# Video: Kinetic Energy

In this lesson, we will learn how to calculate the kinetic energy of objects of different masses, moving at different velocities.

14:47

### Video Transcript

In this video, our topic is kinetic energy. This is energy that’s due to object motion. We’ll come to see that kinetic energy depends on two parameters, on object mass and on object velocity, like we have here with this truck. If each one of those values is nonzero, in other words, the object has some amount of mass and has some amount of velocity, then in that case, the object has some energy due to motion, kinetic energy. Knowing this, if we have an object with a given mass, that’s moving with a given velocity 𝑣, then we can write out this object’s kinetic energy this way. Its kinetic energy is equal to one-half the mass of the object multiplied by its velocity squared. Looking at this equation, something that may stand out is that the object’s velocity is squared while the mass is not. We will get a chance to see just why this is later on. But for now, it’s helpful to see how this difference affects the kinetic energy of different objects.

For example, say we have an object here which has twice the mass of our original object, but the same exact velocity. Using our equation for kinetic energy, we can write out the energy due to motion for each of these objects. For our larger object, that kinetic energy is equal to one-half its mass, two times 𝑚, times 𝑣 squared, whereas for the smaller one it’s just one-half 𝑚 𝑣 squared. So for the larger object, we see this factor of two cancels with the factor of one-half. And this means the kinetic energy, the energy due to motion of our larger mass, is twice as big as that of our smaller mass. Now, remember, this is for the case where the velocities of these two masses are identical.

Now, let’s imagine we add a third moving mass to this scenario. This object has the same mass as our original one. But it has two times the original velocity. If we calculated the kinetic energy of this third object, that energy would be one-half its mass 𝑚 multiplied by its velocity, which is two times 𝑣, quantity, squared. And notice, since the two as well as the 𝑣 are both inside these parentheses, that means that we square both terms. We square two and we square 𝑣. Two squared is four. And 𝑣 squared is 𝑣 squared. So the kinetic energy of our third object winds up being one-half 𝑚 four 𝑣 squared or, written a bit more simply, two times 𝑚𝑣 squared. Let’s take a moment now to compare this result with the kinetic energy of our original object and our object with twice the original mass.

First off, notice that this result we just calculated is four times bigger than the kinetic energy of our original object. So we find that, by doubling velocity, we quadruple, multiply by four, kinetic energy. And even more than that, if we compare this final result with the kinetic energy of our two 𝑚, mass, we see that it’s twice as big. This confirms to us that if we wanna increase an object’s kinetic energy, we can do that more dramatically by increasing velocity then by increasing mass. And the reason for that comes back to the original form of our kinetic energy equation, where we see the object velocity being squared. But the mass is not. Now that we’ve seen a bit about just what this kinetic energy equation means, let’s look into where this equation comes from. This will help us understand just why is it, that an object’s kinetic energy is equal to one-half its mass times its velocity squared.

To help us understand this, imagine we start out with this situation. Say that we have a block of mass 𝑚 on a perfectly smooth flat plane. At first, this block is stationary. It’s not moving. So we can say its initial velocity, we’ll call it 𝑣 sub 𝑖, is zero. But then, we start to apply a force, we can call that force 𝐹, to the block, pushing it to the right steadily. Pushing on the block with this steady force, the block will start to move to the right. And say that we take a second snapshot in time when the block is here. And let’s say further that by the time it reaches this point, the block has a velocity, we’ll call 𝑣 sub 𝑓, its final velocity, and that this acceleration from an initial velocity of zero to a final velocity happens in an amount of time we’ll call 𝑡. All this setup we’re doing, naming all these variables and describing the situation, is to help us to be able to derive this equation for kinetic energy, to see why it is that it has the form it does.

To get there, to arrive at this equation for kinetic energy, we won’t start talking about energy. But instead, we’ll start by talking about work. Recall that the work done on an object is equal to the force we applied to that object multiplied by the distance over which the object moves. When it comes to the force involved here, 𝐹, we can see that, in our scenario, we indeed have a force 𝐹 pushing on our block. And more than that, we see that this force is responsible for accelerating the block since it starts out with a velocity of zero and ends up with a nonzero velocity, 𝑣 sub 𝑓. That is, we’re applying a force to a mass. And we’re causing it to accelerate. This can remind us of one of Newton’s laws of motion, his second law. This law tells us that the net force we applied to an object is equal to that object’s mass multiplied by its acceleration. By this second law then, we can replace the force 𝐹 in our equation for work with the mass of our object multiplied by its acceleration. So work is equal two 𝑚 times 𝑎 times 𝑑 or 𝑚𝑎𝑑.

Now, we’re going to upset this kind of fun equation because we’re just about to substitute in for the acceleration 𝑎. Here’s how we’ll do that. Let’s remember the definition for acceleration. Acceleration is a change in velocity over some amount of time. Another way to write this is to say that it’s equal to a final velocity, 𝑣 sub 𝑓, minus an initial velocity, 𝑣 sub 𝑖, divided by 𝑡. This way of writing acceleration is helpful because, in our scenario, we have a final velocity. We have an initial velocity. And we also have a time. What we’ll do then is replace this acceleration 𝑎 with 𝑣 sub 𝑓 minus 𝑣 sub 𝑖 divided by 𝑡. Before we do though, we can simplify this substitution just a bit further. Remember that 𝑣 sub 𝑖, our object’s initial velocity, is zero. And when we replace 𝑣 sub 𝑖 in this expression with zero, then the expression simplifies to 𝑣 sub 𝑓 divided by 𝑡. And that’s what we’ll substitute in for 𝑎 in our equation for work.

Now, we’ve made great progress. And we’re not too too far away from arriving at our final equation for kinetic energy. Notice we already have 𝑚 and 𝑣 terms in our equation for work. There’s just a bit more to do though. And it involves this term here, our distance 𝑑. Notice that we don’t have a distance 𝑑 marked out in this scenario. But what we do have is velocities 𝑣 sub 𝑓 and 𝑣 sub 𝑖 as well as a time 𝑡. So here’s what we can do. Let’s recall that the average velocity an object has, we’ll symbolize it 𝑣 with a bar over it, that bar means average, is equal to the total displacement of the object, we’ll call that 𝑑, divided by the time it takes to be displaced that distance. And let’s think now for a moment about this average velocity term as it relates to our particular scenario of this block sliding across the plane.

What we have is an object that starts out at rest. And it ends up with some nonzero velocity, we’ve called 𝑣 sub 𝑓. And more than that, we have a constant force 𝐹 pushing on our block to accelerate it at a constant rate to the right. The fact that our force pushing on the block is constant means that the acceleration the block undergoes is constant as well. All throughout the time that it’s moving left to right, it’s speeding up at the same rate. That means that, in this scenario, we can write the block’s average velocity this way. We can say that it’s equal to the sum of the final and the initial velocities of the block divided by two. What we’re basically doing here is taking the average of two numbers. One number is called 𝑣 sub 𝑓. And the other is called 𝑣 sub 𝑖. To find that average, we add these numbers together. And then, we divide them by two.

And notice that, just like before, we can simplify this expression a bit because this term, 𝑣 sub 𝑖, we know to be zero. So really, the average velocity of our block is equal to 𝑣 sub 𝑓 divided by two. And then, by this general equation for average velocity, we can say that 𝑣 sub 𝑓 divided by two is equal to 𝑑 divided by 𝑡. Remember that that’s the right-hand side of this equation here. So let’s erase this average velocity term. And we see that the equation that remains is in terms of values we’re given, 𝑣 sub 𝑓 and 𝑡, and the value we want to solve for, the distance 𝑑. If we now go ahead and multiply both sides of this equation by the time 𝑡, that term cancels out on the right-hand side. And we now have an equation for the distance our block travels in terms of given values, 𝑣 sub 𝑓 and 𝑡. And of course, there’s this factor of one-half in here as well.

So let’s take this expression on the left-hand side and substitute it in for 𝑑 in our equation for work. When we do this, look at what happens. We have a factor of 𝑡 here in the denominator and a factor of 𝑡 here in the numerator. This means that when we multiply through, these factors will cancel out. And now, take a look at this. The work done on our object is equal to 𝑚 times its final velocity squared divided by two. Written another way, that’s equal to one-half its mass times its final velocity squared. And there’s just one last connection to make to tie this equation for work together with this equation for kinetic energy.

The key inside is to realize that the work we do on this block to accelerate it from an initial velocity of zero to a final velocity of 𝑣 sub 𝑓 is energy that’s put into the block from an external supply, whatever it is that’s providing this force 𝐹. But we knew that energy is conserved. If we put energy into the block, it has to go somewhere. And where it goes is an energy due to the block’s motion. It goes into kinetic energy. So the work done on the block by some external source is the supply of energy to the block that gives it motion and therefore kinetic energy. In other words, the work done on the block is equal to its resulting kinetic energy. That’s thanks to the conservation of energy.

By realizing that, we link together work and kinetic energy and then can see how the mathematical equation for kinetic energy is the same as that for work done. And by the way, this 𝑣 here in our equation for kinetic energy is just a more general way of writing 𝑣 sub 𝑓 that we have in our equation for work, though the same term in these two equations are equal. So now we know why it is that, in this equation for kinetic energy, we have this factor of one-half, why it is that 𝑣 is squared and 𝑚 is not. And we also know that standing behind this equation is the concept of work in physics. And we see that kinetic energy and work are closely connected. At this point, let’s get some practice using this relationship through an example.

An object with a mass of 1.25 kilograms has a velocity of 12 metres per second. What is the object’s kinetic energy?

So in this scenario, we have this object. And we’re told that it has a mass, we’ll call it 𝑚, of 1.25 kilograms. And also, it has a velocity that’s nonzero. The velocity, which we’ll call 𝑣, is 12 metres per second. We want to solve for the object’s kinetic energy, that is, its energy due to motion. And to do that, we can recall the mathematical relationship for kinetic energy. An object’s kinetic energy is equal to one-half its mass times its velocity squared. So our object’s kinetic energy, we’ll just call it KE, is equal to one-half its mass, 1.25 kilograms, multiplied by its velocity, 12 metres per second squared. Notice that the units for mass and velocity are in the base units of kilograms and metres per second. That means we’re all set to go. And we can multiply through on the right side of this equation. When we do, we find the result of 90 kilograms metre squared per second squared or 90 joules. That’s how much kinetic energy this object has.

Let’s look at another example involving kinetic energy.

An object that has nine joules of kinetic energy is moving at three metres per second. What is the mass of the object?

Okay, in this situation, we have some object. Let’s say it’s this box here. And we’re told that this box is in motion, moving along with a velocity, we’ll call 𝑣, of three metres per second. Thanks to this motion, this object has a kinetic energy of nine joules. Based on this information, we want to solve for the mass, we can call it 𝑚, of our object. To start out doing that, let’s recall the equation for an object’s kinetic energy in terms of its mass and its velocity. Kinetic energy is equal to one-half 𝑚𝑣 squared. Now in our situation, it’s not the kinetic energy we want to solve for, but rather the mass 𝑚.

To do that, starting with this equation, we can start by multiplying both sides by the factor of two, which cancels out that two with the factor of one-half on the right-hand side. And then, second, we can divide both sides of the equation by 𝑣 squared, the square of this object’s velocity. Doing this cancels out that term also on the right-hand side. Based on our original equation then, an object’s mass is equal to two times its kinetic energy divided by its velocity squared. And that brings us back to solving for the mass of this particular object. That mass is equal to two times the object’s kinetic energy, which is nine joules, divided by its velocity, three metres per second squared. Now, looking at this denominator, we know that because this three metres per second velocity is inside parentheses and being squared, then that means we apply the squared term both to the three as well as to the units, metres per second. What results is this term, nine metres squared per second squared.

Next, if we look at the numerator, we know that two times nine joules is 18 joules. But then, let’s consider this unit joules. Recall that one joule is equal to a newton multiplied by a metre. And then, on top of that, one newton is equal to a kilogram metre per second squared. Making this substitution, we see that a joule can also be written as kilogram metre squared per second squared. But then, with that substitution made, look at this. We have metre squared divided by second squared in both the numerator and denominator of this fraction. So when we calculate this fraction, these units cancel out. We’re left simply with units of kilograms. Knowing that and the fact that 18 divided by nine is equal to two, we understand that our final answer is two kilograms. That’s the mass of this object.

Let’s take a moment now to summarize what we we’ve learned in this lesson. First off, we saw that kinetic energy is energy that’s due to motion. An object needs to be in motion have a nonzero velocity in order to have any kinetic energy. Written as an equation, kinetic energy is equal to one-half an object’s mass times its velocity squared. The form of this equation implies that, for an object with kinetic energy, if we were to double its mass, we would double its kinetic energy. But if we were to double its velocity, we would quadruple its kinetic energy, four times greater. And lastly, as we derive the equation for kinetic energy based on an equation for work, we saw that an object’s kinetic energy is equal to the work done on that object in order to speed it up from being at rest.