Law of Conservation of Energy: According to this principle

The energy is neither created nor destroyed. The energy can be changed from one form to another form. i.e. when there is not any external influence is applied on the particle then the total energy of the particle is always conserved.

Proof of Conservation's Law of Energy: Let us consider, A particle is freely falling from height $h$ under the gravitational acceleration $g$. So the total energy of the particle at different points :

Calculation of Total Energy at Point $A$:

The initial velocity of the particle at point $A$ is $(v_{A})=0$

The kinetic energy of the particle at point $A$ is $(K_{A})=\frac{1}{2}mv_{A}^{2}$

$K_{A}=0\qquad \left( \because v_{A}=0 \right)$

The potential energy of the particle at point $A$ is $(U_{A})=mgh$

The total energy of a particle at point $A$ is $(E_{A})=K_{A}+U_{A}$

Now substitute the value of $K_{A}$ and $U_{A}$ in the above equation. Now the above equation can be written as:

$E_{A}=0+mgh$

Calculation of Total Energy at Point $B$: If a particle travels the distance $x$ and reaches point $B$ then velocity at point $B$ can be calculated by the equation of motion. i.e.

$v_{B}^{2}=v_{A}^{2}+2gx$

But the initial velocity of the particle is zero. i.e $v_{A}=0$ then the above equation can be written as

$v_{B}^{2}=0+2gx$

$v_{B}^{2}=2gx$

The kinetic energy of the particle at point $B$ is $(K_{B})=\frac{1}{2}mv_{B}^{2}$

Now substitute the value of $v_{B}^{2}$ in the above equation. Now the above equation can be written as:

$K_{B}=\frac{1}{2}m \left( 2g x \right)$

$K_{B}=mgx $

The height of the particle at point $B$ is $(h-x)$ then the potential energy at point $B$ is $(U_{B})=mg(h-x)$

The total energy of a particle at point $B$ is $(E_{B})=K_{B}+U_{B}$

Now substitute the value of $K_{B}$ and $U_{B}$ in the above equation. Now the above equation can be written as:

$E_{B}=mgx+mg(h-x)$

Calculation of Total Energy at Point $C$: Now the particle reaches point $C$ by traveling distance $h$ from the initial point $A$. This point $C$ is just before the collision from the surface. So the velocity of the particle at point $C$.

$v_{C}^{2}= v_{A}^{2}+2gh$

$v_{C}^{2}= 0+2gh \qquad \left( v_{A}=0 \right)$

$v_{C}^{2}= 2gh$

The kinetic energy of the particle at point $C$ is $(K_{C})=\frac{1}{2}mv_{C}^{2}$

Now substitute the value of $v_{C}^{2}$ in the above equation. Now the above equation can be written as:

$K_{C}=\frac{1}{2}(2mgh)$

$K_{C}=mgh$

At point $C$ the particle is just above the surface so the height $h$ of the object will be zero. i.e. $h=0$

The potential energy at point $C$ is $(U_{C})=0$

The total energy of the particle at point $C$ is $(E_{C})=K_{C}+U_{C}$

Now substitute the value of $K_{C}$ and $U_{C}$ in the above equation. Now the above equation can be written as:

$E_{C}=mgh+0$

From the above equation $(1)$, equation $(2)$ and equation $(3)$, we get

Now we can conclude that the total energy of the particle is always conserved when there is no external influence applied. This is proof of the law of conservation of Energy.