State and prove law of conservation of energy in case of freely falling body.

Law of conservation of energy states that:−

Energy can neither be created nor be destroyed, but it can transform from one form to another form.

Derivation :

A

$\text{m =}$ m = mass of the body

$\text{g =}$ 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 conserved.

At point A,

AC = height of object from ground$\text{= h}$

Initial speed $\text{u = 0}$

So, Potential energy at A, $\text{P.E. = mgh}$

Kinetic energy at A, $K.E.=\frac{1}{2}m{v}^2=0$

Therefore total energy at A is $\text{TE = P.E + K.E = mgh + 0}$

$T.E.=mgh$

At point B,

Potential energy, $\text{PE = mg (BC) = mg (h – x) = mgh – mgx}$

Use kinematics equation

$v^2=u^2+2gx$

$v^{\prime 2}=0+2gx$

$K.E.=\frac{1}{2}m{v}^{\prime 2}=\frac{1}{2}m\left(2gx\right)$

$K.E.=mgx$

Therefore total energy at B is, ${T.E.}_B=K.E+P.E.=mgx+mg(h-x)=mgh$ $

At point C,

Potential energy, $\text{P.E. = 0}$ P.E. = 0

Kinetic energy, $K.E.=\frac{1}{2}m{v}^2$

Use kinematics equation:

$v^2=u^2+2gh$

$v^2=0+2gh=2gh$

So, $K.E=\frac{1}{2}mg\left(2gh\right)=mgh$

Therefore total energy at B is, $T.E{.}_C=K.E.+P.E.=mgh+0=mgh$

From equations (i), (ii) and (iii) we have,

Therefore, $T.E{.}_A=T.E{.}_B=T.E{.}_C$

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