A block of mass 10kg is attached to a relaxed spring. Other end is attached to ceiling and block is released keeping the spring in vertical position. The spring is such that it has elongation of 10 cm under a force of 100 N. The maximum velocity of block in subsequent motion is :-
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Question

A block of mass 10kg is attached to a relaxed spring. Other end is attached to ceiling and block is released keeping the spring in vertical position. The spring is such that it has elongation of 10 cm under a force of 100 N. The maximum velocity of block in subsequent motion is :-

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G1 Physics · Accepted Answer

The spring constant is k=100/0.10=1000 N/m. The equilibrium extension is mg/k=0.10 m; released from natural length, the amplitude about equilibrium is 0.10 m. Therefore v(max)=ωA=√(k/m)(0.10)=10×0.10=1 m/s. Hence A.