Question

Two blocks of equal masses $m_1=m_2=3kg$, 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 $0.5m{s}^{-2}$. Assuming that the frictional forces on the two blocks are equal, the tension in the string will be

• A. 10 N
• B. 20 N
• C. 40 N
• D. 60 N

Answer 1

The correct option is A

10 N

From the second law of motion, $T-f=m_{1}a........\left(1\right)$
$F-T-f=m_{2}a........\left(2\right)$
Subtracting the two equations, we have
$2T-F=(m_1-m_2)a$
Since $m_1=m_2$, we get $2T-F=0\Rightarrow T=\frac{F}{2}=\frac{20}{2}=10N$
Hence, the correct choice is (a).