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Standard XI

Physics

Tension in a String

Three blocks ...

Question

Three blocks of masses $m_{1}$ , $m_{2}$ and $m_{3}$ are connected by massless strings as shown in the figure and placed on a frictionless table. They are pulled with a force $T_{3} = 40 N$ . The tension $T_{2}$ will be: (Given $m_{1} = 10 k g, m_{2} = 6 k g$ and $m_{3} = 4 k g$ )

20 N

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40 N

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10 N

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32 N

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Solution

The correct option is D $32 N$
Common acceleration for the system is given by
$a_{n e t} = \frac{F_{n e t}}{m_{t o t a l}} = \frac{40}{10 + 6 + 4} = 2 m / s^{2}$

From the FBD of $m_{3}$ , we have

$40 - T_{2} = m_{3} \times a_{n e t}$
$⟹ 40 - T_{2} = 4 \times 2$
$⟹ T_{2} = 40 - 8$
$⟹ T_{2} = 32 N$

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Three blocks of masses m1, m2 and m3 are connected by massless strings as shown in the figure and placed on a frictionless table. They are pulled with a force T3=40 N. The tension T2 will be: (Given m1= 10 kg,m2=6 kg and m3=4 kg)

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