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)  2.0 kg        

B)  4.0 kg

C)  0.2 kg       

D)  0.4 kg

Correct Answer: D

Solution :

 Let the mass of the block B is M. In equilibrium of block B, \[\Rightarrow \]\[T=Mg\] ?(i) If block A does not move, then \[T={{f}_{s}}\] where\[{{f}_{s}}=\]frictional force \[={{\mu }_{S}}N={{\mu }_{s}}mg\] \[\Rightarrow \] \[T={{\mu }_{s}}mg\] ?(ii) Thus, from Eqs. (i) and (ii), we have \[M={{\mu }_{s}}\,mg\]or \[M={{\mu }_{s}}m\] Given: \[{{\mu }_{s}}=0.2,\,m=2\,kg\] \[\therefore \] \[M=0.2\times 2=0.4\,kg\]