We’ve now introduced three values that all have the same unit – kinetic energy, potential energy, and work. Since they have the same unit of measurement, we should be able to use them all together. We know there’s some correlation between these three values, but we don’t know how to use them to describe how an object changes after an action has been completed. This is where the law of conservation of energy comes in. In the following blog post, we’ll explain the law of conservation of energy, the formula that governs it, and common conservation of energy examples.

What is The Law of Conservation of Energy?

The Law of Conservation of Energy states that the energy in a system before an action plus the work done to complete the action will be equal to the energy in the system after the action. This doesn’t always play out perfectly in real life as some energy can be lost to the creation of noise and heat. For the time being, this law will play out perfectly in the problems you’re given in your physics courses.

Principles of the Law of Conservation of Energy

The catchier definition of this law you may have heard before is that energy cannot be created or destroyed. While that is true, it’s missing something. Energy can be transferred from one object to another. This is what allows one pool ball to set another into motion. It effectively tells us that an object or system needs energy from somewhere else in order to change or complete an action. This is also why a perpetual motion machine could never truly exist.

Conservation of Energy Formula and Its Application

The conservation of energy formula can be written as:

Conservation of Energy Formula \sum E_{K1}+\sum E_{P1}+\sum W=\sum E_{K2}+\sum E_{P2}

Recall that E_{K} represents kinetic energy, E_{P} represents potential energy, and W represents work. The reason we’ve added numbers to the subscripts is to show that the left side of the equation represents values from before an action was taken while the right side was after the action was taken. So, before and after something falls or before and after two things collide or something like that. The \sum might be new, though.

\sum is a Greek letter sigma and is used to represent a sum. In this case, it would be the sum of all of the kinetic energies, potential energies, and work going into the system. For example, if you had two moving objects, you would have two kinetic energies to account for. Similarly, if an object has gravitational potential energy and elastic potential energy you would have two potential energies to account for. 

Conservation of Energy Examples

Kinetic and Potential Energy and Work

We discussed kinetic energy, potential energy, and work in depth previously, but let’s check our understanding before we jump into some conservation of energy examples.

First, kinetic energy is energy that comes from motion. It is determined by an object’s mass and velocity and is given by the equation E_{K}=\frac{1}{2}mv^{2}. 

So far, we’ve covered two types of potential energy – gravitational potential energy and elastic potential energy. Gravitational potential energy is determined by an object’s mass, gravitational field strength, and height above ground. It is given by the equation E_{P}=mgh. Elastic potential energy is determined by the spring constant of the elastic object and the distance it has been stretched or compressed. It is given by the equation E_{P}=\frac{1}{2}kx^{2}.

Lastly, work is a measure of the effort needed from an external force to cause an action to occur and it is commonly found with the equation W=Fd\cos\theta. It’s also important to keep in mind that the work put into a system is equal to the change in energy of that system. This is what allows for the law of conservation of energy equation to work.

Example: A Simple Lab Experiment

Let’s consider a fairly simple experiment that can be used to test the law of conservation of energy. You only need a scale, a stopwatch, and a few things you can drop. The idea of this experiment is simply to take the mass of a few different objects with notably different masses. You could use a pen, a notebook, and an empty water bottle – anything that will fall without gathering too much air resistance. We’ll be using some sample data below, but feel free to recreate this experiment to find your own data or try a simulation.

Step 1: Calculate Potential Energy Values

Firstly, you’ll want to take the mass of each object and calculate the potential energy.

Object	Mass	Gravitational Field Strength	Height	E_{P}=mgh	E_{P}
Object 1	0.5\text{ kg}	9.81\text{ m/s}^{2}	3\text{ m}	E_{P}=0.5\text{ kg}\cdot 9.81\text{ m/s}^{2}\cdot 3\text{ m}	14.715\text{ J}
Object 2	2\text{ kg}	9.81\text{ m/s}^{2}	3\text{ m}	E_{P}=2\text{ kg}\cdot 9.81\text{ m/s}^{2}\cdot 3\text{ m}	58.86\text{ J}
Object 3	5\text{ kg}	9.81\text{ m/s}^{2}	3\text{ m}	E_{P}=5\text{ kg}\cdot 9.81\text{ m/s}^{2}\cdot 3\text{ m}	147.15\text{ J}

Step 2: Record Drop Times

Secondly, drop each object from the height you used above and time how long it takes each item to drop. Then, we’ll use this to figure out the average velocity of each object as it falls. You’ll want to drop each item a few times to make sure you have a good average to decrease any potential errors in the data.

Object	Time	v=gt	Average
Object 1: Trial 1	0.80\text{ s}	v=9.81\text{ m/s}^2\cdot 0.80\text{ s}=7.84\text{ m/s}	7.81\text{ m/s}
Object 1: Trial 2	0.77\text{ s}	v=9.81\text{ m/s}^2\cdot 0.77\text{ s}=7.55\text{ m/s}
Object 1: Trial 3	0.82\text{ s}	v=9.81\text{ m/s}^2\cdot 0.82\text{ s}=8.04\text{ m/s}
Object 2: Trial 1	0.79\text{ s}	v=9.81\text{ m/s}^2\cdot 0.79\text{ s}=7.75\text{ m/s}	7.78\text{ m/s}
Object 2: Trial 2	0.81\text{ s}	v=9.81\text{ m/s}^2\cdot 0.81\text{ s}=7.95\text{ m/s}
Object 2: Trial 3	0.78\text{ s}	v=9.81\text{ m/s}^2\cdot 0.78\text{ s}=7.65\text{ m/s}
Object 3: Trial 1	0.76\text{ s}	v=9.81\text{ m/s}^2\cdot 0.76\text{ s}=7.46\text{ m/s}	7.88\text{ m/s}
Object 3: Trial 2	0.80\text{ s}	v=9.81\text{ m/s}^2\cdot 0.80\text{ s}=7.84\text{ m/s}
Object 3: Trial 3	0.84\text{ s}	v=9.81\text{ m/s}^2\cdot 0.85\text{ s}=8.34\text{ m/s}

Step 3: Calculate the Kinetic Energy

Thirdly, we’ll use our known mass and our calculated average velocities to find the final kinetic energy for each object.

Object	Mass	Velocity	E_{K}=\frac{1}{2}mv^{2}	E_{K}
Object 1	0.5\text{ kg}	7.81\text{ m/s}	E_{K}=\frac{1}{2}\cdot 0.5\text{ kg} \cdot (7.81\text{ m/s})^{2}	15.2\text{ J}
Object 2	2\text{ kg}	7.78\text{ m/s}	E_{K}=\frac{1}{2}\cdot 2\text{ kg} \cdot (7.78\text{ m/s})^{2}	60.5\text{ J}
Object 3	5\text{ kg}	7.88text{ m/s}	E_{K}=\frac{1}{2}\cdot 5\text{ kg} \cdot (7.88\text{ m/s})^{2}	155.2\text{ J}

Step 4: Check the Conservation of Energy

Once we’ve collected the above data, we can apply the law of the conservation of energy by comparing the potential and kinetic energy values that we calculated. It is important to realize that we can make this comparison directly from our conservation of energy formula. If we follow our standard problem-solving steps a bit, you’ll soon see why it works.

• E_{K1}=0\text{ J}
• E_{P1}=\text{known}
• W=0\text{ J}
• E_{K2}=\text{known}
• E_{P2}=0\text{ J}

\sum E_{K1}+\sum E_{P1}+\sum W=\sum E_{K2}+\sum E_{P2}

0\text{J}+E_{P1}+0\text{J}=E_{K2}+0\text{J}

E_{P1}=E_{K2}

So, we can see that the potential energy we calculated for each object should be equal to the kinetic energy we calculated. Keep in mind that there was some error in our measured times and calculations. Let’s check the law of conservation of energy.

Object	E_{P1}	E_{K2}	Percent Error
Object 1	14.715\text{ J}	15.2\text{ J}	3.3\%
Object 2	58.86\text{ J}	60.5\text{ J}	2.9\%
Object 3	147.15\text{ J}	155.2\text{ J}	5.5\%

As you can see, we have some error. Specifically, our kinetic energy values aren’t a perfect match to our potential energy values. Given our percent error values are around 3% to 5%, we can therefore assume they were created by inaccurate measurements, rounding errors through the problem, or other human factors. So, we have successfully conducted a lab that proves the Law of the Conservation of Energy.

How Does the Conservation of Energy Apply to Roller Coasters?

Let’s take another example and look at how the conservation of energy applies to roller coasters. In this case, we’ll assume a roller coaster car filled with people has a mass of 700\text{ kg}. On this particular coaster, the car is brought up to its first hill, which is 75\text{ m} tall. Assuming the vehicle is at rest at the top of the hill and no work is done on the car as it falls, what will be its velocity at the bottom of the hill?

Solution

• h_{1}=75\text{ m}
• g=9.81\text{ m/s}^{2}
• v_{1}=0\text{ m/s}
• h_{2}=0\text{ m}
• v_{2}=\text{?}

At first, you may be feeling a little lost on how to solve this one. You probably know you need the formula for the conservation of energy given the context of this article. As for how to use it, we’ll need to break it down by substituting equations for our values.

\sum E_{K1}+\sum E_{P1}+\sum W=\sum E_{K2}+\sum E_{P2}

\frac{1}{2}mv_{1}^{2}+mgh_{1}+\frac{1}{2}kx_{1}^{2}+W=\frac{1}{2}mv_{2}^{2}+mgh_{2}+\frac{1}{2}kx_{2}^{2}

That probably looks quite long and overwhelming because, well, it is. We can simplify it immediately, though. We were told to assume no work was done and no elastic potential energy was ever mentioned, so we can take out both of those values.

\frac{1}{2}mv_{1}^{2}+mgh_{1}=\frac{1}{2}mv_{2}^{2}+mgh_{2}

You can see that this looks a bit better and a bit more manageable. Now, we have a variable matching all of our knowns from our problem statement and the only variable we don’t know is the one we were told to find. You could jump in and rearrange to solve for the final velocity now, or you can simplify a little more. We have a couple of known values of zero, so let’s plug just those in and see what goes away as a result.

\frac{1}{2}m(0\text{ m/s})^{2}+mgh_{1}=\frac{1}{2}mv_{2}^{2}+mg(0\text{ m})

mgh_{1}=\frac{1}{2}mv_{2}^{2}

This is now as simple as we can get. Let’s finish following our usual problem-solving steps to find the answer to this question.

mgh_{1}=\frac{1}{2}mv_{2}^{2}

mv_{2}^{2}=2mgh_{1}

v_{2}^{2}=\frac{2mgh_{1}}{m}

v_{2}=\sqrt{2gh_{1}}

v_{2}=\sqrt{2\cdot 9.81\text{ m/s}^{2}\cdot 75\text{ m}}

v_{2}=38.36\text{ m/s}

Conclusion

The Law of Conservation of Energy is an important one in Physics. It gives engineers the knowledge they need to create things like roller coasters and rocket ships. It also allows us to combine seemingly unrelated ideas like the height of an object and the stretch of spring. Working with this equation can seem overwhelming at first, but it gets easier over time. The more problems you apply it to, the more patterns you’ll see in it, and before no time it will be more helpful than stressful.