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Question

In the figure shown below, coefficient of friction between the 3 kg block and ground varies with time as μ=0.02t and the force acting on the block is given by F=5t. Then, work done by the friction force on the 2 kg block up to 5 sec, from the frame of reference of the 3 kg block is:
[Take g=10 m/s2]


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Solution

From the figure, we can conclude that at t=5 s, external force will be F=25 N.
By assuming that both blocks are moving together, let us calculate the friction required between the two blocks.

Friction between the blocks and the ground at t=5 s will be f1=μ1×N=0.02×5×50=5 N
and force F=5t=5×5=25 N
Hence, acceleration of the system will be a=Ff1M+m=2555=4 m/s2.

For this acceleration, frictional force required on 2 kg block is f2=2×4=8 N and maximum possible friction on this block is f2max=μ2×mg=0.4×20=8 N.
Thus, both blocks will move together till t=5 s and hence work done by friction between the two blocks will be zero in the frame of reference of 3 kg block.
[because there is no relative motion between the blocks]

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