Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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:

4 N

6 N

10 N

zero

Step-by-Step Solution

Analyze the System: Since the blocks are in contact and moving on a frictionless surface under a single external force, they move together with a common acceleration ( $a$ ).
Calculate Acceleration: Consider blocks A and B as a single system.

Total mass $M = m_A + m_B = 2 \text{ kg} + 3 \text{ kg} = 5 \text{ kg}$ .
Net external force $F = 10 \text{ N}$ .
According to Newton's Second Law: $a = \frac{F}{M} = \frac{10}{5} = 2 \text{ m/s}^2$ [Source 32]

Calculate Contact Force: Isolate block B. The only horizontal force acting on it is the contact force exerted by block A ( $F_{AB}$ ).

Apply Newton's Second Law to block B: $F_{AB} = m_B \times a$ $F_{AB} = 3 \text{ kg} \times 2 \text{ m/s}^2 = 6 \text{ N}$
Alternatively, considering block A: $F - F_{BA} = m_A a \implies 10 - F_{BA} = 2(2) \implies F_{BA} = 6 \text{ N}$ . By Newton's Third Law, the force A exerts on B is equal in magnitude. [Source 110, 115]

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