Question

Prove mathematically that the total mechanical energy of a freely falling body in air is conserved.

Answer 1

(i) It states that for a body falling freely the total mechanical energy remains conserved.
(ii) Suppose a ball of mass 'm' falls under the effect of gravity as shown in figure.
Let us find the kinetic and the potential energy of the ball at various points of its free fall. Let the ball fall from point A at a height h above the surface of the earth.

At Point A: At point A, the ball is stationary; therefore, its velocity is zero.
Therefore, kinetic energy, T = 0 and potential energy, U = mgh
Hence, total mechanical energy at point A is 
E = T + U = 0 + mgh = mgh ... (i)
At Point B : Suppose the ball covers a distance x when it moves from A to B. Let v be the velocity of the ball point B. Then by the equation of motion v^2-u^2 = 2aS, we have
v^2 - 0 = 2gx or v^2 = 2gx Therefore,
Kinetic energy, T = 1/2 mv^2 = 1/2 x m x (2gx)
= mgx
And Potential energy, U = mg (h - x)
Hence, total energy at point B is 
E = T + U = mgx + mg(h-x) = mgh ...(ii) 
At Point C : Suppose the ball covers a distance h when it moves from A to C. Let V be the velocity of the ball at point C just before it touches the ground. Then by the equation of motion v^2 - u^2 = 2aS, we have V^2 - 0 = 2gh or V^2 = 2gh.
Therefore,
Kinetic energy,
T = 1/2 mV^2 = 1/2 x m x (2gh) = mgh
and Potential energy, U = 0
Hence, total energy at point E = T + U
= mgh + 0 = mgh ... (iii)
Thus, it is clear from equations (i), (ii) and (iii), that the total mechanical energy of a freely falling ball remains constant.

Answer 2

Answer:

The total mechanical energy of freely falling body is conserved.

Explanation:

Let the initial position of the particle is A. It is made to free fall from it.So, initial velocity of particle is zero. So, the potential energy of a particle is

mgh

where m is the mass of particle, g is the acceleration due to gravity and h is height of the particle from ground.

At point A kinetic energy is zero. So, total energy at point A is kinetic energy + potential energy = mgh

At ground kinetic energy is $1/2mv^{2}$.

By third equation of motion

$v^2 - u^2 = 2 gh$

as u is zero. So,

$v^{2} = 2gh$

$1/2v^{2} = gh$

Multipying by m,

$1/2mv^{2} = mgh$ (1)

Total energy at ground = $1/2mv^{2}$. But from (1) it is equal to mgh that is energy at point A.

So, total energy is conserved.