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Laws of Motion
Question
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Three blocks A, B, and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a 14 N is applied to the 4 kg block, then the contact force between A and B is

  • 2 N

  • 6 N

  • 8 N 

  • 18 N 

Answer

B.

6 N

Given, mA = 4 kg
mB = 2 kg 
=> mC =1 kg

So total mass (M)  = 4+2+1 = 7 kg
Now, F = Ma 
14 = 7a
a=2 m/s2

F-F' = 4a
F' = 14-4x2
F' = 6N

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