Question

Show that total energy of a freely falling body remains constant during its motion.

Answer 1

Hint: When a body falls freely, its kinetic energy increases while its potential energy decreases. But the total mechanical energy is conserved during the free fall.The total mechanical energy of a body at a point during free fall is equal to the sum of kinetic energy and potential energy of the body at that point.

Complete step by step answer:
Consider a body of mass $m$ is in free fall motion as shown below.

Let the initial height of the body from the ground is $H$. The total energy of the body at the point A. The kinetic energy at the height $H$ is,
${\left(K.E.\right)}_A={\displaystyle \frac{1}{2}}m{v}^2$
Here, $v$ is the initial velocity of the body i.e., $v=0$
Therefore, ${\left(K.E.\right)}_A=0$
The potential energy of the body at A is ${\left(P.E.\right)}_A=mgH$
We know that the total energy of the body is $E=K.E+P.E$
$E=0+mgH$
$\Rightarrow E=mgH$
The total energy of the body at the point B.

Apply the kinematic equation when the body at the point B, $v^2-u^2=2aS$
The initial velocity of the body is $u=0$,
The acceleration of the body $a=g$
The height of the body at B from the ground is $h$.
So, the displacement $S=H-h$
Therefore, $v^2=2g\left(H-h\right)$
The kinetic energy at B, ${\left(K.E\right)}_B={\displaystyle \frac{1}{2}}m{v}^2$
${\left(K.E.\right)}_B={\displaystyle \frac{1}{2}}m\cdot 2g\left(H-h\right)$
$\Rightarrow {\left(K.E.\right)}_B=mg\left(H-h\right)$
The potential energy at B, ${\left(P.E.\right)}_B=mgh$
The total energy of the body at point B, $E={\left(K.E.\right)}_B+{\left(P.E.\right)}_B$
Substitute all the required values in the above formula, we got,
$E=mg\left(H-h\right)+mgh$
$\Rightarrow E=mgH$
The total energy of the body at the ground.

Similarly, apply the kinematic equation when the body is at the ground point O just before the rest.
$v^2-u^2=2gH$
$\Rightarrow v=2gH$
The kinetic energy of the body at the ground point O is maximum and is given by
${\left(K.E.\right)}_O={\displaystyle \frac{1}{2}}m\cdot 2gH$
$\Rightarrow {\left(K.E.\right)}_O=mgH$
The potential energy of the body at the ground is ${\left(P.E.\right)}_O=0$
The total energy of the body at the ground is,
$E={\left(K.E.\right)}_O+{\left(P.E.\right)}_O$
$\Rightarrow E=mgH+0$
$\therefore E=mgH$
From above all the three cases, it is found that $E=mgH$.

Hence, the total energy of a body during free fall is constant.

Note: A body is said to be in free fall motion, when gravity is the only force acting on the body. Therefore, no object has truly free-fall motion due to air friction.Kinetic energy is the energy possessed by a body by virtue of its motion. Potential energy is the energy acquired by a body due to its configuration or change in its position.