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Friction Understanding at Molecular Level

Two blocks A ...

Question

Two blocks A and B are separated by some distance and tied by a string as shown in the figure. Mass of block A is 1 kg and mass of block B is 2 kg. The force of friction in both the blocks at $t = 2 s$ is:

A

4 N (\to), 5 N (\leftarrow)

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B

2 N (\to), 5 N (\leftarrow)

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C

0 N (\to), 10 N (\leftarrow)

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D

6 N (\leftarrow), 6 N (\leftarrow)

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Solution

The correct option is C $6 N (\leftarrow), 6 N (\leftarrow)$
For system total frictional force (assuming motion) is,
$F_{r} = μ_{B} m_{B} g + μ_{A} m_{A} g$
$F_{r} = 0.3 \times 2 \times 10 + 0.6 \times 1 \times 10 = 12 N$
Since applied force is more than the total frictional force, system will move and kinetic friction will apply.

Hence, friction on block B is $F_{B} = μ_{B} m_{B} g = 6 N$ towards left
and friction on block A is $F_{A} = μ_{B} m_{B} g = 6 N$ towards left

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