Question

# In the figure shown masses of the blocks $$A$$, $$B$$, and $$C$$ are $$6 kg$$, $$2 kg$$, and $$1 kg$$ respectively. Mass of the spring is negligibly small and its stiffness is $$1000 N/m$$. The coefficient of friction between the block $$A$$ and the table is $$\displaystyle \mu = 0.8.$$ Initially, block $$C$$ is held such that spring is in a relaxed position. The block is released from rest. Find: $$\displaystyle \left ( g= 10 m/s^{2} \right )$$ the maximum distance moved by the block $$C$$.

A
2×102m
B
3×102m
C
4×102m
D
6×102m

Solution

## The correct option is A $$\displaystyle 2\times 10^{-2}m$$Equilibrium position for block $$C$$ is at a distance $$x = mg/k$$ distance below the initial point. $$x = 0.01m$$. If the block $$A$$ doesn't move then the maximum distance $$C$$ can move down is $$0.02m$$ under SHM. When $$C$$ is moved to maximum distance then the tension in the spring is $$1000$$ x $$0.02N=20N$$. If block $$A$$ didn't move then block $$B$$ also didn't move, thus$$\\ mg+k{ x }_{ 2 }-T=0,$$ where $$T$$ is the tension in the string and $${ x }_{ 2 }$$ is elongation in the spring.$$\\ T=mg+k{ x }_{ 2 }=20+20=40N\\$$ For block $$A$$, maximum static friction is $$\mu { m }_{ A }g=48N,$$ thus  block $$\\ A$$ won't move and maximum displacement of the block $$C$$ is $$20 cm.$$Physics

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