Question

Prove that the kinetic energy of a freely falling object on reaching the ground is nothing but the transformation of its initial potential energy.

Answer 1

Let the object be at point A i.e. at height h above the surface of Earth as shown in the figure below.

At point A

The object is stationary i.e. its initial velocity, u = 0.

Kinetic energy, `K = 1/2mv^2 = 1/2m u^2 = 0`

Potential energy, U = mgh ...... (i)

At point B

Let the velocity of the object be v_B and the object has fallen through distance x. 

Using the third equation of motion, we have

`v_B^2 = 0 + 2gx`

⇒ `v_B^2 = 2gx`

Kinetic energy , `K = 1/2mv_B^2 = mgx`

Potential energy , U = mg(h-x) ....(ii)

At point C

Let the velocity of the object be v_C and the object has fallen through a distance h.

Using the third equation of motion, we have

`v_C^2 = 0 + 2gh`

⇒ `v_B^2 = 2gh`

Kinetic energy, `K = 1/2 mv_C^2 = mgh`  ....(iii)

Potential energy, U = 0

From (i) and (iii), we see that the Potential energy of the object at point A has transformed into its kinetic energy at point C. Thus, it can be concluded that the kinetic energy of a freely falling object reaching the ground is nothing but the transformation of its initial potential energy.