Question

energy remains constant in the case of freely falling bodies. proven this statement.

Answer 1

Potential energy = mgh



Kinetic energy = 0 [the velocity is zero as the object is initially at rest]

 Total energy at A = Potential energy + Kinetic energy

Total energy at A = mgh  …(1)

At B,

Potential energy = mgh

= mg(h - x)  [Height from the ground is (h - x)]

Potential energy = mgh - mgx

The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body.

v2 - u2 = 2aS

Here, u = 0, a = g and S = h

Kinetic energy = mgx

Total energy at B = Potential energy + Kinetic energy

Total energy at B = mgh …(2)

At C,


Potential energy = m x g x 0 (h = 0)

Potential energy = 0

The distance covered by the body is h

v2 - u2 = 2aS

Here, u = 0, a = g and S =h

Kinetic energy 

Kinetic energy = mgh
Total energy at C = Potential energy + Kinetic energy

= 0 + mgh

Total energy at C = mgh …(3)

It is clear from equations 1, 2 and 3 that the total energy of the body remains constant at every point. Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body.

Answer 2

• Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another. Let us now prove that the above law holds good in the case of a freely falling body. Let a body of mass 'm' placed at a height 'h' above the ground, start falling down from rest.