Question

A ball of mass $m$ is dropped from a height $h$ on a platform fixed at the top of a vertical spring. The platform is displaced by a distance $x$. The spring constant is (Take acceleration due to gravity $=g$):

• A. $\frac{2mg}{x}$
• B. $\frac{2mgh}{x^2}$
• C. $\frac{2mg(h+x)}{x^2}$
• D. $\frac{2mg(h+x)}{h^2}$

Answer 1

The correct option is C $\frac{2mg(h+x)}{x^2}$

The particle is dropped from height h and the spring be compressed by x.
Loss in PE of the particle $=$ gain in elastic PE of the spring
$mg(h+x)=\frac{1}{2}k{x}^2\Rightarrow k=\frac{2mg(h+x)}{x^2}$