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: \[{{\mu }_{s}}\,=0.2,\,\,m=2kg\] \[\therefore \] \[M=0.2\times 2=0.4\,kg\]You need to login to perform this action.
You will be redirected in
3 sec