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 massless.\[(g=10\,m/{{s}^{2}})\]

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