A) \[(M+m)({{\mu }_{1}}+{{\mu }_{2}})g\]
B) \[(M-m)({{\mu }_{1}}+{{\mu }_{2}})g\]
C) \[(M+m)({{\mu }_{1}}-{{\mu }_{2}})g\]
D) \[(M-m)({{\mu }_{1}}-{{\mu }_{2}})g\]
Correct Answer: A
Solution :
Here, the net external force on the system gets distributed on the two masses in direct proportion to the massed. So, the part \[F'\] of the applied force \[F\] transmitted to the upper block is given by, \[F'=\left( \frac{F-{{\mu }_{2}}(M+m)g}{(M+m)} \right)m\] For the two blocks to have common acceleration, this force on \[m\] must be equal to the maximum possible force of friction on the upper block. If this force exceeds the force of friction, the upper block starts sliding on lower block. Force of friction on upper block, \[{{F}_{f}}={{\mu }_{1}}mg\] Equating \[F'\] and \[{{F}_{f'}}\] and solving, we get \[F=(M+m)({{\mu }_{1}}+{{\mu }_{2}})g\]
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