question_answer

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: \[\left( \text{g}=\text{1}0\text{ m}/{{\text{s}}^{\text{2}}} \right)\]                 

A) 2.0kg            

B) 4.0kg            

C) 0.2kg    

D) 0.4kg

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: \[{{\mu }_{s}}\,=0.2,\,\,m=2kg\] \[\therefore \]          \[M=0.2\times 2=0.4\,kg\]