Kinetic energy

In physics, the kinetic energy of an object is the form of energy that it possesses due to its motion. Whether something is moving slowly or very fast, it has kinetic energy because of that movement.

In classical mechanics, the kinetic energy of a non-rotating object of mass traveling at a speed can be calculated using a special formula. This shows how much energy an object has just because it is moving.

The kinetic energy of an object is equal to the work, or the force needed to get the object moving from rest to its current speed. The same amount of work is done when the object slows down and comes to a stop.

The SI unit for measuring energy is the joule, while in some other systems, the English unit is the foot-pound.

For objects moving close to the speed of light, a different way of calculating kinetic energy is needed, one that includes ideas from relativistic mechanics.

History and etymology

The word "kinetic" comes from an ancient Greek word meaning "motion." Long ago, thinkers like Aristotle talked about energy in two ways: what things are doing and what they could do.

Later, scientists like Gottfried Leibniz and Johann Bernoulli studied how motion relates to energy. They called this energy the "living force” or vis viva. Experiments by Willem 's Gravesande showed that how fast something falls affects how deep it goes into soft material. Émilie du Châtelet helped explain these findings.

In the 1800s, scientists like Thomas Young and William Thomson gave us the modern words we use today for energy in motion.

Overview

Energy comes in many forms, like chemical energy, thermal energy, and electric energy. One important type is kinetic energy, which is the energy of movement. When something moves, it has kinetic energy.

For example, when a cyclist rides a bike, they use energy from food to move. This energy turns into kinetic energy, making the cyclist and bike move. If the cyclist goes uphill, the kinetic energy turns into another type of energy called gravitational potential energy. When they go downhill, this energy turns back into kinetic energy. Kinetic energy can also turn into other forms, like heat if the cyclist uses brakes.

In space, spacecraft use energy to move very fast. This kinetic energy stays the same when they are far from Earth, but when they come back, some of it turns into heat. Kinetic energy can also pass from one object to another, like when a ball hits another ball in billiards.

Kinetic energy for non-relativistic velocity

Treatments of kinetic energy depend upon the relative velocity of objects compared to the fixed speed of light. Speeds experienced directly by humans are non-relativisitic; higher speeds require the theory of relativity.

Kinetic energy of rigid bodies

In classical mechanics, the kinetic energy of a point object (an object so small that its mass can be assumed to exist at one point), or a non-rotating rigid body depends on the mass of the body as well as its speed. The kinetic energy is equal to half the product of the mass and the square of the speed. In formula form:

E k = 1 2 m v 2 {\displaystyle E_{\text{k}}={\frac {1}{2}}mv^{2}} !{\displaystyle E_{\text{k}}={\frac {1}{2}}mv^{2}}

where m {\displaystyle m} !{\displaystyle m} is the mass and v {\displaystyle v} !{\displaystyle v} is the speed (magnitude of the velocity) of the body. In SI units, mass is measured in kilograms, speed in metres per second, and the resulting kinetic energy is in joules.

For example, one would calculate the kinetic energy of an 80 kg mass (about 180 lbs) traveling at 18 metres per second (about 40 mph, or 65 km/h) as

E k = 1 2 ⋅ 80 kg ⋅ ( 18 m/s ) 2 = 12 , 960 J = 12.96 kJ {\displaystyle E_{\text{k}}={\frac {1}{2}}\cdot 80\,{\text{kg}}\cdot \left(18\,{\text{m/s}}\right)^{2}=12,960\,{\text{J}}=12.96\,{\text{kJ}}} !{\displaystyle E_{\text{k}}={\frac {1}{2}}\cdot 80\,{\text{kg}}\cdot \left(18\,{\text{m/s}}\right)^{2}=12,960\,{\text{J}}=12.96\,{\text{kJ}}}

When a person throws a ball, the person does work on it to give it speed as it leaves the hand. The moving ball can then hit something and push it, doing work on what it hits. The kinetic energy of a moving object is equal to the work required to bring it from rest to that speed, or the work the object can do while being brought to rest: net force × displacement = kinetic energy, i.e.,

F s = 1 2 m v 2 {\displaystyle Fs={\frac {1}{2}}mv^{2}} !{\displaystyle Fs={\frac {1}{2}}mv^{2}}

Since the kinetic energy increases with the square of the speed, an object doubling its speed has four times as much kinetic energy. For example, a car traveling twice as fast as another requires four times as much distance to stop, assuming a constant braking force. As a consequence of this quadrupling, it takes four times the work to double the speed.

The kinetic energy of an object is related to its momentum by the equation: E k = p 2 2 m {\displaystyle E_{\text{k}}={\frac {p^{2}}{2m}}} !{\displaystyle E_{\text{k}}={\frac {p^{2}}{2m}}}

where:

• p {\displaystyle p} !{\displaystyle p} is momentum
• m {\displaystyle m} !{\displaystyle m} is mass of the body

For the translational kinetic energy, that is the kinetic energy associated with rectilinear motion, of a rigid body with constant mass m {\displaystyle m} !{\displaystyle m} , whose center of mass is moving in a straight line with speed v {\displaystyle v} !{\displaystyle v} , as seen above is equal to

E t = 1 2 m v 2 {\displaystyle E_{\text{t}}={\frac {1}{2}}mv^{2}} !{\displaystyle E_{\text{t}}={\frac {1}{2}}mv^{2}}

where:

• m {\displaystyle m} !{\displaystyle m} is the mass of the body
• v {\displaystyle v} !{\displaystyle v} is the speed of the center of mass of the body.

The kinetic energy of any entity depends on the reference frame in which it is measured. However, the total energy of an isolated system, i.e. one in which energy can neither enter nor leave, does not change over time in the reference frame in which it is measured. Thus, the chemical energy converted to kinetic energy by a rocket engine is divided differently between the rocket ship and its exhaust stream depending upon the chosen reference frame. This is called the Oberth effect. But the total energy of the system, including kinetic energy, fuel chemical energy, heat, etc., is conserved over time, regardless of the choice of reference frame. Different observers moving with different reference frames would however disagree on the value of this conserved energy.

The kinetic energy of such systems depends on the choice of reference frame: the reference frame that gives the minimum value of that energy is the center of momentum frame, i.e. the reference frame in which the total momentum of the system is zero. This minimum kinetic energy contributes to the invariant mass of the system as a whole.

Derivation

Without vector calculus

The work W done by a force F on an object over a distance s parallel to F equals

W = F ⋅ s . {\displaystyle W=F\cdot s.} !{\displaystyle W=F\cdot s.}

Using Newton's second law

F = m a {\displaystyle F=ma} !{\displaystyle F=ma}

with m the mass and a the acceleration of the object and

s = a t 2 2 {\displaystyle s={\frac {at^{2}}{2}}} !{\displaystyle s={\frac {at^{2}}{2}}}

the distance traveled by the accelerated object in time t, we find with v = a t {\displaystyle v=at} !{\displaystyle v=at} for the velocity v of the object

W = m a a t 2 2 = m ( a t ) 2 2 = m v 2 2 . {\displaystyle W=ma{\frac {at^{2}}{2}}={\frac {m(at)^{2}}{2}}={\frac {mv^{2}}{2}}.} !{\displaystyle W=ma{\frac {at^{2}}{2}}={\frac {m(at)^{2}}{2}}={\frac {mv^{2}}{2}}.}

With vector calculus

The work done in accelerating a particle with mass m during the infinitesimal time interval dt is given by the dot product of force F and the infinitesimal displacement dx

F ⋅ d x = F ⋅ v d t = d p d t ⋅ v d t = v ⋅ d p = v ⋅ d ( m v ) , {\displaystyle \mathbf {F} \cdot d\mathbf {x} =\mathbf {F} \cdot \mathbf {v} dt={\frac {d\mathbf {p} }{dt}}\cdot \mathbf {v} dt=\mathbf {v} \cdot d\mathbf {p} =\mathbf {v} \cdot d(m\mathbf {v} )\,,} !{\displaystyle \mathbf {F} \cdot d\mathbf {x} =\mathbf {F} \cdot \mathbf {v} dt={\frac {d\mathbf {p} }{dt}}\cdot \mathbf {v} dt=\mathbf {v} \cdot d\mathbf {p} =\mathbf {v} \cdot d(m\mathbf {v} )\,,}

where we have assumed the relationship p = m v and the validity of Newton's second law. (However, also see the special relativistic derivation below.)

Applying the product rule we see that:

d ( v ⋅ v ) = ( d v ) ⋅ v + v ⋅ ( d v ) = 2 ( v ⋅ d v ) . {\displaystyle d(\mathbf {v} \cdot \mathbf {v} )=(d\mathbf {v} )\cdot \mathbf {v} +\mathbf {v} \cdot (d\mathbf {v} )=2(\mathbf {v} \cdot d\mathbf {v} ).} !{\displaystyle d(\mathbf {v} \cdot \mathbf {v} )=(d\mathbf {v} )\cdot \mathbf {v} +\mathbf {v} \cdot (d\mathbf {v} )=2(\mathbf {v} \cdot d\mathbf {v} ).}

Therefore (assuming constant mass so that dm = 0), we have

v ⋅ d ( m v ) = m 2 d ( v ⋅ v ) = m 2 d v 2 = d ( m v 2 2 ) . {\displaystyle \mathbf {v} \cdot d(m\mathbf {v} )={\frac {m}{2}}d(\mathbf {v} \cdot \mathbf {v} )={\frac {m}{2}}dv^{2}=d\left({\frac {mv^{2}}{2}}\right).} !{\displaystyle \mathbf {v} \cdot d(m\mathbf {v} )={\frac {m}{2}}d(\mathbf {v} \cdot \mathbf {v} )={\frac {m}{2}}dv^{2}=d\left({\frac {mv^{2}}{2}}\right).}

Since this is a total differential (that is, it only depends on the final state, not how the particle got there), we can integrate it and call the result kinetic energy:

E k = ∫ v 1 v 2 p ⋅ d v = ∫ v 1 v 2 m v ⋅ d v = m v 2 2 | v 1 v 2 = 1 2 m ( v 2 2 − v 1 2 ) . {\displaystyle E_{\text{k}}=\int _{v_{1}}^{v_{2}}\mathbf {p} \cdot d\mathbf {v} =\int _{v_{1}}^{v_{2}}m\mathbf {v} \cdot d\mathbf {v} ={mv^{2} \over 2}{\bigg \vert }_{v_{1}}^{v_{2}}={1 \over 2}m(v_{2}^{2}-v_{1}^{2}).} !{\displaystyle E_{\text{k}}=\int _{v_{1}}^{v_{2}}\mathbf {p} \cdot d\mathbf {v} =\int _{v_{1}}^{v_{2}}m\mathbf {v} \cdot d\mathbf {v} ={mv^{2} \over 2}{\bigg \vert }_{v_{1}}^{v_{2}}={1 \over 2}m(v_{2}^{2}-v_{1}^{2}).}

This equation states that the kinetic energy (E_k) is equal to the integral of the dot product of the momentum (p) of a body and the infinitesimal change of the velocity (v) of the body. It is assumed that the body starts with no kinetic energy when it is at rest (motionless).

Rotating bodies

Main article: Rotational energy

If a rigid body Q is rotating about any line through the center of mass then it has rotational kinetic energy ( E r {\displaystyle E_{\text{r}}\,} !{\displaystyle E_{\text{r}}\,} ) which is simply the sum of the kinetic energies of its moving parts, and is thus given by:

E r = ∫ Q v 2 d m 2 = ∫ Q ( r ω ) 2 d m 2 = ω 2 2 ∫ Q r 2 d m = ω 2 2 I = 1 2 I ω 2 {\displaystyle E_{\text{r}}=\int _{Q}{\frac {v^{2}dm}{2}}=\int _{Q}{\frac {(r\omega )^{2}dm}{2}}={\frac {\omega ^{2}}{2}}\int _{Q}{r^{2}}dm={\frac {\omega ^{2}}{2}}I={\frac {1}{2}}I\omega ^{2}} !{\displaystyle E_{\text{r}}=\int _{Q}{\frac {v^{2}dm}{2}}=\int _{Q}{\frac {(r\omega )^{2}dm}{2}}={\frac {\omega ^{2}}{2}}\int _{Q}{r^{2}}dm={\frac {\omega ^{2}}{2}}I={\frac {1}{2}}I\omega ^{2}}

where:

• ω is the body's angular velocity
• r is the distance of any mass dm from that line
• I {\displaystyle I} !{\displaystyle I} is the body's moment of inertia, equal to ∫ Q r 2 d m {\textstyle \int _{Q}{r^{2}}dm} !{\textstyle \int _{Q}{r^{2}}dm} .

(In this equation the moment of inertia must be taken about an axis through the center of mass and the rotation measured by ω must be around that axis; more general equations exist for systems where the object is subject to wobble due to its eccentric shape).

Kinetic energy of systems

A system of bodies may have internal kinetic energy due to the relative motion of the bodies in the system. For example, in the Solar System the planets and planetoids are orbiting the Sun. In a tank of gas, the molecules are moving in all directions. The kinetic energy of the system is the sum of the kinetic energies of the bodies it contains.

A macroscopic body that is stationary (i.e. a reference frame has been chosen to correspond to the body's center of momentum) may have various kinds of internal energy at the molecular or atomic level, which may be regarded as kinetic energy, due to molecular translation, rotation, and vibration, electron translation and spin, and nuclear spin. These all contribute to the body's mass, as provided by the special theory of relativity. When discussing movements of a macroscopic body, the kinetic energy referred to is usually that of the macroscopic movement only. However, all internal energies of all types contribute to a body's mass, inertia, and total energy.

Fluid dynamics

In fluid dynamics, the kinetic energy per unit volume at each point in an incompressible fluid flow field is called the dynamic pressure at that point.

E k = 1 2 m v 2 {\displaystyle E_{\text{k}}={\frac {1}{2}}mv^{2}} !{\displaystyle E_{\text{k}}={\frac {1}{2}}mv^{2}}

Dividing by V, the unit of volume:

E k V = 1 2 m V v 2 q = 1 2 ρ v 2 {\displaystyle {\begin{aligned}{\frac {E_{\text{k}}}{V}}&={\frac {1}{2}}{\frac {m}{V}}v^{2}\\q&={\frac {1}{2}}\rho v^{2}\end{aligned}}} !{\displaystyle {\begin{aligned}{\frac {E_{\text{k}}}{V}}&={\frac {1}{2}}{\frac {m}{V}}v^{2}\\q&={\frac {1}{2}}\rho v^{2}\end{aligned}}}

where q {\displaystyle q} !{\displaystyle q} is the dynamic pressure, and ρ is the density of the incompressible fluid.

Frame of reference

The speed, and thus the kinetic energy of a single object is frame-dependent (relative): it can take any non-negative value, by choosing a suitable inertial frame of reference. For example, a bullet passing an observer has kinetic energy in the reference frame of this observer. The same bullet is stationary to an observer moving with the same velocity as the bullet, and so has zero kinetic energy. By contrast, the total kinetic energy of a system of objects cannot be reduced to zero by a suitable choice of the inertial reference frame, unless all the objects have the same velocity. In any other case, the total kinetic energy has a non-zero minimum, as no inertial reference frame can be chosen in which all the objects are stationary. This minimum kinetic energy contributes to the system's invariant mass, which is independent of the reference frame.

The total kinetic energy of a system depends on the inertial frame of reference: it is the sum of the total kinetic energy in a center of momentum frame and the kinetic energy the total mass would have if it were concentrated in the center of mass.

This may be simply shown: let V {\displaystyle \textstyle \mathbf {V} } !{\displaystyle \textstyle \mathbf {V} } be the relative velocity of the center of mass frame i in the frame k. Since

v 2 = ( v i + V ) 2 = ( v i + V ) ⋅ ( v i + V ) = v i ⋅ v i + 2 v i ⋅ V + V ⋅ V = v i 2 + 2 v i ⋅ V + V 2 , {\displaystyle v^{2}=\left(v_{i}+V\right)^{2}=\left(\mathbf {v} _{i}+\mathbf {V} \right)\cdot \left(\mathbf {v} _{i}+\mathbf {V} \right)=\mathbf {v} _{i}\cdot \mathbf {v} _{i}+2\mathbf {v} _{i}\cdot \mathbf {V} +\mathbf {V} \cdot \mathbf {V} =v_{i}^{2}+2\mathbf {v} _{i}\cdot \mathbf {V} +V^{2},} !{\displaystyle v^{2}=\left(v_{i}+V\right)^{2}=\left(\mathbf {v} _{i}+\mathbf {V} \right)\cdot \left(\mathbf {v} _{i}+\mathbf {V} \right)=\mathbf {v} _{i}\cdot \mathbf {v} _{i}+2\mathbf {v} _{i}\cdot \mathbf {V} +\mathbf {V} \cdot \mathbf {V} =v_{i}^{2}+2\mathbf {v} _{i}\cdot \mathbf {V} +V^{2},}

Then,

E k = ∫ v 2 2 d m = ∫ v i 2 2 d m + V ⋅ ∫ v i d m + V 2 2 ∫ d m . {\displaystyle E_{\text{k}}=\int {\frac {v^{2}}{2}}dm=\int {\frac {v_{i}^{2}}{2}}dm+\mathbf {V} \cdot \int \mathbf {v} _{i}dm+{\frac {V^{2}}{2}}\int dm.} !{\displaystyle E_{\text{k}}=\int {\frac {v^{2}}{2}}dm=\int {\frac {v_{i}^{2}}{2}}dm+\mathbf {V} \cdot \int \mathbf {v} _{i}dm+{\frac {V^{2}}{2}}\int dm.}

∫ v i 2 2 d m = E i {\textstyle \int {\frac {v_{i}^{2}}{2}}dm=E_{i}} !{\textstyle \int {\frac {v_{i}^{2}}{2}}dm=E_{i}} is the kinetic energy in the center of mass frame, and ∫ d m = M {\textstyle \int dm=M} !{\textstyle \int dm=M} is the total mass. ∫ v i d m {\textstyle \int \mathbf {v} _{i}dm} !{\textstyle \int \mathbf {v} _{i}dm} is the total momentum in the center of mass frame, which is by definition zero. Substituting, we get:

E k = E i + M V 2 2 . {\displaystyle E_{\text{k}}=E_{i}+{\frac {MV^{2}}{2}}.} !{\displaystyle E_{\text{k}}=E_{i}+{\frac {MV^{2}}{2}}.}

Thus the kinetic energy of a system is lowest to center of momentum reference frames, i.e., frames of reference in which the center of mass is stationary (either the center of mass frame or any other center of momentum frame). In any different frame of reference, there is additional kinetic energy corresponding to the total mass moving at the speed of the center of mass. The kinetic energy of the system in the center of momentum frame is a quantity that is invariant (all observers see it to be the same).

Rotation in systems

It sometimes is convenient to split the total kinetic energy of a body into the sum of the body's center-of-mass translational kinetic energy and the energy of rotation around the center of mass (rotational energy):

E k = E t + E r {\displaystyle E_{\text{k}}=E_{\text{t}}+E_{\text{r}}} !{\displaystyle E_{\text{k}}=E_{\text{t}}+E_{\text{r}}}

where:

• E_k is the total kinetic energy
• E_t is the translational kinetic energy
• E_r is the rotational energy or angular kinetic energy in the rest frame

Thus the kinetic energy of a tennis ball in flight is the kinetic energy due to its rotation, plus the kinetic energy due to its translation.

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Relativistic kinetic energy

See also: Mass in special relativity and Tests of relativistic energy and momentum

When an object moves very fast, close to the speed of light, we need special rules from physics to find its kinetic energy. In these situations, energy and momentum mix together in a special way, similar to how time and space mix in space-time.

The total energy of a moving object can be found using a special formula. If the object is standing still, this formula tells us its rest energy. By subtracting this rest energy from the total energy, we can find the kinetic energy of the object.

At everyday speeds, this kinetic energy formula simplifies to the one we use in basic physics. But for objects moving close to the speed of light, the full formula must be used.

Kinetic energy in quantum mechanics

Further information: Hamiltonian (quantum mechanics)

In quantum mechanics, things like kinetic energy are shown using special math tools called operators. For a tiny part of matter called a particle, the kinetic energy operator is part of something called the Hamiltonian. It is written using another operator called momentum.

In a simpler picture of quantum mechanics, the momentum operator uses a special math symbol. The kinetic energy operator can then be written using this symbol.

For a group of tiny parts called electrons, the total kinetic energy can be found by adding up each electron’s share. This uses a math description of where the electrons are and how they move.

There is also a way to find kinetic energy using just the density of electrons, which tells how many electrons are in each spot, without needing the full math description of their positions.