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A block is suspended by an ideal spring ...

A block is suspended by an ideal spring of force constant force F and if maximum displacement of block from its initial mean position of rest is `x_(0)` then

A

`F/K lt delta lt (2F)/(K)`

B

`delta=(2F)/(K)`

C

`delta=F//K`

D

Increase in potential energy of the spring is `1/2 K delta^(2)`

Text Solution

Verified by Experts

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, then work done by F & further loss of gravitational potential energy of the block is used to increase the potential energy of this spring.
`=1/2K((mg)/(K) + delta)^(2)-(m^(2)g^(2))/(2 K)`
From this equation , `delta=(2F)/(K)`.
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