Question

Prove the law of conservation of energy.

Answer 1

m = mass of the body
g = acceleration due to gravity
Suppose a body is at point A initially. Then it falls freely, during the motion, the mechanical energy of the body is conserved.
At point A,
AC = height of object from ground = h
Initial speed, u = 0
So, P.E. = mgh
$K.E.=\frac{1}{2}m{v}^2=\frac{1}{2}m(0{)}^2=0$
T.E. = P.E + K.E = mgh + 0
$(T.E.{)}_A=mgh\dots \dots (i)$
At point B, P.E. = mg (BC) = mg (h - x) = mgh - mgx
Use kinematic's equation:
$v^2=u^2+2gx$
$\Rightarrow v^{{}^{\prime }2}=0+2gx$
$K.E.=\frac{1}{2}m{v}^{{}^{\prime }2}=\frac{1}{2}m\left(2gx\right)$ $=mgx$
$(T.E.{)}_B=P.E.+K.E.=mgh-mgx+mgx$ $=mgh\dots \dots \left(ii\right)$
At point C,
P. E. = 0
$K.E.=\frac{1}{2}m{v}^2$
Use kinematic's equation:
$v^2=u^2+2gh$
$\Rightarrow v^{{}^{\prime }2}=2gh$
So, $K.E.=\frac{1}{2}m\left(2gh\right)=mgh$
$(T.E.{)}_C=K.E.+P.E.=mgh+0=mgh\dots \dots (iii)$
$(T.E.{)}_A=(T.E.{)}_B=(T.E.{)}_C$
Hence, the total energy of the body is conserved during free fall.