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 the block from its initial position of rest is `delta`, then

Options

A. (a) `F/kltdeltagt(2F)/(k)`
B. (b) `delta=(2F)/(k)`
C. (c) Work done by force F is equal to `Fdelta`
D. (d) Increase in energy stored in the spring is `1/2kdelta^2`

Answer

B, C

Explanation

Let the 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 strain energy of this spring.
`(Fdelta+mgdelta)=[1/2k((mg)/(k)+delta)^2-(m^2g^2)/(2k)]` (i)
From Eq. (i) `delta=2F//k`
Hence, option (b) is correct.
Since a constant force is applied on the block to pull it down, during this process, work done by force, `W=Fdelta`. Hence, option (c) is also correct.
Increase in energy of the spring is equal to
`[1/2k((mg)/(k)+delta)^2-(m^2g^2)/(2k)]`
From this equation, increase in energy is not equal to `1/2kdelta^2`.
Hence, option (d) is wrong.