Question

A stone is dropped from a height $h$ . It's potential energy at any instant is nine times it's kinetic energy. When the velocity of the stone becomes doubles, what will be the new ratio of the new $KE$ and $PE$ at that instant?

Answer 1

Let at a particular distance the height is $x$ and the stone is dropped from the height $h$.

The potential energy is given as,

$PE=9\times KE$

$mg\left(h-x\right)=9\times mgx$

$x=\frac{h}{10}$

The new kinetic energy is given as,

$K{E}_n=4mgx$

$K{E}_n=4mg\times \frac{h}{10}$

$K{E}_n=\frac{2}{5}mgh$

The potential energy is given as,

$P{E}_n=\frac{3}{5}mgh$

The required ratio is given as,

$r=\frac{2}{3}$