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Relative Velocity

Three blocks ...

Question

Three blocks 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 $30 N$
common acceleration $a = \frac{60}{10 + 20 + 30} = 1 m / s^{2}$
so from FBD $T_{1} = m_{1} a = 10 N$
$(T_{2} - T_{1}) = m_{2} a = 20 N$
$(T_{3} - T_{2}) = m_{3} a = 30 N$
from equations substituting value we will get $T_{2} = 30 N$

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