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Second Law of Motion

Three block a...

Question

Three block are connected as shown in figure, on a horizontal frictionless table and pulled to the right with a force $T_{3} = 60 N$ . If $m_{1} = 10 k g, m_{2} = 20 k g$ and $m_{3} = 30 k g$ , the tension $T_{2}$ is -

A

10 N

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B

20 N

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C

30 N

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D

60 N

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Solution

The correct option is C $30 N$
Net force on system (10+20+30)= $60 K g$ is $T_{3} = 60 N$ so common acceleration is $a = \frac{60 N}{60 K g} = 1 m s^{2}$
Now treating $m_{1} + m_{2}$ we get net force on it as $T_{2} = m a s s \times a c c e l e r a t i o n = 30 K g \times 1 m s^{2} = 30 N$
Option C is correct.

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