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 \[(g=10\text{ }m/{{s}^{2}})\]

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