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