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Question

Three-blocks A,B and C of mass 4 kg, 2 kg and 1 kg respectively are in contact of 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 will be


A
6 N
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B
8 N
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C
18 N
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D
2 N
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Solution

The correct option is A 6 N

Writing force equations (assuming common acceleration to be a),
14N=4a
NN1=2a
N1=1×a

Solving, we get N=6 N

Alternate solution:
Since all the three blocks are going to move with common acceleration, we can consider them as one system.
a=Net external forceTotal mass=147=2m/s2

Now if you consider blocks B and C together as a system, the net external force acting on it is contact force N.


Therefore N=a(2+1)=6 N

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