Question

What is the derivation for conservation of energy for free falling object

Answer 1

½mv² = mgh
→½v²=gh
→v²=2gh
→v=√2gg
hope it helped

Answer 2

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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