A horizontal force 10 N is applied to a block A as shown in figure. The mass of blocks A and B are 2 kg and 3 kg respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is: (a) Zero (b) 4 N (c) 6 N (d) 10 N

Answer: (c). a = (F)/(mA + mB) = (10)/(2 + 3) = 2 m/s² Force of A on B gives B its acceleration: N = mB a = 3 × 2 = 6 N. So the answer is (c) 6 N.

A horizontal force 10 N is applied to a block A as shown in figure… | NEET 2024

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Connected Bodies & Pulleys

A horizontal force $10\text{ N}$ is applied to a block $A$ as shown in figure. The mass of blocks $A$ and $B$ are $2\text{ kg}$ and 3 $\text{kg}$ respectively. The blocks slide over a frictionless surface. The force exerted by block $A$ on block $B$ is:

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Answer: (c)

$a = \dfrac{F}{m_A + m_B} = \dfrac{10}{2 + 3} = 2\text{ m/s}^2$
Force of A on B gives B its acceleration: $N = m_B a = 3 \times 2 = 6\text{ N}$ .
So the answer is (c) 6 N.

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