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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.
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Solution
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 vB 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 vC 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.
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