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Question

A block is suspended by an ideal spring of force constant K. If the block is pulled down by applying a constant force F and if maximum displacement of block from its initial position of rest is δ then

A
FK<δ2FK
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B
δ=2FK
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C
δ=F/K
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D
Increase in potential energy of the spring is 12Kδ2
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Solution

The correct option is D δ=2FK
Let mass of the block hanging from the spring be m. Then initial elongation of the spring will be equal to mg/K. When the force F is applied to pull the block down, the work done by F and further loss of gravitational potential energy of the block is used to increase the potential energy of this spring
Hence
=12K(mgK+δ)2m2g22K
from this equation δ=2FK

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