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 $\delta$ then

• A. $\frac{F}{K}<\delta \frac{2F}{K}$
• B. $\delta =\frac{2F}{K}$
• C. $\delta =F/K$
• D. Increase in potential energy of the spring is $\frac{1}{2}K{\delta }^2$

Answer 1

Answer: (B)

$\delta =\frac{2F}{K}$

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
$=\frac{1}{2}K{(\frac{mg}{K}+\delta )}^2-\frac{m^2{g}^2}{2K}$
from this equation $\delta =\frac{2F}{K}$