Lesson Explainer: Kinetic Energy Mathematics

In this explainer, we will learn how to calculate the kinetic energy of a moving particle of mass π‘š that moves with a velocity of magnitude 𝑣.

An object that is moving is said to have kinetic energy. The faster the object moves, the more kinetic energy it has.

The kinetic energy of an object can be increased by doing work on it. If a force acts on an object as that object moves along a path, we say that the force does work on the object. If the force also causes the object to accelerate, then the speed of the object will increase, so its kinetic energy will also increase.

Imagine throwing a bowling ball down the lane of a bowling alley. When you throw the ball, you exert a force on it. That force causes the ball to accelerate; hence, it does work on the ball, increasing its kinetic energy.

The kinetic energy of an object can also be decreased by doing work on it. If the force that acts on an object acts in the opposite direction to the motion of the object, the object will decelerate. The speed of the object will decrease, so its kinetic energy will decrease.

Consider an object of mass π‘š, which is moving with a velocity ⃑𝑣. Recall that the speed of an object is defined by the magnitude of its velocity. We often represent this more simply by using 𝑣 to represent speed; hence, ‖‖⃑𝑣‖‖=𝑣.

The kinetic energy of such an object can be defined in the following way.

Definition: Kinetic Energy

The kinetic energy, 𝐸K, of an object with mass π‘š and velocity ⃑𝑣 is given by 𝐸=12π‘šπ‘£,K where 𝑣 represents the speed at which the object is moving.

Looking at the above definition, we can see that the kinetic energy depends linearly on the mass of an object and depends on the square of the speed of the object.

Before moving forward, let us consider how to obtain the simplified formula shown above by using a vector definition of kinetic energy: 𝐸=12π‘šο€Ήβƒ‘π‘£β‹…βƒ‘π‘£ο….K

For this definition, we can see that we are taking the dot product of the velocity vector with itself. In general, this dot product can be represented as ⃑𝑣⋅⃑𝑣=‖‖⃑𝑣‖‖‖‖⃑𝑣‖‖(πœƒ).cos

Here, πœƒ represents, the angle between the two vectors; however, since both vectors are the same, they point in the same direction, and hence πœƒ=0: ⃑𝑣⋅⃑𝑣=‖‖⃑𝑣‖‖‖‖⃑𝑣‖‖(0)=‖‖⃑𝑣‖‖.cos

We now have that taking the dot product of a vector with itself can be found by squaring the magnitude of the vector. In our case, the magnitude of the velocity vector is its speed, and so 𝐸=12π‘šο€Ήβƒ‘π‘£β‹…βƒ‘π‘£ο…=12π‘šβ€–β€–βƒ‘π‘£β€–β€–=12π‘šπ‘£.K

Based on this definition, you will notice that we can calculate the kinetic energy of an object when given only its mass and its speed. Even if the direction in which the object is moving is unknown, the kinetic energy can still be found.

It should also be apparent that the kinetic energy is a nonnegative scalar, since both the mass and the speed of an object are themselves nonnegative scalars.

Let us take a look at some examples.

Example 1: Finding the Kinetic Energy of a Moving Body

A body of mass 940 g is moving at 4 m/s. Determine its kinetic energy.

Answer

In this question, we are given values for the mass and speed of an object and asked to find its kinetic energy. The value for the speed is given in the standard unit of metres per second, but the value for the mass is given in grams. Let’s start by converting this into the more standard unit of kilograms.

There are 1β€Žβ€‰β€Ž000 g in 1 kg, so we have to divide the value of the mass by 1β€Žβ€‰β€Ž000: 940 g = 0.94 kg.

Now we can substitute these values for the mass and speed of the object into the formula for the kinetic energy of an object: 𝐸=12π‘šπ‘£πΈ=12Γ—0.94Γ—(4/)𝐸=7.52β‹…/.KKKkgmskgms

The units kilogram-square metres per second squared are the same as the standard unit for energy, joules, which has the symbol J. So the kinetic energy of the object is 7.52 J.

In the next example, we will see that this relationship can be used in reverse. When given the kinetic energy of an object along with the object’s mass or speed, the remaining quantity can be found.

Example 2: Finding the Speed of a Moving Object given its Kinetic Energy

Given that the kinetic energy of a moving bullet of mass 135 kg, at a certain instant, was 7β€Žβ€‰β€Ž000 joules, determine its speed.

Answer

In this question, we are given values for the mass and kinetic energy of an object and asked to find its speed. First, let’s take the formula for the kinetic energy of an object and rearrange it to make 𝑣 the subject: 𝐸=12π‘šπ‘£2πΈπ‘š=π‘£ο„ž2πΈπ‘š=𝑣𝑣=ο„ž2πΈπ‘š.KKKK

Note that since we are dealing with speedβ€”a scalar quantityβ€”we will take the positive root of this square root. Now, let’s substitute in the values for 𝐸K and π‘š: 𝑣=ο„‘ο„£ο„£ο„ 2Γ—7000𝑣=700.

Since the kinetic energy of the bullet was given in joules and its mass was given in kilograms, this value for the speed is in metres per second, so the speed of the bullet is 700 m/s.

In the previous example, it is worth noting that in order to find an object’s speed using its mass and kinetic energy, we took the square root of some quantity.

Ordinarily, square roots have two valid solutions, both a positive and a negative solution. In order to understand why our answer was not 𝑣=Β±700, we must remember that speed is defined as a nonnegative scalar. This means that when finding the speed of an object, as in the method above, we can ignore the negative solution.

For our next examples, we will be working with velocity vectors. Recall that we will now need to use the vector form of the kinetic energy equation: 𝐸=12π‘šο€Ήβƒ‘π‘£β‹…βƒ‘π‘£ο…=12π‘šβ€–β€–βƒ‘π‘£β€–β€–.K

In practice, velocities will often be given in forms such as ⃑𝑣=π‘Žβƒ‘π‘–+𝑏⃑𝑗.

When trying to find the kinetic energy, taking the dot product of the vector with itself can be quickly found using a useful shortcut. The calculation reduces to taking the sum of the squares of the coefficients for each of the perpendicular unit vectors: ⃑𝑣⋅⃑𝑣=‖‖⃑𝑣‖‖=ο€»βˆšπ‘Ž+𝑏=π‘Ž+𝑏.

The following examples will use this simple shortcut.

Example 3: Finding the Kinetic Energy of a Moving Body given Its Velocity Components

A body of mass 500 g is moving at a constant velocity ⃑𝑣=ο€Ί2βƒ‘π‘–βˆ’3⃑𝑗/cms, where ⃑𝑖 and ⃑𝑗 are two perpendicular unit vectors. Find its kinetic energy.

Answer

In this question, we are given the mass of an object and its velocity as a vector in component form. We can use the formula 𝐸=12π‘šβƒ‘π‘£β‹…βƒ‘π‘£K to find the kinetic energy of the object. Let’s substitute in the values: 𝐸=12Γ—500Γ—ο€Ί2βƒ‘π‘–βˆ’3⃑𝑗⋅2βƒ‘π‘–βˆ’3⃑𝑗𝐸=12Γ—500Γ—13𝐸=3250.KKK

In this case, we do not need to do any unit conversions. Since the value for the mass was given in grams and the value for the speed was given in centimetres per second, the value for the energy is in units of ergs, with the symbol erg. So the kinetic energy of the body is 3β€Žβ€‰β€Ž250 erg.

1 erg is equal to 10 J, so we could also express this answer in joules by multiplying the value in ergs by 10. We could also have gotten an answer in joules by converting the values we were given for the mass and the velocity into the standard units of kilograms and metres per second respectively. The kinetic energy of the object in joules is 3.25Γ—10οŠͺ J.

When given the kinetic energy and the velocity of an object, we can use our formula to find its mass. As shown in example 2 of this explainer, this is demonstrated with a simple rearrangement of our formula: 𝐸=12π‘šβ€–β€–βƒ‘π‘£β€–β€–2𝐸‖‖⃑𝑣‖‖=π‘š.KK

This reverse method will be explored in the next example, but before doing this, we should take care to note that the reverse method cannot be used to find the velocity of an object when given its kinetic energy and mass. Let us attempt to rearrange our formula to understand why this is the case: 𝐸=12π‘šβ€–β€–βƒ‘π‘£β€–β€–2πΈπ‘š=β€–β€–βƒ‘π‘£β€–β€–ο„ž2πΈπ‘š=‖‖⃑𝑣‖‖.KKK

At this point, it becomes clear that we can only find the magnitude of our velocity (i.e., speed) using this information. Since kinetic energy itself is not a scalar quantity, it does not capture any information regarding the direction of movement of the object it describes.

For this reason, we cannot regain the directional information using only 𝐸K and π‘š (which are both scalar quantities). Although the magnitude of the velocity can be uniquely defined based on this information (as shown in example 2), there are an infinite number of possibilities for the direction of travel of the object. This means the velocity vector cannot be uniquely defined.

Example 4: Finding the Mass of a Body given Its Kinetic Energy and Its Velocity in Vector Form

A body is moving at a constant velocity ⃑𝑣=ο€Ί250βƒ‘π‘–βˆ’250⃑𝑗/cms, where ⃑𝑖 and ⃑𝑗 are two perpendicular unit vectors. Given that the kinetic energy of the body is 4.8 J, find the mass of the body.

Answer

To solve this question, we can use the formula for the kinetic energy, 𝐸K, of a particle given its mass, π‘š, and its velocity, ⃑𝑣: 𝐸=12π‘šβƒ‘π‘£β‹…βƒ‘π‘£.K

Since we want to find the mass, let’s first rearrange the formula to make π‘š the subject: 2𝐸⃑𝑣⋅⃑𝑣=π‘š.K

Here, we have a vector appearing in the denominator of a fraction. It is not possible to divide a scalar by a vector, but notice that we actually have a dot product of a vector with itself in the denominator. The result of a dot product is a scalar, so here we are actually dividing by a scalar.

To make sure we get a mass in kilograms, let’s convert ⃑𝑣 into units of metres per second. There are 100 cm in 1 m, so ⃑𝑣=ο€Ί2.5βƒ‘π‘–βˆ’2.5⃑𝑗/ms.

Now let’s substitute these values into the equation: π‘š=2Γ—4.8ο€Ί2.5βƒ‘π‘–βˆ’2.5⃑𝑗⋅2.5βƒ‘π‘–βˆ’2.5βƒ‘π‘—ο†π‘š=2Γ—4.812.5π‘š=0.768.

So the mass of the body is 0.768 kg or, equivalently, 768 g.

In practice, you may encounter systems with more than one moving object. When working with such systems, it may be helpful to instead use relative velocities.

The formula for kinetic energy behaves as expected when inputting relative velocities. In certain cases, it may even cut down on calculation, as finding the kinetic energy of one moving body relative to another moving body can give a clearer picture of the energy that may be transferred if the two objects were to collide. Let us take a look at one such example.

Example 5: Finding the Kinetic Energy of a Shell Fired Toward a Moving Tank given Their Velocities

A cannon fired a shell of mass 16 kg at 285 m/s toward a tank that was moving at 72 km/h in a straight line directly toward the cannon. Determine the kinetic energy of the shell relative to the motion of the tank.

Answer

In order to find the kinetic energy of the shell relative to the tank, we must first find the relative speed of the shell. The shell and the tank are moving toward each other, so to find the relative speed of the shell, we just add the speed of the tank to the speed of the shell. They are given in different units, however, so let’s convert the speed of the tank into units of metres per second: 72 km/h is the same as 72β€Žβ€‰β€Ž000 m/h, and since there are 3β€Žβ€‰β€Ž600 seconds in an hour, 72000/=(72000Γ·3600)/mhms, which is 20 m/s.

So the relative speed of the shell to the tank is 285/+20/=305/msmsms. We can now just use the kinetic energy formula to find the kinetic energy of the shell relative to the tank: 𝐸=12π‘šπ‘£πΈ=12Γ—16Γ—305𝐸=744200.KKK

Since the mass was in units of kilograms and the speed was in units of metres per second, this value for the energy is in units of joules, so the kinetic energy of the shell is 744β€Žβ€‰β€Ž200 J.

Finally, here are the key points for the information presented in this explainer.

Key Points

  • Kinetic energy is a nonnegative scalar quantity.
  • The kinetic energy, 𝐸K, of an object with mass π‘š and speed 𝑣 is given by 𝐸=12π‘šπ‘£.K
  • If instead we are given the velocity of an object, ⃑𝑣, the kinetic energy can be found using 𝐸=12π‘šο€Ήβƒ‘π‘£β‹…βƒ‘π‘£ο…=12π‘šβ€–β€–βƒ‘π‘£β€–β€–.K
  • When given the kinetic energy and the speed (or velocity) of an object, we can find its mass.
  • When given the kinetic energy and the mass of an object, we can find its speed, but we cannot obtain the direction of motion. This means the velocity cannot be found.

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