What do a car driving along the highway, a book falling to the ground, and a rocket shooting off into space all have in common? These are all objects in motion, and thus they all have kinetic energy. Any object in motion has kinetic energy, which means that the object can do work on another object. A passenger riding in a car driving along the highway is moving along with the car because the car in motion is exerting force on the passenger, bringing the passenger into motion as well. In this article, we will define kinetic energy and discuss the relationship between kinetic energy and work. We will develop a formula that describes kinetic energy and talk about the differences between kinetic energy and potential energy. We will also mention the types of kinetic energy and go over some examples.
Definition of Kinetic Energy
Using Newton's second law with force and acceleration vectors to describe the motion of an object can sometimes be difficult. Vectors can complicate equations since we have to consider both their magnitude and direction. For physics problems that are difficult to solve using force and acceleration vectors, it is a lot easier to use energy instead. Kinetic energy is the capability of an object in motion to do work. There are different types of kinetic energy such as thermal and electric kinetic energy, but in this article, we will focus on mechanical kinetic energy. The SI unit of kinetic energy is the joule, which is abbreviated with![]()
. A joule is a newton-meter, or![]()
. Kinetic energy is a scalar quantity, which makes it easier to work with than a vector. The translational kinetic energy of an object depends on the mass and speed of the object and is given by the following formula:
$$ K = \frac{1}{2} m \vec{v}^2 $$
We will discuss how we got to this equation in greater detail in the next section. From the equation, we see that the kinetic energy of an object can only be a positive quantity or zero if the object is not moving. It does not depend on the direction of motion.
Kinetic energy: the capability of an object in motion to do work.
Let's quickly review what work is so that we can better understand kinetic energy. For this article, we will focus only on constant forces acting on objects; we will cover varying forces in a different article. The work done on an object is the scalar product of the force vector acting on the object and the displacement vector.
Work: the scalar product of the force vector acting on the object and the displacement vector.
We can find the work done on an object by taking the scalar product of the force and the displacement:
$$ W = \vec{F} \cdot \vec{d} $$
If we just take the component of the force vector that is parallel to the displacement vector, we can write our formula like this:
$$ W = Fd \cos{\theta}$$
In the above equation, \(F\) is the magnitude of the force vector, \(d\) is the magnitude of the displacement vector, and \(\theta\) is the angle between the vectors. Notice that the work, like kinetic energy, is a scalar quantity.
Now that we have reviewed what work is, we can discuss how kinetic energy relates to work. As stated above, kinetic energy is the capability of an object in motion to do work. The magnitude of the change in the kinetic energy of an object is the total work done on the object:
$$ \begin{aligned} W &= \Delta K \\ &=K_2 - K_1 \end{aligned}$$
The variables \(K_1\) and \(K_2\) in this equation represent the initial kinetic energy and the final kinetic energy respectively. We can think of the equation for kinetic energy, \(K = \frac{1}{2} m \vec{v}^2 \), as the work done to bring an object from rest to its current speed.
Only the component of the force that is parallel to the displacement vector changes the kinetic energy. If the object has a force component that is perpendicular to the displacement vector, that force component can change the direction of motion without doing work on the object. For example, an object in uniform circular motion has constant kinetic energy, and the centripetal force that is perpendicular to the direction of motion keeps the object in uniform circular motion.
Consider a \(12\,\mathrm{kg}\) block that is pushed with constant force a distance of \(10\,\mathrm{m}\) at an angle of \(\theta = 35^{\circ}\) with respect to the horizontal. What is the change of kinetic energy of the block? Take the magnitude of the force from the push to be \(50\,\mathrm{N}\) and the magnitude of the friction force to be \(25\,\mathrm{N}\).
Fig. 1: A block being pushed across a surface
The change in kinetic energy is equal to the net work done on the object, so we can use the forces to find the net work. The normal force and the force from gravity are perpendicular to the displacement vector, so the work done by these forces is zero. The work done by the friction force is in the direction opposite to that of the displacement vector and is thus negative.
$$ \begin{aligned} W_f &= F_f d \cos(\theta) \\ &= -(25\,\mathrm{N})(10\,\mathrm{m}) \cos(180^{\circ}) \\ &= -250\,\mathrm{J} \end{aligned}$$
The component of the pushing force vector that is perpendicular to the displacement vector does no work on the block, but the component that is parallel to the displacement vector does positive work on the block.
$$ \begin{aligned} W_p&= F_p d \cos(\theta) \\ &= (50\,\mathrm{N})(10\,\mathrm{m}) \cos(35^{\circ}) \\ &= 410\,\mathrm{J} \end{aligned}$$
Thus the change in kinetic energy is:
$$ \begin{aligned} \Delta K &= W_{net} \\ &= W_g + W_n + W_f + W_p \\ &= 0\,\mathrm{J} + 0\,\mathrm{J} - 250\,\mathrm{J} + 410\,\mathrm{J} \\ &= 160\,\mathrm{J} \end{aligned}$$
Developing a Formula for Kinetic Energy
How did we get to the formula relating kinetic energy to work? Consider an object that has a constant force applied to it moving horizontally. We can then use the constant acceleration formula and solve for the acceleration:
$$ \begin{aligned} \vec{v}_2^2 &= \vec{v}_1^2 + 2 \vec{a}_x \vec{d} \\ \vec{a}_x &= \frac{\vec{v}_2^2 - \vec{v}_1^2}{2 \vec{d}} \end{aligned}$$
In this equation, \(\vec{v}_1\) and \(\vec{v}_2\) are the initial and final velocities, \(\vec{d}\) is the distance traveled, and \(\vec{a}_x\) is the acceleration in the direction of the displacement. Now we can multiply both sides of the equation by the mass of the object:
$$ m \vec{a}_x = \frac{m \left(\vec{v}_2^2 - \vec{v}_1^2\right)}{2 \vec{d}} $$
We recognize the left-hand side of this equation as the net force in the direction of the displacement. So, equating the left-hand side to the net force and then multiplying the distance to that side we get:
$$ \vec{F} \cdot \vec{d} = \frac{1}{2}m \vec{v}_2^2 - \frac{1}{2} m \vec{v}_1^2 $$
We can now identify the work done on the object and the final and initial kinetic energies:
$$W = K_2 - K_1$$
This equation shows us how the work done on an object is equal to the change in kinetic energy that it experiences.
So far we have only discussed the relationship between kinetic energy and work when a constant force is being applied to the object. We will discuss their relationship when there is a varying force in a later article.
Types of Kinetic Energy
We've talked in this article about translational kinetic energy. Two other types of kinetic energy are rotational kinetic energy and vibrational kinetic energy. For now, we do not need to worry about vibrational kinetic energy, but we will discuss a bit about rotational kinetic energy.
The rotational kinetic energy of a rotating, rigid body is given by:
$$K = \frac{1}{2} I \vec{\omega}^2$$
In this equation, \(I\) is the moment of inertia of the rigid body and \(\vec{\omega}\) is its angular speed. The change in rotational kinetic energy is the work done on the object, and it is found by multiplying the angular displacement, \(\Delta \theta\), and the net torque, \(\tau\):
$$ \begin{aligned} W &= \Delta K \\ &= \tau \Delta \theta \end{aligned}$$
We go into greater detail about rotational systems in the section on rotational motion.
Kinetic Energy and Potential Energy
We have discussed how kinetic energy is only dependent upon the mass of the object and its velocity. Potential energy is energy that is related to the position of the system and its internal configuration. The total mechanical energy of a system can be found by taking the sum of the kinetic and potential energies. If there are only conservative forces working on a system, then the total mechanical energy is conserved.
A quick example of this is a ball in freefall from a certain height, \(h\). We will ignore air resistance and take gravity as the only force acting on the ball. At height \(h\), the ball has gravitational potential energy. As the ball falls, the gravitational potential energy decreases until the ball hits the ground at which point it is now zero. The kinetic energy of the ball increases as it falls because its velocity is increasing. The total mechanical energy of the system remains the same at any point.
Fig. 2: Total mechanical energy of a ball in freefall.
Consider a \(1000.0\,\mathrm{kg}\) car traveling with a velocity of \(15.0\,\frac{\mathrm{m}}{\mathrm{s}}\). How much work is required for the car to accelerate to \(40\,\frac{\mathrm{m}}{\mathrm{s}}\)?
Remember that the work is equivalent to the change in kinetic energy. We can find the initial and final kinetic energies to calculate the work required. The initial kinetic energy and final kinetic energy are given by:
$$ \begin{aligned} K_1 &= \frac{1}{2} m \vec{v}_1^2 \\ &= \frac{1}{2}\left(1000.0\,\mathrm{kg}\right)\left(15.0\,\frac{\mathrm{m}}{\mathrm{s}}\right)^2 \\ &= 1.13 \times 10^5\,\mathrm{J} \\ \\ K_2 &= \frac{1}{2} m \vec{v}_2^2 \\ &= \frac{1}{2}\left(1000.0\,\mathrm{kg}\right)\left(40\,\frac{\mathrm{m}}{\mathrm{s}}\right)^2 \\ &= 8 \times 10^5\,\mathrm{J} \end{aligned}$$
Then we find the work required by finding the difference between the initial and final kinetic energies:
$$ \begin{aligned} W &= K_2 - K_1 \\ &= 8 \times 10^5\,\mathrm{J} - 1.13 \times 10^5\,\mathrm{J} \\ &= 6.87 \times 10^5\,\mathrm{J} \end{aligned}$$
Two identical sleds cross the same distance along frictionless ice. One sled is traveling with a velocity twice that of the other sled. How much greater is the kinetic energy of the sled traveling faster?
Fig. 3: Identical sleds traveling with one traveling with twice the velocity of the other.
The kinetic energy of the slower sled is given by \(K_s=\frac{1}{2}m\vec{v}^2\), and that of the faster sled is \(k_f=\frac{1}{2}m\left(2\vec{v}\right)^2 = 2m\vec{v}^2\). Taking the ratio of these, we find:
$$ \begin{aligned} \frac{K_f}{K_s} &= \frac{2m\vec{v}^2}{\frac{1}{2}m\vec{v}^2} \\ &= 4 \end{aligned}$$
Thus \(K_f = 4K_s\), so the kinetic energy of the faster sled is four times greater than that of the slower sled.
Kinetic Energy - Key takeaways
- Kinetic energy is the capability of an object in motion to do work.
- The formula for the kinetic energy of an object is given by \(K=\frac{1}{2}m\vec{v}^2\).
- The work done on an object is the change in kinetic energy. The work of each force can be found by taking the scalar product of the force vector and the displacement vector.
- Translational, rotational, and vibrational are all types of kinetic energy.
- Potential energy is energy related to the position and internal configuration of the system.
- Taking the sum of the kinetic energy and potential energy gives you the total mechanical energy of a system.
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