Prove the energy remains coserved for a freely falling mass m.
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Prove the energy remains conserved for a freely falling mass m.
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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.
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