Question

Show mathematically that energy of a ball of mass m falling freely from a height 'h' remains conserved at every point on its downward motion.​

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Question =Show mathematically that energy of a ball of mass m falling freely from a height 'h' remains conserved at every point on its downward motion.

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Apurbo Satapathy

Mar 21, 2015

Show that total energy is conserved if the ball of mass m is dropped from a height h

Show that total energy is conserved if the ball of mass ‘m’ is dropped from a height ‘h’

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Amarnathreddy M

Law of Conservation of Energy The law of conservation of energy is the fundamental law, law of conservation of energy says that the energy can neither be created nor destroyed, the sum total energy existing in all forms in the universe remains constant. Energy can only be transformed from one form to another.

Principle of Conservation of Mechanical Energy, which states that the energy can neither be created nor destroyed; it can only be transformed from one state to another. Orthe total mechanical energy of a system is conserved if the forces doing the work on it are conservative.Consider any two points A and B in the path of a body falling freely from a certain height H as in the above figure.Total mechanical energy at A:M.EA = mgH + ½ mvA2 , here vA = 0 ⇒M.EA = mgHTotal mechanical energy at B:M.EB = mg(H - h) +½ mvB2 ⇒M.EB = mg(H - h) + ½ m ( u2 +2gh), where u = 0 ⇒M.EB = mgH - mgh + ½ m ( 02 +2gh) ⇒M.EB = mgH - mgh + ½ m ×2gh ⇒M.EB = mgH - mgh + mgh ⇒M.EB = mgHTotal mechanical energy at C:As the body reaches the ground its height from the ground becomes zero.M.EC = mgH + ½ mvC2 , here H = 0⇒M.EC = 0 + ½ mvC2, but vC2 = 2gH ⇒M.EC = ½ m× 2gH ⇒M.EC = mgHHence the total mechanical energy at any point in its path is Constant i.e., M.EA = M.EB = M.EC = mgHAccording to the Principle of Conservation of Mechanical Energy, we can say that for points A and B, the total mechanical energy is constant in the path travelled by a body under the action of a conservative force, i.e., the total mechanical energy at A is equal to the total mechanical energy at B.Note:Work done by a conservative force is path independent. It is equal to the difference between the potential energies of the initial and final positions and is completely recoverable.”

As the work done by a conservative force depends on the initial and final position, we can say that work done by a conservative force in a closed path is zero as the initial and final positions in a closed path are the same.

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Answer 2

$\huge{\underline{\underline{\sf{\purple{Required\:Answer:}}}}}$

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,

$\displaystyle{\sf{\:\:\:\:\:\:\:\:K.E.\:at\:B\:=\:0}}$

$\displaystyle{\sf{\:\:\:\:\:\:\:\:P.E.\:at\:B\:=mgh+0=mgh.}}$

$\displaystyle{\sf{\:\:Total\:energy\:at\:B\:=mgh+0=mgh.}}$

If the body fαlls from B to C through α distαnce x, the body possesses both P.E. and K.E.

Now,

$\displaystyle{\sf{\:\:\:\:\:\:\:\:P.E.\:at\:C\:=mg(h-x).}}$

Let v be the velocity αcquired by the body when it reαches C .

$\displaystyle{\sf{ \therefore \:\:\:\: v^{2} - 0 = 2gx \:\:\: or v^{2} = 2gx}}$

$\displaystyle{\sf{\:\:\:\: K.E.\:at\:C\:=\: \dfrac{1}{2}mv^{2} = \dfrac{1}{2}m \times 2gx = mgx.}}$

So totαl energy αt C = mg(h - x) + mgx = mgh

When the body reαches th initiαl position A

$\displaystyle{\sf{\:\:\:\:\:\:\:P.E.\:at\:A\:=\:0}}$

Let V be the velocity of the body αt A. V is given by

$\displaystyle{\sf{\therefore \:\:\:\:\: K.E.\:at\:=\:\dfrac{1}{2}mV^{2}\:=\: \frac{1}{2}m \times 2gh = mgh}}$

$\displaystyle{\sf{\therefore\:\:Total\:energy\:at\:A\:=}}$${\boxed{\sf{\orange{mgh.}}}}$

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