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NEET PHYSICSLAWS OF MOTIONEasy

Question

Three blocks A, B, and C of masses 4 kg4 \text{ kg}, 2 kg2 \text{ kg}, and 1 kg1 \text{ kg} respectively, are in contact on a frictionless surface, as shown. If a force of 14 N14 \text{ N} is applied to the 4 kg4 \text{ kg} block, then the contact force between A and B is:

A

2 N2 \text{ N}

B

6 N6 \text{ N}

C

8 N8 \text{ N}

D

18 N18 \text{ N}

Step-by-Step Solution

  1. System Acceleration: Since the blocks are in contact and moving together under a single force on a frictionless surface, they share a common acceleration aa. According to Newton's Second Law (F=maF = ma) [NCERT Class 11, Physics Part I, Sec 4.5]: a=Total ForceTotal Mass=144+2+1=147=2 m/s2a = \frac{\text{Total Force}}{\text{Total Mass}} = \frac{14}{4 + 2 + 1} = \frac{14}{7} = 2 \text{ m/s}^2
  2. Contact Force Analysis: To find the contact force between block A and block B (FABF_{AB}), we can analyze the forces acting on the combined system of blocks B and C. The force FABF_{AB} is the external force responsible for accelerating masses B and C. FAB=(mB+mC)×aF_{AB} = (m_B + m_C) \times a FAB=(2+1)×2=3×2=6 NF_{AB} = (2 + 1) \times 2 = 3 \times 2 = 6 \text{ N} Alternatively, analyzing block A: The net force on A is FappliedFBA=mAaF_{applied} - F_{BA} = m_A a. Since FBA=FABF_{BA} = F_{AB} (Newton's Third Law [NCERT Class 11, Physics Part I, Sec 4.6]), 14FAB=4(2)FAB=6 N14 - F_{AB} = 4(2) \Rightarrow F_{AB} = 6 \text{ N}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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