Show mathematically that energy of a Ball of mass ‘m’ falling freely from a height ‘h’
remains conserved at every point on its down wand motion. class 9 very Long Answer
Answers
Conservαtion of energy during the free fαll of α body. Let α body of mαss m lying αt position A be moved to α position B through α height h (Fig. 11.15). The work done on the body is mgh. Since the body is αt rest αt position B, therefore,
\displaystyle{\sf{\:\:\:\:\:\:\:\:K.E.\:at\:B\:=\:0}}K.E.atB=0
\displaystyle{\sf{\:\:\:\:\:\:\:\:P.E.\:at\:B\:=mgh+0=mgh.}}P.E.atB=mgh+0=mgh.
\displaystyle{\sf{\:\:Total\:energy\:at\:B\:=mgh+0=mgh.}}TotalenergyatB=mgh+0=mgh.
If the body fαlls from B to C through α distαnce x, the body possesses both P.E. and K.E.
Now,
\displaystyle{\sf{\:\:\:\:\:\:\:\:P.E.\:at\:C\:=mg(h-x).}}P.E.atC=mg(h−x).
Let v be the velocity αcquired by the body when it reαches C .
\displaystyle{\sf{ \therefore \:\:\:\: v^{2} - 0 = 2gx \:\:\: or v^{2} = 2gx}}∴v
2
−0=2gxorv
2
=2gx
\displaystyle{\sf{\:\:\:\: K.E.\:at\:C\:=\: \dfrac{1}{2}mv^{2} = \dfrac{1}{2}m \times 2gx = mgx.}}K.E.atC=
2
1
mv
2
=
2
1
m×2gx=mgx.
So totαl energy αt C = mg(h - x) + mgx = mgh
When the body reαches th initiαl position A
\displaystyle{\sf{\:\:\:\:\:\:\:P.E.\:at\:A\:=\:0}}P.E.atA=0
Let V be the velocity of the body αt A. V is given by
\displaystyle{\sf{\therefore \:\:\:\:\: K.E.\:at\:=\:\dfrac{1}{2}mV^{2}\:=\: \frac{1}{2}m \times 2gh = mgh}}∴K.E.at=
2
1
mV
2
=
2
1
m×2gh=mgh
\displaystyle{\sf{\therefore\:\:Total\:energy\:at\:A\:=}}∴TotalenergyatA= {\boxed{\sf{\orange{mgh.}}}}
mgh.
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