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