State the principle of conservation of energy and prove it for free falling body
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let a body of mass m on the ground which has no energy ,
as h = 0 so no mgh( potential energy) and
v = 0 so no 1/2 kv square (kinetic energy)
take it to a high building, now body has potential energy as it has height but no kinetic energy bcoz velocity is still zero, it is at rest
now when we throw body from building, its height is decreasing so potential energy ( mgh) is decreasing but it has velocity so it has now kinetic energy..
so one energy changed its form from potential energy to kinetic energy
as h = 0 so no mgh( potential energy) and
v = 0 so no 1/2 kv square (kinetic energy)
take it to a high building, now body has potential energy as it has height but no kinetic energy bcoz velocity is still zero, it is at rest
now when we throw body from building, its height is decreasing so potential energy ( mgh) is decreasing but it has velocity so it has now kinetic energy..
so one energy changed its form from potential energy to kinetic energy
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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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