Two blocks A and B of masses \[{{m}_{A}}=1kg\]and \[{{m}_{B}}=3kg\]are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is : (Take \[g=10\text{ }m/{{s}^{2}}\]) |
[JEE Main 10-4-2019 Afternoon] |
A) 16 N
B) 40 N
C) 12 N
D) 8 N
Correct Answer: A
Solution :
[a] |
\[{{a}_{A\max }}=\mu g=2m/{{s}^{2}}\] |
\[F-8=4\times 2\] |
\[F=16N\] |
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