Show that when a body is dropped from a certain height, the sum of its kinetic energy at any instant during its fall is constant.

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 ‘H’ 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 + 𝑚𝑔𝐻 = 𝑚𝑔𝐻            … (i)

Suppose during fall, the body is at position B. The body has fallen at a distance 𝑥 from its initial position. If velocity of body at B is v, then from formula 𝑣² = 𝑢² + 2𝑎𝑠, we have

𝑣² = 0 + 2g𝑥 = 2g𝑥

∴ Kinetic energy of body at point B = {1\over2} mv²

=  {1\over2} 𝑚 × 2g𝑥 = 𝑚g𝑥

Potential energy of body at point B = 𝑚𝑔 (𝐻-𝑥)

∴ Total energy of body at point B = Kinetic energy + Potential energy

= 𝑚𝑔𝑥 + 𝑚𝑔 (𝐻 – 𝑥) = 𝑚𝑔𝐻               … (𝑖𝑖)

Now suppose the body is at point C, just above the surface of earth (𝑖.𝑒., just about to strike the earth). Its potential energy is zero.

The height by which the body falls = 𝐻

If 𝑣 is velocity of body at C, then from formula

𝑣² = 𝑢² + 2𝑎𝑠

We have 𝑢 = 0, 𝑎 = 𝑔, 𝑠 = 𝐻

So, v² = 0 + 2𝑔ℎ = 2𝑔ℎ

∴ Kinetic energy of body at position C = {1\over2} mv²

= {1\over2} m× 2𝑔𝐻 = 𝑚𝑔𝐻

∴ Total energy of body at C

= Kinetic energy + Potential energy

= 𝑚𝑔𝐻 + 0 = 𝑚𝑔𝐻                      … (𝑖𝑖𝑖)

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.