Imagine you are holding a ball on the terrace of your building. You can see the ball drop when you let go of the ball. It is the gravitational force acting on the ball that makes the ball drop. But where did the ball get the energy from? Did you feed the ball? Is there any energy transformation? Read on to understand how Conservation of Mechanical Energy works in such scenarios.
Principle of conservation of mechanical energy
Let's first understand what mechanical energy is.
Mechanical energy is the sum of a system's kinetic and potential energy that can be used to do useful work.
Energy exists in many forms; however, all energy can be classified as either kinetic or potential energy. The conservation of mechanical energy is based on the principle of the law of conservation of energy. According to the law of conservation of energy, energy cannot be created or destroyed; it can only be converted from one form to another.
Let's look at an example to find the mechanical energy of a system.
A pilot of an aeroplane fires a shot weighing \(0.1\, \mathrm{kg}\) with a velocity of \(300\, \mathrm{m\, s^{-2}}\). Calculate the mechanical energy of the shot when it is at the height of \(700 \, \mathrm{m}\) above the ground.
Solution:
From the statement of the problem you know: \(m=0.1\, \mathrm{kg}\); \(v=300\, \mathrm{m\,s^{-1}}\); \(h=700\, \mathrm{m}\); and \(g=9.8\, \mathrm{m\,s^{-2}}\).
You want to calculate the mechanical energy, which is the sum of kinetic and potential energies. So
\[\begin{align} \text{Mechanical energy} &= \frac{1}{2}mv^2 + mgh \\ &= \frac{1}{2} (0.1\,\mathrm{kg} )(300 \, \mathrm{m\,s^{-1}} )^2+(0.1\,\mathrm{kg} ) (9.8\, \mathrm{m\,s^{-2}})(700\, \mathrm{m} )\\ &=4500\, \mathrm{J} +686\, \mathrm{J} \\ &=5186\, \mathrm{J}. \end{align} \]
Law of conservation of mechanical energy
Let's go back to example at the start, where you have dropped the ball from the terrace of your building.
Before you drop the ball, the ball is not in motion. So, the kinetic energy of the ball is zero. It has only the gravitational potential energy due to height.
When you drop the ball, it starts moving down and gets velocity. Now since the ball has velocity, it also has kinetic energy.
As the ball gets closer and closer to the ground, its height from the ground level decreases so does its potential energy. The ball's velocity keeps increasing as it moves down, and so does its kinetic energy.
When the ball finally touches the ground, it has only kinetic energy, and its potential energy becomes zero.
What do you observe in this case? You can see that the potential energy is converted into kinetic energy. From this perspective, let's define the law of conservation of mechanical energy.
In an isolated, friction-free system, the total mechanical energy is always conserved. If energy disappears in one form, it reappears in another form in an equivalent amount.
Please note in our dropping ball example, you have neglected the air resistance to the ball's motion under gravitational force. If the frictional force is considered, some mechanical energy is converted into thermal energy.
Conservation of mechanical energy formula
By now, you already know that the system's total energy is conserved and is constant, let's say, \(C\). Let \(\text{KE}_\text{initial}\) and \(\text{PE}_\text{initial}\) be the initial kinetic and potential energy of the system, and \(\text{KE}_\text{final}\) and \(\text{PE}_\text{final}\) be the final kinetic and potential energy of the system. As per the law of conservation of mechanical energy
\[\text{KE}_\text{initial} + \text{PE}_\text{initial} = \text{KE}_\text{final} + \text{PE}_\text{final} = C.\]
You will understand how the total energy at any point is constant in the next section of this article.
Conservation of mechanical energy equation
Let's look at why the ball which is falling from the terrace has constant energy at every point of its motion. Consider a ball falling freely under gravity, as shown in the figure below.
Consider the ball of mass \(m\) dropped from point \(A\) at a height \(h\) above the ground. Let's see the total energy of the ball in three different cases.
Case 1:
At point \(A\), the potential energy of the ball is
\[\text{PE}_\text{A}=mgh,\]
and the kinetic energy of the ball is
\[\text{KE}_\text{A}=0.\]
Therefore, the total energy of the ball at point \(A\) is given by
\[\begin {align}\text{PE}_\text{A}+ \text{KE}_\text{A}& =mgh+0\\ &=0.\end{align} \]
Case 2:
As the ball falls, its potential energy decreases, but its kinetic energy increases. Let \(v\) be the velocity of the ball at point \(B\) at a distance \(x\) from the \(A\).
At point \(B\), the potential energy of the ball is
\[\text{PE}_\text{B}=mg(h-x),\]
and the kinetic energy of the ball is
\[\text{KE}_\text{B}=\frac{1}{2}mv^2.\]
Therefore, the total energy of the ball at point \(B\) is given by
\[\begin {align}\text{PE}_\text{B}+ \text{KE}_\text{B}& =mg(h-x)+\frac{1}{2}mv^2 \\ &=mgh-mgx+\frac{1}{2}m(2gx) \\ &=mgh-mgx+mgx\\&=mgh.\end{align} \]
Note that when the ball is in motion from point \(A\) to point \(B\), its initial velocity \(u=0\, \mathrm{m\,s^{-1}}\) and the displacement \(s=x\). By substituting it in one of the equations of motion, \(v^2+u^2=2gs\), you get \(v^2=2gx\).
Case 3:
When the ball falls to the ground at point \(C\), \(h=0\). Let \(v\) be the velocity of the ball when it reaches the ground. Then the potential energy of the ball is
\[\text{PE}_\text{C}=0,\]
and the kinetic energy of the ball is
\[\text{KE}_\text{C}=\frac{1}{2}mv^2.\]
Therefore, the total energy of the ball at point \(C\) is given by
\[\begin {align}\text{PE}_\text{C}+ \text{KE}_\text{C}& =0+\frac{1}{2}mv^2\\ &=\frac{1}{2}m(2gh)\\&=mgh.\end{align} \]
Note that when the ball is in motion from point \(A\) to point \(C\), its initial velocity \(u=0\, \mathrm{m\,s^{-1}}\) and the displacement \(s=h\), which is the height of the building. By substituting it in one of the equations of motion, \(v^2+u^2=2gs\), you get \(v^2=2gh\).
From all three cases, you can see that the total energy of the ball remains constant (that is, it is always \(mgh\) in this case).
Conservation of mechanical energy examples
Let's look at an example based on the conservation of mechanical energy for a freely falling body.
A body of mass \(2\, \mathrm{kg}\) falling freely under gravity takes \(6\, \mathrm{s}\) to reach the ground. Calculate the kinetic and potential energies of the body when the body has travelled \(3\, \mathrm{s}\).
Solution:
Let the body be dropped from a height \(h\) above the ground. The initial velocity is \(u=0 \, \mathrm{m\,s^{-1}}\), and the acceleration is provided by gravity is \(a=g=9.8\, \mathrm{m\,s^{-2}}\). By using the equation of motion
\[s=ut+\frac{1}{2}at^2 ,\]
where \(s\) is displacement, you get
\[\begin{align} h&=(0 \, \mathrm{m\, s^{-1}})(6\, \mathrm{s})+\frac{1}{2}(9.8\, \mathrm{m\,s^{-2}})(6\, \mathrm{s})^2\\ &=176.4\, \mathrm{m}.\end{align}\]
From the conservation of mechanical energy, you know that the total energy of the body is equal to the potential energy at height \(176.4 \, \mathrm{m} \) (because initially, the kinetic energy will be zero). That is
\[\begin{align} \text{Total energy}&=mgh\\&=(2\, \mathrm{kg})(9.8\, \mathrm{m\,s^{-2}})(176.4\, \mathrm{m})\\&=3457.44\, \mathrm{J}. \end{align}\]
Let \(v\) be the velocity of the body when it falls for \(t=3\, \mathrm{s}\). Using the equation of motion, you get
\[\begin{align}v&=u+at\\ &=0\, \mathrm{m\,s^{-1}}+(9.8\, \mathrm{m\,s^{-2}})(3\, \mathrm{s}) \\ &=29.4\, \mathrm{m\,s^{-1}}.\end{align}\]
The kinetic energy of the body is \[\begin {align}\text{KE}&=\frac{1}{2}mv^2\\&=\frac{1}{2}(2\, \mathrm{kg})(29.4\, \mathrm{m\,s^{-1}})^2\\ &=864.36\, \mathrm{J}.\end{align}\]
Thus the potential energy of the body is given by \[\begin{align} \text{PE} &=\text{Total energy} - \text{KE}\\ &=3457.44\, \mathrm{J} -864.36\, \mathrm{J} \\&=2593.08\,\mathrm{J}.\end{align}\]
Now let's consider a very interesting scenario when the body is sliding down an inclined plane.
A basketball of mass \(0.2\, \mathrm{kg}\) is rolling down a smooth plane inclined at angle \(30^\circ\) with the horizontal. The basketball starts from rest at point \(B\), and reaches point \(A\) with a velocity of \(2\, \mathrm{m\,s^{-1}}\). Find the distance from \(A\) to \(B\).
Solution:
Fig. 2 - Basketball rolling down an inclined surface
When the basketball rolls down from \(B\) to \(A\), there will be a decrease in potential energy and an increase in kinetic energy.
The decrease in potential energy is given by
\[\begin{align}\text{Decrese in PE}&=mgh\\ &=mgx_m\sin30^\circ \\& =(0.2\, \mathrm{kg})(9.8\, \mathrm{m\,s^{-2}})(x_m\sin30^\circ)\\&=0.98x_m \, \mathrm{J}.\end{align}\]
From figure 2, that the distance from \(A\) to \(B\) is labelled \(x_m\) and the vertical distance moved by the basketball is \(x_m\sin30^\circ \). So you can substitute \(h=x_m\sin30^\circ\).
The increase in kinetic energy is given by \[\begin{align}\text{Increase in KE}&=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ &=\frac{1}{2}(0.2\, \mathrm{kg})(2\,\mathrm{m\,s^{-1}})^2-\frac{1}{2}(0.2\,\mathrm{kg})(0)^2 \\& =0.4\,\mathrm{J}.\end{align}\]
By applying the law of conservation of mechanical energy, you can say that decrease in potential energy is equal to an increase in kinetic energy. That is \[\begin{align}\text{Decrease in PE}&=\text{Increase in KE} \\0.98x_m &=0.4\\ x_m& \approx 0.4\, \mathrm{m}\end{align} \]
The distance from \(A\) to \(B\) is approximately \(0.4\,\mathrm{m}\).
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