Question

So that the mechanical energy of a free falling body is conserved

Answer 1

The gravitational potential energy of a freely falling body is converted into its kinetic energy, so that the mechanical energy is conserved.

Mechanical energy is the sum of gravitational potential energy and the kinetic energy. Kinetic energy could be translational , vibrational or rotational.

freely falling body:

   use the equations of motion of kinetics.  Let the height from which an object is dropped be H. Its height at any time be h.   Here u = 0.

    v = u + g t = g t
    v² = u² + 2  g s = 2 g (H - h).

=> 1/2 m v² = m g (H - h) = m g H - m g h
=> m g h + 1/2 m v² = m g H

   The two terms on the LHS represent PE and KE. So their sum is a constant and is conserved in the gravitational field.

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

∴ 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 !

________________________________________