Prove that mechanical energy of a freely falling body is mgh throughout the free fall

Dear Student,
 

Law of Conservation of energy states that energy can neither be created nor destroyed but can be changes from one form to another.

Now in the case of a freely falling body the energy conservation is potential to kinetic energy and their sum is always constant.

We have divided the situation into three different cases

(1) when the body is at its highest point

here the body has a finite potential energy which would be maximum at this point. The kinetic energy of the body will be zero. So, the total energy of the body will be

E = PE + 0 = mgh  (1)

 

(2) when the body is in between the highest point and the ground at a height 's'. Here the body has a finite kinetic energy as well as potential energy. So, the total energy of the body will be

E = PE + KE = mgs + (1/2)mv²

now as

v² - 0 = 2g(h - s)

or

v² = 2g(h-s)

so,

E = mgs + (1/2)m[2g(h-s)]

or

E = mgs + [mgh - mgs]

so,

E = mgh  (2) 

 

(3)when the body just touches the ground. It has zero potential energy and maximum kinetic energy, so the total energy would be

E = 0 + KE = (1/2)mv'²

here

v'² - 0 = 2gh

or

v'² = 2gh

thus,

E = (1/2).m(2gh)

so, we have

E = mgh  (3)

 

thus, from (1), (2) and (3) we infer that the total energy of the body remains constant and thus law of conservation of energy has been proven.