# Conservation of Mechanical Energy

Conservation of Mechanical Energy

We can see many situations in which energy is transferred to or from objects and systems, similar to money being transferred between accounts. In each situation, we assume that the energy that was involved can be accounted for, that is, energy cannot appear or disappear. Energy obeys the law of conservation of energy, which is concerned with the total energy of the system.

The total energy ‘E’ of a system can change only by amounts of energy that are transferred to or from the system.

If a system is isolated from its environment, there can be no transfer of energy to or from it. In this case, the law of conservation of energy states that “The total energy ‘E’ of an isolated system cannot change”.

Energy transformation may be going on within an isolated system, say, between kinetic energy and potential energy or kinetic energy and thermal energy. However, the sum total of all the forms of energy in the system remains unchanged.

Let us consider the example of an ideal simple pendulum (frictionless). We can see that the mechanical energy of this system is a combination of its kinetic energy and gravitational potential energy. As the pendulum swings back and forth, a constant exchange between the kinetic energy and potential energy takes place. When the bob attains its maximum height, the potential energy of the system is the highest whereas the kinetic energy is zero. At the mean position, the kinetic energy is the highest and the potential energy is zero. Between these two extreme points, we see that the system possesses both kinetic and potential energy, the sum of which is constant. These observations tell us a lot about the conservation of mechanical energy. But how can we prove it for every other system? In this section, we shall learn more about the conservation of mechanical energy using a suitable example.

Conservation of Mechanical Energy: It states that the Mechanical Energy of an isolated system remains constant without friction.

According to the principle of conservation of mechanical energy, the total mechanical energy of a system is conserved i.e., the energy can neither be created nor be destroyed, it can only be internally converted from one form to another, if the forces doing work on the system are conservative in nature.

We define the change in potential energy of the system, corresponding to a conservative internal force, as

$${{v}_{f}}\,\,-\,{{u}_{i}}\,\,=\,\,-\,\,W\,\,=\,\,\,\,\int\limits_{i}^{f}{\overrightarrow{F}}.\overrightarrow{dr}$$

Where,

W = Work done by the internal forces on the system as the system passes from the initial configuration “i “to the final configuration “f “.

We cannot define potential energy corresponding to non – conservative forces. Suppose only conservative internal force operate within the system and the potential energy (U) is define, corresponding to these forces. There are no external forces or the work done by them is zero.

Therefore, Uf – Ut = – W = – (K.Ef – K.Et)

Uf + K.Ef = Ut + K.Et

The sum of the kinetic energy and the potential energy is called the total mechanical energy. The above equation shows that the total mechanical energy of the system remains constant. If the internal forces are conservative and the external forces do not do any work. This is called the principle of Conservation of energy. The total mechanical energy is not constant for non – conservative forces such as frictional forces.

How to find the Conservation of Mechanical Energy?

Problem: A 7.82 kg block is connected to a spring with a spring constant of 672 N/m. It oscillates back and forth on a frictionless horizontal plane. The potential energy of this block at maximum extension is 32.3J, then find the maximum velocity of this block?

Solution: Since we are oscillating on a frictionless surface, we can use the conservation of mechanical energy: (K.E + P.Eg + P.Es)i = (K.E + P.Eg + P.Es)f

Note that the gravitational potential energy is P.Eg = mgy. Since y is constant, we can just see it to zero for both the initial and final position, hence P.Et – P.Ef = 0. Let the initial position be the maximum extent where xi = xmax and vi = 0 (released from rest) and the final position be at xf = 0 where vf = vmax, then (K.E + P.Es)i = (K.E + P.Es)f

½ mvᵢ² + P.Esᵢ = ½ mvf² + ½ kf²

0 + P.Esᵢ = ½ mvmax² + 0

$${{v}_{\max }}\,\,=\,\,\sqrt{\frac{2\times P.{{E}_{si}}}{m}}\,\,=\,\,\sqrt{\frac{2(32.3\,J)}{7.82\,kg}}\,\,=\,\,2.87\,\,m/\sec$$

Therefore, maximum velocity of the block is 2.87 m/sec.