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,
${\left( {K.E.} \right)_A} = \dfrac{1}{2}m{v^2}$
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, ${\left( {K.E} \right)_B} = \dfrac{1}{2}m{v^2}$
${\left( {K.E.} \right)_B} = \dfrac{1}{2}m \cdot 2g\left( {H - h} \right)$
$\Rightarrow {\left( {K.E.} \right)_B} = mg\left( {H - h} \right)$
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
${\left( {K.E.} \right)_O} = \dfrac{1}{2}m \cdot 2gH$
$\Rightarrow {\left( {K.E.} \right)_O} = mgH$
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.