Physics, asked by Pradhyumna1234, 1 year ago

State and prove law of conservation of energy In case of a free fall

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Answered by navikaur
5

HERE IS IT ...HOPE IT WILL HELP U..


(i) It states that for a body falling freely the total mechanical energy remains conserved.
(ii) Suppose a ball of mass ‘m’ falls under the effect of gravity as shown in figure.
Let us find the kinetic and the potential energy of the ball at various points of its free fall. Let the ball fall from point A at a height h above the surface of the earth.



At Point A: At point A, the ball is stationary; therefore, its velocity is zero.
Therefore, kinetic energy, T = 0 and potential energy, U = mgh
Hence, total mechanical energy at point A is
E = T + U = 0 + mgh = mgh … (i)
At Point B : Suppose the ball covers a distance x when it moves from A to B. Let v be the velocity of the ball point B. Then by the equation of motion v^2-u^2 = 2aS, we have
v^2 - 0 = 2gx or v^2 = 2gx Therefore,
Kinetic energy, T = 1/2 mv^2 = 1/2 x m x (2gx)
= mgx
And Potential energy, U = mg (h - x)
Hence, total energy at point B is
E = T + U = mgx + mg(h-x) = mgh …(ii)
At Point C : Suppose the ball covers a distance h when it moves from A to C. Let V be the velocity of the ball at point C just before it touches the ground. Then by the equation of motion v^2 - u^2 = 2aS, we have V^2 - 0 = 2gh or V^2 = 2gh.
Therefore,
Kinetic energy,
T = 1/2 mV^2 = 1/2 x m x (2gh) = mgh
and Potential energy, U = 0
Hence, total energy at point E = T + U
= mgh + 0 = mgh … (iii)
Thus, it is clear from equations (i), (ii) and (iii), that the total mechanical energy of a freely falling ball remains constant.


Pradhyumna1234: Slipped
Answered by Anonymous
4

Important Formulas :


Potential energy U = mgh

Kinetic energy K = 1/2 mv²

Total energy = K + U


At A

U = mgh


K = 0 since v = 0

Total energy = K + U

                     = mgh + 0

                     = mgh


At B


U = m g h

h = h - x


U = mg ( h - x )

K = 1/2 m v²


By laws of motion:

v² = 2 gh

= > v²= 2 gx


K = mgx


K + U = mg( h - x ) + mgx

         = mgh - mgx + mgx

         = mgh


At C


K = 1/2 m v²

  = 1/2 m × 2 gh [ v² = 2 gh ]

  = mgh



U = 0 since h = 0


Hence K + U = mgh + 0

                     = mgh



Observations :


In all cases ,

K + U = mgh


There fore the total mechanical energy remains constant.

This verifies the Law of Conservation of Energy and also proves that the energy of a body during free fall is conserved !


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