Lesson Explainer: Kinetic Energy Physics

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

Kinetic energy (KE) is the energy an object has as a result of movement. This amount of energy increases with both the mass and the velocity of the object, and if it is not moving, it has no KE.

Definition: Kinetic Energy

The equation is KE=12π‘šπ‘£, where π‘š is mass and 𝑣 is velocity.

Let’s see what calculating kinetic energy looks like by determining it for a thrown watermelon in a vacuum, where gravity and friction are not affecting it. The watermelon has a mass of 10 kg and travels to the right with a constant velocity of 10 m/s.

Putting the mass and velocity into the equation for kinetic energy, we have KEkgms=12(10)(10/).

Note that the entirety of the 𝑣 is squared. This means that negative velocities will become positive when squared. This, combined with it being impossible to have negative mass, means that kinetic energy is always positive, making it a scalar quantity.

Velocity being squared also means that it will give units of square metres per second squared (m2/s2). When multiplied with kilograms, though, we have kgmsΓ—/, which is the SI unit of energy: joules. Crunching the numbers (don’t forget the factor of 12 out front), we see that the kinetic energy for this melon is 12(10)ο€Ή100/=500.kgmsJ

We see that while both mass and velocity increase KE, velocity 𝑣 is squared. This means that doubling the mass will double the kinetic energy, but doubling the velocity will quadruple the energy. Let’s demonstrate this with more watermelons. Compared to the previous melon, the one on the left has the same velocity, but twice the mass, and the one on the right has the same mass, but twice the velocity.

Let’s compare the kinetic energy of each, starting with the larger watermelon on the left. The larger melon’s mass is 20 kg and its velocity is 10 m/s. Putting them into the equation gives KEkgms=12(20)(10/), which becomes 12(20)ο€Ή100/=1000.kgmsJ

So the larger melon has 1β€Žβ€‰β€Ž000 J of kinetic energy, twice as much as the original melon’s 500 J.

Now let’s look at the smaller melon, the one on the right. Its mass is 10 kg and its velocity is 20 m/s. Putting them into the equation gives KEkgms=12(10)(20/), which becomes 12(10)ο€Ή400/=2000.kgmsJ

The smaller melon has 2β€Žβ€‰β€Ž000 J of kinetic energy, quadruple that of the original melon’s 500 J. So let’s have a look at an example.

Example 1: Finding Kinetic Energy given Mass and Velocity

An object with a mass of 1.25 kg has a velocity of 12 m/s. What is the object’s kinetic energy?


Let’s look at the KE equation again: KE=12π‘šπ‘£.

The mass of the object is 1.25 kg, and the velocity is 12 m/s. Putting these into the equation gives KEkgms=12(1.25)(12/), which simplifies to 12(1.25)ο€Ή144/=90.kgmsJ

Don’t forget about the factor of 12 out front when multiplying through! This object thus has a kinetic energy of 90 J.

Sometimes a problem will provide the kinetic energy outright but is looking for some other variable. In these cases, it becomes necessary to use algebra to isolate the variable. Let’s look an another watermelon, but this time we are only given its kinetic energy and velocity.

We are then tasked with finding its mass. It becomes necessary to manipulate the kinetic energy equation, so let’s look at how to do that, starting with the base equation: KE=12π‘šπ‘£.

We want to have mass all by itself. We can start by multiplying both sides by 2 to eliminate the factor of 12 on the right side: KEΓ—2=12π‘šπ‘£Γ—2, which simplifies to 2=π‘šπ‘£.KE

We then need to get rid of π‘£οŠ¨ on the right side. This can be done by dividing both sides by π‘£οŠ¨ as follows: 2𝑣=π‘šπ‘£π‘£.KE

The π‘£οŠ¨ divides to 1 on the right side, leaving us just π‘š: 2𝑣=π‘š.KE

Now we can begin putting our known variables into the equation to solve it. We know the KE is 800 J and the velocity is βˆ’16 m/s, so the equation will look like 2(800)(βˆ’16/)=π‘š.Jms

Though the velocity is negative, the square cancels it out, making it positive. The squaring also applies to the units for velocity, turning m/s into m2/s2: 2(800)256/=π‘š.Jms

Let’s then take the product of the factor of 2 and the kinetic energy (just multiplying them together along the top): 1600256/=π‘š.Jms

The units of joules can also be expressed as kgmsΓ—/: 1600Γ—/256/=π‘š.kgmsms

Since we have m2/s2 on both the top and the bottom, they cancel out, leaving behind only the kg. We can then easily divide to give 1600256=6.25.kgkg

The mass of the watermelon is 6.25 kg.

Let’s look at another example.

Example 2: Finding Mass given Kinetic Energy and Velocity

An object that has 9 J of kinetic energy is moving at 3 m/s. What is the mass of the object?


We seek to isolate the mass of the object, given its kinetic energy and velocity. In the example, above, we had already solved the kinetic energy equation in terms of mass as follows: 2𝑣=π‘š.KE

The kinetic energy of the object is 9 J, and the velocity is 3 m/s. Putting these into the equation yields 2(9)(3/)=π‘š.Jms

Crunching the numbers gives 18(9/)=π‘š.Jms

The units that joules are in are kgmsΓ—/, meaning that the m2/s2 units on the left side of the equation are canceled through division, leaving behind the kg. Thus, the final equation looks like 18Γ—/(9/)=2.kgmsmskg

The mass of the object is 2 kg.

Other times, kinetic energy and mass are given, but not velocity. Finding the velocity from the equation takes some extra steps compared to finding mass, since there is a square root to deal with.

You cannot solve for π‘£οŠ¨: you must solve for 𝑣 by itself, which requires taking a square root to make it just 𝑣. Let’s look at another watermelon to see how this would work.

We are given the kinetic energy of this melon, 1β€Žβ€‰β€Ž080 J, and its mass, 15 kg, but not its velocity. Let’s begin by writing out the equation for kinetic energy again: KE=12π‘šπ‘£.

We want just 𝑣 on one side of the equation, so we can begin by getting rid of the 12 term on the right side by multiplying both sides by 2: KEΓ—2=12π‘šπ‘£Γ—2, which simplifies to 2=π‘šπ‘£.KE

Dividing both sides by π‘š will get rid of the π‘š on the right side: 2π‘š=π‘šπ‘£π‘š,KE which becomes 2π‘š=𝑣.KE

We are not done yet though! Again, we cannot just take π‘£οŠ¨ as the velocity term, we need to get rid of the square. Taking the square root of both sides is what we need to do: ο„ž2π‘š=βˆšπ‘£.KE

The square root then cancels the square of 𝑣, which makes the equation ο„ž2π‘š=𝑣.KE

Now we can put our values for kinetic energy and mass into the equation to find the velocity. The kinetic energy for this watermelon is 1β€Žβ€‰β€Ž080 J and its mass is 15 kg: ο„Ÿ2(1080)15=𝑣.Jkg

Let’s take the product of the factor of 2 and the kinetic energy. We also expand out the unit for joules, kgmsΓ—/: ο„Ÿ2160Γ—/15=𝑣.kgmskg

Dividing gives ο„ž2160/15=𝑣.ms

This leaves behind the units m2/s2: √144/=𝑣.ms

When the square root is taken, it eliminates the squares of units as well, leaving behind m/s: the unit of velocity: 12/=𝑣.ms

The velocity of this watermelon is thus 12 m/s.

Let’s look at another example.

Example 3: Finding Velocity given Mass and Kinetic Energy

An object has 75 J of kinetic energy and a mass of 1.5 kg. What is the velocity of the object?


Since we are seeking the velocity of the object, we need to isolate it in the kinetic energy equation. We did this in the example above, giving us the equation ο„ž2π‘š=𝑣.KE

This object has a kinetic energy of 75 J and a mass of 1.5 kg. Putting these into the equation gives ο„Ÿ2(75)1.5=𝑣.Jkg

The unit of joules is kgmsΓ—/. Since all the units for kinetic energy and mass are inside the square root, the kg can cancel out while still inside, leaving behind only m2/s2. Let’s also multiply the factor of 2 inside the square root: ο„Ÿ150Γ—/1.5=π‘£βˆš100/=𝑣.kgmskgms

Taking the square root of m2/s2 gives just m/s: exactly the unit that velocity needs to be in. Thus, 10/=𝑣.ms

The velocity of the object is 10 m/s.

The kinetic energy equation uses two common terms in general kinematic equations, velocity and mass. Relating these terms to other equations or between different objects is commonly done in physics because of how prevalent these terms are.

Observe the two watermelons below.

The humongous melon has a known mass and velocity (and thus its kinetic energy can be found), but the tiny melon only has a known velocity. We, however, have another relation: the kinetic energy of the tiny melon is 3 times that of the humongous melon.

We can use this relation to find the kinetic energy and thus the mass of the smaller melon! Let’s write out the relation in equation form: KEKETH=3, where KEH is the kinetic energy of the humongous melon and KET is the kinetic energy of the tiny melon.

Note that the tiny melon having 3 times the KE as the humongous melon means the same as the humongous melon having 13the KE of the tiny melon. So we can also write the equation as 13=.KEKETH

We can then find the kinetic energy of the humongous melon and put it into the equation that will be the most convenient. Here’s what the separate kinetic energy equation will look like: KEHHH=12π‘šπ‘£.

The humongous melon’s mass is 40 kg and its velocity is 6 m/s. We put these into the equation as follows: KEkgmsH=12(40)(6/).

Multiplying the 12 into 40 kg and squaring the 6 m/s makes the equation KEkgmsH=(20)ο€Ή36/.

Multiplying these gives the kinetic energy in joules: KEJH=720.

So the kinetic energy of the humongous watermelon, KEH, is 720 J. We can then put this back into our first equation above as follows: KEKEKEJTHT=3=3(720), which becomes KEJT=2160.

We know the kinetic energy of the tiny melon, 2β€Žβ€‰β€Ž160 J! All we need to do now is solve for its mass. Its kinetic energy equation is KETTT=12π‘šπ‘£.

We multiply both sides by 2 to remove the factor of 12: 2=π‘šπ‘£.KETTT

Then, we divide both sides by 𝑣T to remove the 𝑣T on the right side: 2𝑣=π‘š.KETTT

Now let’s put in our values for kinetic energy, 2β€Žβ€‰β€Ž160 J, and velocity, 32 m/s: 2(2160)(32/)=π‘š.JmsT

Multiplying the factor of 2 and expanding the unit for joules, 4320Γ—/(32/)=π‘š.kgmsmsT

Then, applying the square gives 4320Γ—/1024/=π‘š.kgmsmsT

Dividing causes the m2/s2 unit to cancel out, leaving behind only kg. Rounding to 2 decimal places, this results in 43201024=4.22.kgkg

The tiny melon has a mass of 4.22 kg.

Let’s look at another example.

Example 4: Finding the Velocity of a Car Using Comparative Kinetic Energy

A motorcycle of mass 250 kg moving at 32 m/s has four times as much kinetic energy as a 640 kg mass car. What is the velocity of the car?


Let’s begin by writing a relation between the kinetic energy of the motorcycle and that of the car. We know that the motorcycle has 4 times the energy of the car: KEKEMC=4, where KEM is the kinetic energy of the motorcycle and KEC is the kinetic energy of the car.

We can then determine the value of KEM, since we already know the mass and velocity of the motorcycle: KEMMM=12π‘šπ‘£.

The mass of the motorcycle, π‘šM, is 250 kg and its velocity, 𝑣M, is 32 m/s. Thus, KEkgmsM=12250Γ—(32/), which is 12250Γ—ο€Ή1024/=128000.kgmsJ

Now we know that the kinetic energy of the motorcycle, KEM, is 128β€Žβ€‰β€Ž000 J. We can then relate this to our first equation with the kinetic energy of the car so that 128000=4.JKEC

Let’s now look at the kinetic energy of the car: KECCC=12π‘šπ‘£.

This time, we only know the mass of the car, π‘šC, which is 640 kg, but not the velocity, 𝑣C. Let’s put the mass into the equation and simplify as much as we can: 12(640)𝑣=(320)𝑣.kgkgCC

Now we can put this kinetic energy equation back into the equation above: 128000=4(320)𝑣.JkgC

Multiplying the 320 kg by 4, we have 128000=(1280)𝑣.JkgC

We then can isolate 𝑣 by dividing both sides by 1β€Žβ€‰β€Ž280 kg. Let’s also expand out the unit for joules so we can see what is canceling in the next few steps: 128000Γ—/1280=(1280)𝑣1280.kgmskgkgkgC

The masses on the right cancel out, and the kg units on the left cancel out as well, leaving 100/=𝑣.msC

Now all we have to do is remove the square of the 𝑣C by taking the square root of both sides: √100/=βˆšπ‘£,msC which simplifies to 10/=𝑣.msC

Thus, the velocity of the car is 10 m/s.

Despite being less than half the mass, the motorcycle traveling at just over thrice the speed had four times the kinetic energy of the car. This once again shows how much more influence the squared velocity term has on the kinetic energy compared to the mass.

Since kinetic energy uses velocity, it can be related to other kinematic equations such as those that use time or acceleration.

Let’s look at an example.

Example 5: The Relationship between Distance and Kinetic Energy

A body starts moving from rest with a constant acceleration. Which of the following shows the relationship between the distance (𝑑) covered by the body and its kinetic energy (KE)?


The graphs presented are in terms of distance 𝑑 and kinetic energy, so lets observe the relationship between distance and KE.

The body described in this problem has a constant acceleration, meaning its velocity is increasing constantly. When velocity is increasing, KE increases at an exponential rate, due to the π‘£οŠ¨ relationship KE=12π‘šπ‘£.

Now let’s look at the full equation for distance 𝑑 when an object is accelerating: 𝑑=π‘₯+𝑣𝑑+12π‘Žπ‘‘, where π‘₯ is initial position, π‘£οŠ¦ is initial velocity, 𝑑 is time, and π‘Ž is acceleration.

The initial position is 0, so we can eliminate it from the equation. Similarly, the initial velocity of the object is 0, since it is starting from rest. This makes the distance equation simplify to 𝑑=12π‘Žπ‘‘.

Now we see that both KE and 𝑑 are based on squared relationships: the similarity between the equations is apparent. As 𝑑 increases, the distance increases based on a square. As 𝑣 increases, the KE increases based on a square.

The increase of 𝑣 is dependent on the increase of 𝑑, and thus, neither KE nor 𝑑 increases faster than the other. Looking at A, 𝑑 is increasing faster than KE, so it cannot be this one.

Looking at C, KE is constant while distance increases. This cannot be true though, since the object is moving with a constant acceleration, so it must have an increasing velocity, which means its kinetic energy is similarly constantly increasing.

Looking at D, KE is increasing much faster than 𝑑. The kinetic energy of an object cannot increase proportionally faster than distance, since they are both dependent on velocity in the same way.

The correct answer is then B, with KE and 𝑑 increasing steadily relative to each other.

Key Points

  • The formula for kinetic energy is KE=12π‘šπ‘£οŠ¨.
  • Kinetic energy is due to motion and is a scalar quantity.
  • Doubling the mass doubles KE, and doubling the velocity quadruples KE.

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