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Question

In the diagram shown below the coefficient of friction between 2 kg and 4 kg block is 0.1 and between 4 kg and ground is 0.2. Find the force of friction on 2 kg block at time t=0.5 second.


A
2 N towards left
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B
1 N towards right
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C
1 N towards left
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D
0 N
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Solution

The correct option is B 1 N towards right
The FBD of the blocks are as shown


From the FBD we have
N1=2g=20 N
Maximum available value of friction between 2 kg and 4 kg block is
(f1)max=μN1=0.1×20=2 N

and, N2=N1+4g=20+40=60 N
Similarly, maximum available value of friction between 4 kg and ground is
(f2)max=μN2=0.2×60=12 N

Now, it is given that F=2t towards left
So, at t=0.5 s,F=2×0.5=1 N
F@ t=0.5 s<(f1)max, the whole system is at rest, the tension will not develop in the string and the friction force acting on 2 kg block will be 1 N towards right.

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