The mechanical energy (kinetic energy + potential energy) of a freely falling object remains constant. It may be shown by calculation as follows:
suppose a body of mass m falls from point A, which is at height ‘Hโ from the surface of earth. Initially at point A, kinetic energy is zero and the body has only potential energy.
Total energy of body at point A
= Kinetic energy + Potential energy
= 0 + ๐๐๐ป = ๐๐๐ปย ย ย ย ย ย โฆ (i)
Suppose during fall, the body is at position B. The body has fallen at a distance ๐ฅ from its initial position. If velocity of body at B is v, then from formula ๐ฃ2 = ๐ข2 + 2๐๐ , we have
๐ฃ2 = 0 + 2g๐ฅ = 2g๐ฅ
โด Kinetic energy of body at point B = {1\over2} mv2
=ย {1\over2} ๐ ร 2g๐ฅ = ๐g๐ฅ
Potential energy of body at point B = ๐๐ (๐ป-๐ฅ)
โด Total energy of body at point B = Kinetic energy + Potential energy
= ๐๐๐ฅ + ๐๐ (๐ป โ ๐ฅ) = ๐๐๐ปย ย ย ย ย ย ย ย โฆ (๐๐)
Now suppose the body is at point C, just above the surface of earth (๐.๐., just about to strike the earth). Its potential energy is zero.
The height by which the body falls = ๐ป
If ๐ฃ is velocity of body at C, then from formula
๐ฃ2 = ๐ข2 + 2๐๐
We have ๐ข = 0, ๐ = ๐, ๐ = ๐ป
So, v2 = 0 + 2๐โ = 2๐โ
โด Kinetic energy of body at position C = {1\over2} mv2
= {1\over2} mร 2๐๐ป = ๐๐๐ป
โด Total energy of body at C
= Kinetic energy + Potential energy
= ๐๐๐ป + 0 = ๐๐๐ปย ย ย ย ย ย ย ย ย ย ย โฆ (๐๐๐)
Thus, we see that the sum of kinetic energy and potential energy of freely falling body at each point remains constant.
Thus, under force of gravity, the total mechanical energy of body remains constant.