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Laws of Motion question

2015 · Q138
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Laws of Motion question

2015 · Q138

NEETPhysicsLaws of MotionMCQ+4 / −1
Three blocks A, B and C, of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is
AIPMT 2015 Cancelled Paper Physics - Laws of Motion Question 24 English
  1. A
    8 N
  2. B
    18 N
  3. C
    2 N
  4. D
    6 N
View written solutionFree

Correct answer: D

AIPMT 2015 Cancelled Paper Physics - Laws of Motion Question 24 English Explanation
Here, MA = 4 kg, MB = 2 kg, MC = 1 kg, F = 14 N

Net mass, M = MA + MB + MC = 4 + 2 + 1 = 7 kg

Let a be the acceleration of the system.
Using Newton’s second law of motion,

F = Ma
14 = 7a    $ \therefore $ a = 2 ms–2

Let F' be the force applied on block A by block B i.e. the contact force between A and B. Free body diagram for block A

Again using Newton’s second law of motion,
F – F' = 4a
14 – F' = 4 × 2 $ \Rightarrow $ 14 – 8 = F' $ \therefore $ F' = 6 N
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