Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

Three blocks A, B, and C of masses $4 \text{ kg}$ , $2 \text{ kg}$ , and $1 \text{ kg}$ respectively, are in contact on a frictionless surface, as shown. If a force of $14 \text{ N}$ is applied to the $4 \text{ kg}$ block, then the contact force between A and B is:

$2 \text{ N}$

$6 \text{ N}$

$8 \text{ N}$

$18 \text{ N}$

Step-by-Step Solution

System Acceleration: Since the blocks are in contact and moving together under a single force on a frictionless surface, they share a common acceleration $a$ . According to Newton's Second Law ( $F = ma$ ) [NCERT Class 11, Physics Part I, Sec 4.5]: $a = \frac{\text{Total Force}}{\text{Total Mass}} = \frac{14}{4 + 2 + 1} = \frac{14}{7} = 2 \text{ m/s}^2$
Contact Force Analysis: To find the contact force between block A and block B ( $F_{AB}$ ), we can analyze the forces acting on the combined system of blocks B and C. The force $F_{AB}$ is the external force responsible for accelerating masses B and C. $F_{AB} = (m_B + m_C) \times a$ $F_{AB} = (2 + 1) \times 2 = 3 \times 2 = 6 \text{ N}$ Alternatively, analyzing block A: The net force on A is $F_{applied} - F_{BA} = m_A a$ . Since $F_{BA} = F_{AB}$ (Newton's Third Law [NCERT Class 11, Physics Part I, Sec 4.6]), $14 - F_{AB} = 4(2) \Rightarrow F_{AB} = 6 \text{ N}$ .

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