# Conservation of Mechanical Energy

Here we discuss the conservation of mechanical energy where we define the potential and kinetic energy in our Earth-body system.

## Potential Energy

We know that the weight of a body is the downward force of gravity. A ball is released from a particular height in Figure 1.

The ball falls under the action of gravitational force of magnitude $mg$ where $m$ is the mass of the ball. The work done by the gravitational force between the heights ${y_1}$ and ${y_2}$ is,

\begin{align*} W &= -mg({y_2} - {y_1})\\ {\rm{or,}}\quad W &= mg{y_1} - mg{y_2} = U_1 - U_2 = -(U_2 - U_1) = -\Delta U \tag{1} \label{1} \end{align*}

where $U_1$ and $U_2$ are the gravitational potential energies at height $y_1$ and $y_2$ respectively. Therefore, the gravitational potential energy $U$ at any height $y$ is $mgy$ which is the work done by the gravitational force from height $y$ to $y=0$. Notice that the initial potential energy $U_1 = mg{y_1}$ is the total work done by the gravitational force from $y = {y_1}$ to $y = 0$ and this is similar for the final potential energy $U_2 = mg{y_2}$. So the potential energy is in fact the work done.

Note that $\Delta U$ is the final potential energy $U_2$ minus the initial potential energy $U_1$ but the work done by the gravitational force is the initial potential energy minus the final potential energy. The ball in Figure 1 falls under the action of gravity and therefore $y_1 > y_2$ and the work done by the gravitational force is positive as it should be. Eq. \eqref{1} tells us that the work done by the gravitational force is the negative of change in potential energy.

## Kinetic Energy

Consider that a body moves through a displacement $\vec s$ along positive x-direction under the action of a constant net force $\vec F$. Also note that ${v_{ix}}$ is the initial velocity and ${v_{fx}}$ is the final velocity of the body. Newton's second law gives,

${F} = m{a_x}$

We have ${v_{fx}}^2 = {v_{ix}}^2 + 2{a_x}s{\rm{ }}$. Therefore, putting the value of $a_x$ from ${v_{fx}}^2 = {v_{ix}}^2 + 2{a_x}s{\rm{ }}$ in the above equation, we get,

\begin{align*} F &= m\left( {\frac{{{v_{fx}}^2 - {v_{ix}}^2}}{{2s}}} \right)\\ {\rm{or,}}\quad Fs &= \frac{1}{2}m{v_{fx}}^2 - \frac{1}{2}m{v_{ix}}^2\\ {\rm{or,}}\quad W &= \frac{1}{2}m{v_{fx}}^2 - \frac{1}{2}m{v_{ix}}^2 \tag{2} \label{2} \end{align*}

In Eq. \eqref{2} the quantity $\frac{1}{2}m{v^2}$ which is the half of the product of mass and velocity squared is called kinetic energy of the body. It means the kinetic energy of a moving body with velocity $v$ at any instant of time is $\frac{1}{2}m{v}^2$. So, $\frac{1}{2}m{v_{ix}}^2$ is the initial kinetic energy and $\frac{1}{2}m{v_{fx}}^2$ is the final kinetic energy. Therefore, ${K_1} = \frac{1}{2}m{v_{ix}}^2$ and ${K_2} = \frac{1}{2}m{v_{fx}}^2$, therefore

$W = {K_2} - {K_1} = \Delta K \tag{3} \label{3}$

Eq. \eqref{3} is also called work-energy theorem. You can conclude from Eq. \eqref{3} that the work done by a net or resultant force on a body is equal to the change in kinetic energy of the body.

The above derivation of work done as the change in kinetic energy is based on a constant force applied but the work-energy theorem given by Eq. \eqref{3} or Eq. \eqref{2} is true even if the force is not constant or the motion is not always along a straight line.

## Conservation of Mechanical Energy

In Figure 1 a ball falls under the action of gravity from a particular height. We neglect the air resistance on the ball. And from Eq. \eqref{1} and Eq. \eqref{3} we have

\begin{align*} \Delta K &= - \Delta U\\ {\rm{or,}}\quad {K_2} - {K_1} &= - ({U_2} - {U_1}) = {U_1} - {U_2}\\ {\rm{or,}}\quad {K_1} + {U_1} &= {K_2} + {U_2} \tag{4} \label{4} \end{align*}

The sum of the initial kinetic energy and the initial potential energy is equal to the sum of the final kinetic energy and the final potential energy. The sum of the kinetic energy and the potential energy of a body at any point on its path is called mechanical energy. Hence the total mechanical energy of the ball is constant or conserved as suggested by Eq. \eqref{4}.

In Figure 2 a ball falls from the height $y_1$ and the potential energy of the ball at height $y_1$ is ${U_1} = mg{y_1}$. The velocity of the ball at height $y_1$ is zero, so the kinetic energy is zero i.e. ${K_1} =0$. As the ball falls its potential energy decreases and kinetic energy increases. Therefore, ${U_1} > {U_2} > {U_3}$ and ${K_1} < {K_2} < {K_3}$ but from Eq. \eqref{4} we have ${K_1} + {U_1} = {K_2} + {U_2} = {K_3} + {U_3}$. The conversation of mechanical energy tells us that the sum of kinetic and potential energy of the ball at every point on its path is constant or conserved that is, the potential energy lost is recovered as the kinetic energy and the sum remains constant.

In our case the force which does work is the gravitational force and it is a conservative force. If a conservative force does work, the mechanical energy is always conserved if no external force does work. For example, friction is not a conservative force as the work done by the friction is never stored as potential energy or recovered back as kinetic energy. What if an external force does work other than the conservative force? If ${W_{{\rm{ext}}}}$ is the external work done by the air resistance on the ball in Figure 2, the work done by this force is equal to the change in total mechanical energy. If ${E_1} = {K_1} + {U_1}$ and ${E_2} = {K_2} + {U_2}$, the work done by the external force is ${W_{{\rm{ext}}}} = {E_2} - {E_1} = \Delta E$. Or we can write Eq. \eqref{4} including the work done by the external force as, ${K_1} + {U_1} + {W_{{\rm{ext}}}} = {K_2} + {U_2} \tag{5} \label{5}$

The force the air resistance exerts on the ball is in opposite direction of displacement so it does negative work. Therefore, the final mechanical energy is lesser than the initial mechanical energy and $\Delta E$ is negative.

Mechanics