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. Work done by force F is equal to $F\delta$
• D. Increases in energy stored in spring is $\frac{1}{2}k{\delta }^2$

Answer 1

The correct option is B $\delta =\frac{2F}{K}$

If the mass of the hanging block be 'm' then elongation of spring is $\frac{mg}{k}$.
Due to the applied force the additional stretching is $\delta$
$\therefore F\delta +mg\delta =\frac{1}{2}K{(\frac{mg}{K}+\delta )}^2-\frac{m^2{g}^2}{2K}=\frac{1}{2}K(\frac{m^2{g}^2}{K^2}+{\delta }^2+\frac{2mg\delta }{K})-\frac{m^2{g}^2}{2K}=\frac{1}{2}K{\delta }^2+mg\delta \Rightarrow \delta =\frac{2F}{K}$