# LAQ : State the law of conservation of mechanical energy. Show that the total mechanical energy of a freely falling body is conserved.

Question

State the law of conservation of mechanical energy. Show that the total mechanical energy of a freely falling body is conserved.

In a conservative force system, the total mechanical energy i.e. the sum of kinetic and potential energy is always constant – this is known as the law of conservation of mechanical energy.

The above law is valid for a freely falling body under gravity.

Let a body of mass m is raised to a vertical height h above the ground and it is kept rest at A. At point A, the energy of the body is entirely potential and it is EA = mgh.

Now, when the body falls vertically downloads, its velocity goes on increasing and when the body is at point B (when it falls a vertical height x from A) let its velocity become VB. From the relation v2 = u2 + 2gh

we get, vB2 = 02 + 2gx  ∴ vB2 = 2gx

At point B, the energy of the body is partly potential and partly kinetic.

∴ Total energy at B is EB = mg(h-x) + 1\over 2mvB2

= mg(h-x) + 1\over 2m2gx

= mgh – mgx + mgx

= mgh

Again, when the body reaches point C (just before hitting the ground) it possesses kinetic energy only and if the velocity of the body at that point C be then v then vc2 = 2gh

∴ Total Energy at C, Ec = 1\over 2 mv2

= 1\over 2m2gh

= mgh

∴ EA = EB = EC = mgh

i.e. total mechanical energy at all points during the free fall of a body under gravity remains constant. Hence the principle of conservation of energy is valid for a freely falling body.

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