What does law of conservation of energy verify

Dear student,
To verify, conservation of energy, let us consider an object of mass m is initially at a point A (distance of h from ground), and dropped from it, we will calculate the total energy at initial point A, final point C, and an intermediate point B.

m = mass of body

g = Acceleration due to gravity

Suppose a body is at point A initially. Then it falls freely, during the motion mechanical energy of the body is conserve.

At point A,

AC = height of object from ground

       = h

Initial speed u = 0

So, P.E. = mgh

$$\begin{gathered} P.E. = mgh \\ K.E. =\frac{1}{2}m{v}^2=0 \\ Total energy= TE = P.E + K.E = mgh + 0 \end{gathered}$$

(TE)_A = mgh ... (i)
 

At point B,
at a distance x from A,

let velocity be v' at B
Use kinematics equation
$$\begin{gathered} v^2-u^2=2gx \\ \Rightarrow v{\text{'}}^2=2gx \\ K.E.= \frac{1}{2}mv{\text{'}}^2=mgx \end{gathered}$$

$$\begin{gathered} Total energy at B, \\ {\left(TE\right)}_B = P.E. + K.E = mgh – mgx + mgx \end{gathered}$$
(TE)_B = mgh  ...(ii)
 

At point C,

P.E. = 0

let velocity be v at C

Use kinematics equation

$$\begin{gathered} v^2-u^2=2gh \\ \Rightarrow v^2=2gh \\ K.E. =\frac{1}{2}m{v}^2=mgh \end{gathered}$$

$Total energy at C= {(T.E.)}_C = K.E + P.E. = mgh +0$
(TE)_C = mgh ...(iii)
 

(TE)_A =(TE)_B=(TE)_C
i.e. total energy is same at all these three point.

Hence, the total energy of the body is conserve during free fall. And law of conservation of energy verified.

Regards,