The coefficient of static friction, \[{{\mu }_{s}}\] between block A of mass 2 kg and the table as shown in the figure, is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and mass less:\[(g=10\text{ }m/{{s}^{2}})\] [AIPMT (S) 2004] |
A) 2.0 kg
B) 4.0 kg
C) 0.2 kg
D) 0.4 kg
Correct Answer: D
Solution :
Key Idea: The tension in the string is equal to static frictional force between block A and the surface. Let the mass of the block B is M. |
In equilibrium, |
\[T-Mg=0\] |
\[\Rightarrow \] \[T=mg\] ...(i) |
If blocks do not move, then |
\[T={{f}_{s}}\] |
where \[{{f}_{s}}=\] frictional force \[={{\mu }_{s}}R={{\mu }_{s}}\,\,mg\] |
\[\therefore \] \[T={{\mu }_{s\,\,\,}}mg\] ...(ii) |
Thus, from Eqs. (i) and (ii), we have |
\[mg={{\mu }_{s}}\,\,\,mg\] |
or \[M={{\mu }_{s}}\,\,m\] |
Given: \[{{\mu }_{s}}\,=0.2,\,\,m=2kg\] |
\[\therefore \] \[M=0.2\times 2=0.4\,kg\] |
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