Question

A body of mass $m$ dropped from a certain height strikes a light vertical fixed spring of stiffness $k$. The height of its fall before touching the spring if the maximum compression of the spring is equal to $\frac{3mg}{k}$ is

• A. $\frac{3mg}{2k}$
• B. $\frac{2mg}{k}$
• C. $\frac{3mg}{4k}$
• D. $\frac{mg}{4k}$

Answer 1

The correct option is A $\frac{3mg}{2k}$

Change in height of the particle is $h+x$
where $x$ is compression in the spring

Applying law of conservation of energy
work done by gravity = work stored in spring in the form of potential energy

$mg(h+x)=\frac{1}{2}k{x}^2$ ( $\because x=\frac{3mg}{k}$ maximum compression)
$(h+x)=\frac{k{x}^2}{2mg}$
$h=\frac{k{x}^2}{2mg}-x=\frac{k(3mg/k{)}^2}{2mg}-\frac{3mg}{k}h=\frac{9mg}{2k}-\frac{3mg}{k}=\frac{3mg}{2k}$