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, \[{{\mu }_{s}}=0.2,\,m=2\,kg\] \[\therefore \] \[M=0.2\times 2=0.4\,kg\]You need to login to perform this action.
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