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Question

Two blocks of masses 2 kg and 4 kg are connected by a light string and kept on horizontal surface. A force of 16 N is acted on 4kg block horizontally as shown in figure. Besides it is given that coefficient of friction between 4~kg and ground is 0.3 and between 2 kg block and ground is 0.6. Then frictional force between 2 kg block and ground is

A
12 N
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B
4 N
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C
6 N
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D
zero
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Solution

The correct option is B 4 N
Let's first find the overall acceleration of system (2 kg and 4 kg blocks)
Limiting friction can act between the surfaces of blocks are
1. (Fs)max between mass 2 kg block and ground is 0.6×2×10=12 N.
2. Similarly, (Fs)max between mass 4 kg block and ground is 0.3×4×10=12 N.
Total limiting friction which can be act between blocks and ground is 12 N+12 N=24 N which is greater than the force applied 16 N. Hence the blocks will not move. So the acceleration of system is zero and it is in equilibrium.
Let's consider the FBD of 4 kg block and 2 kg block,


From FBD of 4 kg we can calculate the tension in string which is equal to T=1612=4 N
Hence from the FBD of 2 kg block,
Fs=4 N



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