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

• A. $\frac{F}{k}<\delta>\frac{2 F}{k}$
• B. $\delta=\frac{2 F}{k}$
• C. Work done by force $F$ is equal to $F \delta$
• D. Increase in energy stored in the spring is $\frac{1}{2} k \delta^{2}$

Answer 1

Answer: (B), (C)

Let the mass of the block hanging from the spring be $m$. Then initial elongation of the spring will be equal to $m g / k$. 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.
$(F \delta+m g \delta)=\left[\frac{1}{2} k\left(\frac{m g}{k}+\delta\right)^{2}-\frac{m^{2} g^{2}}{2 k}\right]$ ...(i)
From Eq. (i), $\delta=2 F / k$
Hence, option (2) is correct.
Since a constant force is applied on the block to pull it down, during this process, work done by force, $W=F \delta$.
Hence, option (3) is also correct.
Increase in energy of the spring is equal to
$\left[\frac{1}{2} k\left(\frac{m g}{k}+\delta\right)^{2}-\frac{m^{2} g^{2}}{2 k}\right]$
From this equation, increase in energy is not equal to $\frac{1}{2} k \delta^{2}$.
Hence, option (4) is wrong.