Question

prove that the energy remains constant in case of a freely falling body

Answer 1

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.
In this case we have to show that the total energy (potential energy + kinetic energy) of the body at A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy.
At A, 
Potential energy = mgh
Kinetic energy = 1/2 mv² = 1/2 * m* 0Kinetic energy = 0 [the velocity is zero as the object is initially at rest]
\ Total energy at A = Potential energy + Kinetic energy
= mgh + 0 
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
Kinetic energy = 1/2 mv²
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 = 2aSHere, u = 0, a = g and S = x
v² - 0 = 2gx 
Here, u = 0, a = g and S = x
v² -  0 = 2gx 
v² = 2gx 
kinetic energy = 1/2 mv²
= 1/2 m2gx   
Kinetic energy = mgx
 Total energy at B = Potential energy + Kinetic energy
= mgh - mgx + mgx
Total energy at B = mgh …(2)
At C,
Potential energy = m x g x 0 (h = 0)
Potential energy = 0Kinetic energy = 1/2 mv²
The distance covered by the body is hv2 - u2 = 2aS
Here, u = 0, a = g and S = h
v² - 0 = 2gh 
v² = 2gh 
kinetic energy = 1/2 mv² 
= 1/2 m * 2gh 
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. i know its too big but this is anser make it short if u want!!!

Answer 2

The total energy of an object in "free fall" or "free rise" is; 
E = (1/2)mv^2 + mgy 

The time derivative is; 
dE/dt = mv(dv/dt) + mg(dy/dt) 

Now dv/dt is the acceleration, which is always -g (if up is positive). 
And dy/dt is the velocity which may be positive (rising object) or negative(falling object). But if its negative it has to be negative in both terms. So we can write; 
dE/dt = mv(-g) + mgv 

Thus dE/dt =0 , and E=constant