6 kg$ and $m_{3} = 4 kg$ the tension $T_{2}$ will be

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20 N

40 N

10 N

32 N

answer is D.

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Detailed Solution

$a = \frac{F}{m_{1} + m_{2} + m_{3}} = \frac{40}{20} = 2 m / s^{2}$

$T_{2} = (m_{1} + m_{2}) a = (16) 2 = 32 N$

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