 |
| Law of Conservation of Energy |
|
| Let us see what is actually taking place in the following examples: |
| |
 Steam engine: The coal burns. Heat due to the combustion of coal converts water into steam. The expansive force exerted by the steam on the piston of the engine moves the locomotive. |
| |
| Chemical energy is converted to heat energy and heat energy is converted to the expansive force of steam. This transforms to kinetic energy when the locomotive moves. |
| |
 Hydroelectric power plant: Water stored in a reservoir is made to fall on turbines which are kept at a lower level and which in turn are connected to coils of an a.c. generator. |
| |
| Potential energy of the water in the reservoir changes to kinetic energy and kinetic energy of the falling water to kinetic energy of the turbines, which in turn changes to electrical energy. |
| |
| Thus, it is clear that whenever energy in one form disappears, an equivalent amount of energy in another form reappears, so that the total energy remains constant. |
| |
|
| 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 at A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy. |
| |
 |
| Body of Mass 'm' Placed at a Height 'h' |
| |
| At A, |
| |
| Potential energy = mgh |
| |
Kinetic energy |
| |
| Kinetic energy = 0 [the velocity is zero as the object is initially at rest] |
| |
| |
| \ Total energy at A = Potential energy + Kinetic energy |
| |
 |
| |
| Total energy at A = mgh …(1) |
| |
| At B, |
| |
| Potential energy = mgh |
| |
| = mg(h - x) [Height from the ground is (h - x)] |
| |
| Potential energy = mgh - mgx |
| |
Kinetic energy  |
| |
| The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body. |
| |
| v2 - u2 = 2aS |
| |
| Here, u = 0, a = g and S = x |
| |
 |
| |
 |
| |
| |
 |
| |
 |
| |
| Kinetic energy = mgx |
| |
 Total energy at B = Potential energy + Kinetic energy |
| |
 |
| |
| Total energy at B = mgh …(2) |
| |
| At C, |
| |
| Potential energy = m x g x 0 (h = 0) |
| |
| Potential energy = 0 |
| |
Kinetic energy  |
| |
| The distance covered by the body is h |
| |
| v2 - u2 = 2aS |
| |
| Here, u = 0, a = g and S = h |
| |
 |
| |
 |
| |
Kinetic energy  |
| |
 |
| |
| Kinetic energy = mgh |
| |
 Total energy at C = Potential energy + Kinetic energy |
| |
| = 0 + mgh |
| |
| Total energy at C = mgh …(3) |
| |
| It 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. |
| |
