A) \[\frac{3}{5}m\]
B) \[\frac{4}{5}m\]
C) \[\frac{6}{5}m\]
D) \[\frac{3}{2}m\]
Correct Answer: A
Solution :
As long as the block of mass m remains stationary, the block of mass M released from rest comes down by \[\frac{2Mg}{K}\] (before coming it rest momentarily again). Thus the maximum extension in the spring is\[x=\frac{2Mg}{K}\] ?(1) For block of mass m to just move up the incline \[kx=mg\sin \theta +\mu \,mg\cos \theta \] ...(2) \[2Mg=mg\times \frac{3}{5}+\frac{3}{4}mg\times \frac{4}{5}\]or\[M=\frac{3}{5}m\]
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