Two blocks of equal masses $m_{1} = m_{2} = 3 k g$ , connected by a light string, are placed on a horizontal surface which is not frictionless. If a force of 20 N is applied in the horizontal direction on a block, the acceleration of each block is $0.5 m s^{- 2}$ . Assuming that the frictional forces on the two blocks are equal, the tension in the string will be

10 N

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20 N

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40 N

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60 N

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Solution

The correct option is A
10 N

From the second law of motion, $T - f = m_{1} a . . . . . . . . (1)$
$F - T - f = m_{2} a . . . . . . . . (2)$
Subtracting the two equations, we have
$2 T - F = (m_{1} - m_{2}) a$
Since $m_{1} = m_{2}$ , we get $2 T - F = 0 \Rightarrow T = \frac{F}{2} = \frac{20}{2} = 10 N$
Hence, the correct choice is (a).

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The correct option is A 10 N From the second law of motion, T−f=m1a........(1) F−T−f=m2a........(2) Subtracting the two equations, we have 2T−F=(m1−m2)a Since m1=m2, we get 2T−F=0⇒T=F2=202=10 N Hence, the correct choice is (a).