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Question

Prove the law of conservation of energy.


Solution


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.=12mv2=12m(0)2=0
T.E. = P.E + K.E = mgh + 0
(T.E.)A=mgh(i)
At point B, P.E. = mg (BC) = mg (h - x) = mgh - mgx
Use kinematic's equation:
v2=u2+2gx
v2=0+2gx
K.E.=12mv2=12m(2gx) =mgx
(T.E.)B=P.E.+K.E.=mghmgx+mgx =mgh(ii)
At point C,
P. E. = 0
K.E.=12mv2
Use kinematic's equation:
v2=u2+2gh
v2=2gh
So, K.E.=12m(2gh)=mgh
(T.E.)C=K.E.+P.E.=mgh+0=mgh(iii)
(T.E.)A=(T.E.)B=(T.E.)C
Hence, the total energy of the body is conserved during free fall.

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