Question

A particle is released from a height of $h$ . At a certain height, its kinetic energy is two times its potential energy. Find the height and speed of the particle.

Answer 1

Hint: Use the law of conservation of energy. Equate the total energy of the system to the sum of the kinetic energy and the potential energy of the particle at the instant where the kinetic energy is twice the potential energy of the particle. Use this relation and find the height of the particle at that instant. Use the calculated value of height in the equation that relates potential and kinetic energy.

Complete step by step solution:
Let $x$ be the distance from the ground to the particle at the instant when the kinetic energy equals twice the potential energy. The total energy of the particle will be a constant according to the law of conservation of energy. As the potential energy at the maximum height possible $\left( h \right)$ is the highest energy of the particle( kinetic energy is zero at maximum height), it can be taken to be equal to the total energy of the particle.
Therefore, the total energy of the particle $E$ is given by
$E = mgh$
Where $m$ is the mass of the particle,
$g$ is the acceleration due to gravity.
It is given that the kinetic energy of the system is twice the potential energy of the system,
$\therefore KE = 2PE$
The potential energy at this instant,
$PE = mgx$
Therefore, applying the law of conservation of energy at a distance $x$ from the ground, we get
$E = KE + PE$
By substituting the relation between $KE$ and $PE$ , and by substituting the equation for $PE$ , we get
$mgh = 3mgx$
Canceling the common terms on both sides, we get the distance as
$x = \dfrac{h}{3}$
This is the height at which the kinetic energy equals the potential energy.
Now to calculate the kinetic energy, we use the relation between kinetic energy and potential energy and we substitute the calculated value $x$ in the equation for potential,
$KE = \dfrac{{2mgh}}{3}$
By substituting the equation for kinetic energy, we get
$\dfrac{1}{2}m{v^2} = \dfrac{{2mgh}}{3}$
Canceling $m$ from both sides and taking square root on both sides give
$v = 2\sqrt {\dfrac{{gh}}{3}} $

Therefore, the calculated value of the velocity at the instant is $2\sqrt {\dfrac{{gh}}{3}} $.

Note:
Another way to find the height is by dividing the energy into three equal parts. When the particle just starts to fall, it has maximum potential energy, and zero kinetic energy. When the body reaches a distance of $\dfrac{2}{3}h$ , the kinetic energy will be $\dfrac{1}{3}E$ . And when the body reaches a height $\dfrac{1}{3}h$ , the kinetic energy will be $\dfrac{2}{3}E$ . Thus, the height will be $\dfrac{1}{3}h$ so that $KE = 2PE$ .