Principle of Conservation of Energy

This principle of conservation of energy states that in an isolated system, the total amount of energy remains constant over time. This means that energy cannot be created or destroyed; it only changes form. For example, the energy used in one form will always be balanced by an equal amount of energy transformed into other forms. Therefore, the overall energy in the system stays the same.

For a practical understanding, consider the mechanical energy conservation through an example of a body falling due to gravity.

Suppose \( m \) is the mass of an object held at a certain height \( h \) above the ground. At the starting point \( A \), the body’s energy configuration is:

• Kinetic Energy (\(\text{K.E.}\)) at \( A \) is zero because it is not moving.
• Potential Energy (\(\text{P.E.}\)) is calculated by \( mgh \), where \( g \) is the acceleration due to gravity.
• Total Energy (\(\text{T.E.}\)) which is the sum of kinetic and potential energies, is

$ \displaystyle \begin{array}{l}\text{T}\text{.E}\text{.}=\text{ K}\text{.E}\text{. + P}\text{.E}\text{.}\\\Rightarrow \text{T}\text{.E}\text{.}=0+mgh\end{array}$

$$ E_1 = mgh \quad \text{(1)} $$

Principle of conservation of energy during free fall

Consider a scenario where a body is allowed to fall freely due to gravity, and upon reaching the ground at point C, it achieves a velocity \( v \).

Using the motion equation:

$$ v^2 – u^2 = 2as $$

$$ v^2 – 0 = 2(gh) $$

$$ v^2 = 2gh \quad \text{(2)} $$

At point C, the kinetic energy (K.E.) of the body is calculated as:

$$ \text{K.E.} = \frac{1}{2} mv^2 $$

$$ = \frac{1}{2} m(2gh) = mgh $$

Given that point C is at the ground level, the potential energy (P.E.) there is zero:

$$ \text{P.E.} = mgh – mgh = 0 $$

Therefore, the total energy of the body, which is the sum of its kinetic and potential energies, remains:

$$ E_2 = mgh + 0 = mgh \quad \text{(3)} $$

If we observe the body passing point B with a velocity \( v_1 \) where the height AB equals \( x \), the equations are:

$$ v_1^2 – u^2 = 2as $$

$$ v_1^2 – 0 = 2(gx) $$

$$ v_1^2 = 2gx \quad \text{(4)} $$

At point B, the kinetic energy of the body is:

$$ \text{K.E.} = \frac{1}{2} mv_1^2 $$

$$ = \frac{1}{2} m(2gx) = mgx $$

Considering the height of point B above the ground is \( h-x \), the potential energy at B is:

$$ \text{P.E.} = mg(h – x) $$

Summing up, the total energy at point B is:

$$ E_3 = mgx + mg(h – x) = mgh \quad \text{(5)} $$

Combining all these observations from points A, B, and C, we see:

$$ E_1 = E_2 = E_3 = mgh $$

This demonstrates that the total energy of the body during free fall remains constant at all positions. The form of energy, however, keeps on changing. At A, the energy is entirely potential energy and at C, it is entirely kinetic energy. In between A and C, energy is partially potential and partially kinetic. This variation of energy is shown in figure below. Total energy stays constant at \(mgh\), throughout.

Thus in an isolated system, where only conservative forces cause energy changes, the kinetic energy and potential energy can change, but the mechanical energy of the system (which is sum of K.E. and P.E.) cannot change.

We can, therefore, equate the sum of K.E. and P.E. at one instant to the sum of K.E. and P.E. at another instant—without considering intermediate states.

This law has been found to be valid in every situation. No violation, whatsoever, of this law has ever been observed. Hence what remains constant in our universe is the energy.

Some other examples of conservation of energy include:

1. Vibrations of a simple pendulum

In the given figure, OA is normal position of rest of a simple pendulum. When the bob of the pendulum is displaced to B, through a height \( h \), it is given PE = \( mgh \), where \( m \) is mass of the bob. On releasing the bob at B, it moves towards A. PE of the bob is being converted into KE. On reaching A, the entire PE. has been converted into KE. The bob, therefore, cannot stop at A.

On account of inertia, it overshoots the position A and reaches C at the same height \( h \) above A. The entire KE. of the bob at A is converted into PE. at C. The whole process repeats itself. This is known as vibrations of a simple pendulum.

The pendulum vibrates about the equilibrium position OA. At extreme positions B and C, the bob is momentarily at rest. Therefore its K.E. = 0. The entire energy at B & C is potential energy. At A, there is no height and hence no potential energy. The entire energy at A is kinetic energy.

2. Motion of a small spherical ball over a watch glass

The figure below shows a watch glass or a large concave mirror embedded in clay. Consider a tiny spherical ball placed at the edge B at a height \( h \) above the center A. If \( m \) is the mass of the ball, P.E. of the ball held at B = \( mgh \).

When the ball is released at B, it starts rolling down the mirror. P.E. of the ball is being converted into K.E. At the bottom A, the velocity of the ball is maximum as the entire P.E. has been converted into K.E. The ball cannot stop at A on account of inertia. It goes over to the other edge C. The velocity of the ball goes on decreasing and so does its K.E. At C, K.E. = 0, and P.E. is maximum. The entire process is repeated, and thus the ball keeps on rolling over the mirror about A.

3. Vibrations of an elastic spring

As discussed already in kinetic energy that total energy is conserved in the case of an elastic spring.

Note: In all the above examples, we have neglected the loss of energy due to air resistance/friction, etc. If we were to take into account these opposing forces, K.E. would go on decreasing as it appears in the form of heat energy. But total energy (including the heat energy) would remain constant.

Example: A rubber ball falls on a floor from a height of \( 19.6 \, m \). Calculate the velocity with which it strikes the ground. To what height will the ball rebound if it loses \( 25\% \) of its energy on striking the ground?

Sol. Here, \( h_1 = 19.6 \, m \), \( v_1 = ? \)

Loss of energy = \( 25\% \) of \( h_1 \), \( ? \)

On striking the floor, K.E. = P.E.

$$ \frac{1}{2} mv_1^2 = mgh_1 $$

$$ v_1^2 = 2gh_1 = 2 \times 9.8 \times 19.6 $$

$$ v_1 = 19.6 \, m/s $$

As the body loses \( 25\% \) of its energy, the final energy \( E_2 = 75\% \) of the initial energy

$$ mgh_2 = \frac{75}{100} (mgh) $$

$$ h_2 = \frac{3}{4} h_1 = \frac{3}{4} \times 19.6 = 14.7 \, m $$