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| 1. |
Derivation of law of conservation of energy |
| Answer» Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another.Let us now prove that the above law holds good in the case of a freely falling body.Let a body of mass \'m\' placed at a height \'h\' above the ground, start falling down from rest.In this case we have to show that the total energy (potential energy + kinetic energy) of the body A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy.Body of mass m placed at a height hAt A,Potential energy = mghKinetic energy = 0 [the velocity is zero as the object is initially atrest]\xa0Total energy at A = Potential energy + Kinetic energy.Total energy at A = mgh …1At B,Potential energy = mgh= mg(h - x) [Height from the ground is (h-x)]Potential energy = mgh - mgxKinetic energy = ½ mv2The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body.Here, u=0, a=g and s=xKinetic energy = mgx\xa0Total energy at B = Potential energy + Kinetic energyTotal energy at B = mgh …2At C,Potential energy = m x g x 0Potential energy = 0Kinetic energy = ½ mv2The freely falling body has covered the distance h.Here, u=0, a=g and s=hKinetic energy = ½ mv2Kinetic energy = mgh\xa0Total energy at C = Potential energy + Kinetic energyTotal energy at C = mgh …3It is clear from equations 1, 2 and 3 that the total energy of the body remains constant at every point. Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body. | |