Physics Book - User contributions [en]

Relativistic Momentum

2026-10-30T05:36:39Z

Lheiman07:

This page gives the relativistic definition of linear momentum and compares it to the traditional definition of linear momentum.

==The Main Idea==

The [[Linear Momentum]] of an object is traditionally defined as <math>\vec{p} = m \vec{v}</math>. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to [[Newton's Second Law: the Momentum Principle]], and in certain situations, [[Conservation of Momentum|the momentum of a system is conserved]]. However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:

<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>

where <math>\vec{p}</math> is the momentum of the particle, <math>m</math> is mass, <math>\vec{v}</math> is the velocity of the particle, <math>v</math> is the magnitude of the velocity (the speed of the particle), and <math>c</math> is the speed of light (about <math> 3 * 10^8 </math> m/s).

The factor <math>\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} </math> appears in many other relativistic formulas, so it has been given a name (the Lorentz factor) and a symbol (<math>\gamma</math>). This turns the relativistic momentum formula into

<math> \vec{p} = \gamma m \vec{v} </math>.

The Lorentz factor is a ratio and therefore has no units.

Note: sometimes, particularly in older textbooks, the factor <math>\gamma * m</math> is considered the "relativistic mass" of the particle. This would reduce the relativistic momentum formula to <math> \vec{p} = m_{rel} * \vec{v} </math>. However, this way of thinking is no longer favored, since the mass of a moving particle cannot be measured independently from its momentum, and there is therefore no evidence to suggest that a particle's mass increases as its speed approaches the speed of light. It is recommended that you use the formula <math> \vec{p} = \gamma m \vec{v} </math>, keeping <math>\gamma</math> and <math>m</math> separate and using the particle's rest mass (traditional notion of mass) for m.

===A Mathematical Model===

Let us mathematically analyze the equations

<math> \vec{p} = \gamma m \vec{v} </math>

and

<math>\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} </math>.

One observation worth noting is that in the limit where the speed of the particle <math>v</math> is much smaller than the speed of light <math>c</math>, the fraction <math>\frac{v^2}{c^2}</math> is approximately 0 and so <math>\gamma</math> is approximately 1. This explains why the traditional equation <math>\vec{p} = m \vec{v}</math> works very well for objects moving at everyday speeds; the relativistic definition and the traditional definition agree when <math>\gamma</math> is 1. The relativistic definition should be used if <math>v > \frac{1}{10} c</math>; for slower particles, the error caused by using the traditional formula is negligible.

Another observation worth noting is that in the limit where the speed of the particle <math>v</math> approaches the speed of light <math>c</math>, the fraction <math>\frac{v^2}{c^2}</math> approaches 1 and so <math>\gamma</math> approaches infinity. This has far-reaching implications. An object of non-zero mass traveling at the speed of light would have infinite momentum according to the relativistic definition, so an infinite impulse must be applied to a particle at rest to change its speed to the speed of light. As real forces are always finite in both magnitude and duration, no particle of nonzero mass can ever achieve the speed of light. Let us analyze what happens if a constant force is applied to a particle at rest for a long time. The particle's momentum will increase linearly over time as a result of the force, as described by Newton's second law. At first, this means its speed increases linearly over time as well. However, as the particle enters the relativistic regime, <math>\gamma</math> begins to increase as well, meaning that the particle's speed doesn't have to increase as much to sustain the steady increase in momentum. Finally, when the particle's speed is nearly the speed of light, even the tiniest increases in <math>v</math> cause <math>\gamma</math> to skyrocket, meaning that the particle hardly speeds up while the particle's momentum continues to steadily increase. The speed of the particle would asymptotically approach the speed of light over time if the force continues to be applied.

===Relativistic Kinetic Energy and Total Energy===

In classical mechanics, the kinetic energy of an object is defined as

:<math>K = \frac{1}{2}mv^2</math>

However, this equation becomes inaccurate when an object moves at speeds comparable to the speed of light. Similar to momentum, kinetic energy must be modified to account for relativistic effects.

The relativistic definition of total energy is

:<math>E = \gamma mc^2</math>

where <math>\gamma</math> is the Lorentz factor, <math>m</math> is the rest mass of the particle, and <math>c</math> is the speed of light.

An important distinction is that this equation represents the TOTAL energy of a particle, not just its kinetic energy. Even when a particle is completely at rest, it possesses energy due to its mass. This is called rest energy and is described by Einstein's famous equation:

:<math>E_0 = mc^2</math>

The relativistic kinetic energy is the difference between the total energy and the rest energy:

:<math>K = E - E_0</math>

Substituting the equations for total energy and rest energy gives

:<math>K = \gamma mc^2 - mc^2</math>

Factoring out <math>mc^2</math> gives the relativistic kinetic energy equation:

:<math>\boxed{K = (\gamma - 1)mc^2}</math>

This equation demonstrates that an object's kinetic energy increases as its velocity increases and the Lorentz factor becomes larger.

'''Relationship Between Kinetic Energy and Total Energy'''

The total energy of a relativistic particle can be expressed as the sum of its rest energy and kinetic energy:

:<math>E = K + E_0</math>

:<math>E = (\gamma - 1)mc^2 + mc^2</math>

Simplifying:

:<math>\boxed{E = \gamma mc^2}</math>

This relationship is important because it demonstrates that the total energy of a particle consists of both its rest energy and its kinetic energy.

'''What Happens at Different Velocities?'''

There are two important limiting cases to consider:

* '''When an object is at rest:''' Its velocity is zero, meaning <math>\gamma = 1</math>. Therefore, its kinetic energy is zero, but it still possesses rest energy equal to <math>mc^2</math>.

* '''When an object approaches the speed of light:''' The Lorentz factor increases dramatically. Consequently, the kinetic energy and total energy increase without bound as the velocity approaches the speed of light. This is another reason why an object with nonzero rest mass cannot reach the speed of light.

'''Connection to Classical Kinetic Energy'''

At everyday speeds, where <math>v \ll c</math>, the relativistic kinetic energy equation approaches the classical kinetic energy equation.

In this limit, the Lorentz factor can be approximated as

:<math>\gamma \approx 1 + \frac{v^2}{2c^2}</math>

Substituting this approximation into the relativistic kinetic energy equation gives

:<math>K = (\gamma - 1)mc^2</math>

:<math>K \approx \left(1 + \frac{v^2}{2c^2} - 1\right)mc^2</math>

Simplifying:

:<math>K \approx \frac{1}{2}mv^2</math>

This demonstrates that classical kinetic energy is simply an approximation of relativistic kinetic energy when an object's speed is much smaller than the speed of light.

==Examples==

===Simple===

Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a speed of <math> .97c </math> 
 What is the magnitude of the momentum of the proton?

Solution:

:<math> \frac{v}{c} = \frac{.97c}{c}</math> <math> = .97 </math>
:<math> \gamma = \frac{1}{\sqrt{1-(.97)^2}} = 4.1135 </math>
::Plug values in
:<math> p = \gamma * m*v = (4.1135)*(1.7*10^{-27})*(.97*(3*10^8)) = 2.035 * 10^{-18} </math> kg*m/s

===Middling===

Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a velocity <math> <1 * 10^7 , 2 * 10^7 , 3 * 10^7> </math> m/s. 
 What is the momentum of the proton?

Solution:

:<math> \left\vert \overrightarrow{v} \right\vert = \sqrt{(1*10^7)^2 + (2*10^7)^2 +(3*10^7)^2} </math> m/s <math> = 3.7 * 10^7 </math> m/s.
:<math> \frac{\left\vert \overrightarrow{v} \right\vert}{c} = \frac{3.7 * 10^7 m/s}{3 * 10^8 m/s}</math> <math> = .12 </math>
:<math> \gamma = \frac{1}{\sqrt{1-(.12)^2}} = 1.007 </math>
:<math> \overrightarrow{p} = \gamma * m * \overrightarrow{v} </math>
::Plug values in
:<math> \overrightarrow{p} = (1.007)*(1.7*10^{-27})*<1 * 10^7 , 2 * 10^7 , 3 * 10^7> = <1.7 * 10^{-20}, 3.4 * 10^{-20}, 5.1 *10^{-20}> </math> kg*m/s

===Middling===

A proton has a rest mass of <math>1.67 \times 10^{-27}</math> kg and is traveling at <math>0.80c</math>, where <math>c = 3.00 \times 10^8</math> m/s.

'''(a)''' Calculate the Lorentz factor of the proton.

'''(b)''' Calculate the relativistic kinetic energy of the proton.

'''(c)''' Calculate the total relativistic energy of the proton.

'''Solution:'''

'''(a)''' First, calculate the Lorentz factor using

:<math>\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}</math>

Substituting the given velocity:

:<math>\gamma = \frac{1}{\sqrt{1-\frac{(0.80c)^2}{c^2}}}</math>

:<math>= \frac{1}{\sqrt{1-0.64}}</math>

:<math>= \frac{1}{\sqrt{0.36}}</math>

:<math>\boxed{\gamma = 1.667}</math>

'''(b)''' Calculate the relativistic kinetic energy using

:<math>K = (\gamma - 1)mc^2</math>

Substituting the known values:

:<math>K = (1.667-1)(1.67 \times 10^{-27})(3.00 \times 10^8)^2</math>

:<math>K = (0.667)(1.67 \times 10^{-27})(9.00 \times 10^{16})</math>

:<math>\boxed{K = 1.00 \times 10^{-10} \ J}</math>

'''(c)''' Calculate the total relativistic energy using

:<math>E = \gamma mc^2</math>

:<math>E = (1.667)(1.67 \times 10^{-27})(3.00 \times 10^8)^2</math>

:<math>\boxed{E = 2.50 \times 10^{-10} \ J}</math>

We can verify this answer by adding the kinetic energy and rest energy:

:<math>E = K + mc^2</math>

:<math>E = 1.00 \times 10^{-10} + 1.503 \times 10^{-10}</math>

:<math>\boxed{E = 2.50 \times 10^{-10} \ J}</math>

This demonstrates that the total relativistic energy of a particle is the sum of its rest energy and relativistic kinetic energy.

===Difficult===

Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving at a speed of <math>.95c</math>. A force of magnitude <math>2 * 10^{-17}</math>N acts on it for .5s in the direction of the proton's velocity, causing it to speed up. What is its new speed at the end of the .5s time interval? Use <math>3*10^8</math> for <math>c</math>.

Solution:

Let us first find the initial momentum of the particle:

<math> \vec{p} = \gamma m \vec{v} </math>

<math> \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-.95^2}} = 3.203 </math>

<math> \vec{p} = 3.203 * 1.7 * 10^{-27} * (.95 * 3 * 10^8) = 1.55 * 10^{-18} </math> kg*m/s

Now let us find the momentum at the end of the impulse using the [[Impulse and Momentum|impulse-momentum theorem]].

<math> \vec{p}_f = \vec{p}_i + \vec{J} </math>

Since the impulse is in the same direction as the velocity, this vector addition becomes scalar (regular) addition.

<math> \vec{p}_f = 1.55 * 10^{-18} + .5 * 2 * 10^{-17} = 1.155 * 10^{-17} </math> kg*m/s

Now let us solve for the final speed using

<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>

<math> 1.155 * 10^{-17} = \frac{1}{\sqrt{1-\frac{v^2}{(3 * 10^8)^2}}} * 1.7 * 10^{-27} * v </math>

It is impossible to algebraically solve for <math>v</math>, but using an equation solver yields

<math>v = 2.997 * 10^8 </math> m/s.

This demonstrates that when a particle is moving at nearly the speed of light, even a large impulse cannot cause it to meet or exceed the speed of light. An impulse of <math>10^{-17}</math> N*s may not seem very large, but it increased the particle's momentum by approximately a factor of 10.

==Connectedness==

Relativistic momentum has applications in quantum physics, particle physics, and astrophysics. In these branches of physics, particles routinely travel at relativistic speeds, and often, their momenta are directly measurable by the particles' interactions with matter. In industry, relativistic momentum is mostly used by space flight institutions.

==History==

In 1905, [[Albert Einstein]] published his Special Theory of Relativity. This theory set the "speed limit" for the universe at the speed of light and redefined many properties such as momentum and energy for fast-moving objects. As particles' speeds approach the speed of light, the laws of physics change even for concepts as fundamental as space and time. After this discovery, physicists were better able to understand the motion of fast-moving particles. Einstein published the more comprehensive General Theory of Relativity in 1915, which allowed differently accelerating frames of reference to be reconciled. Some aspects of relativity have since been used in a variety of technologies, including GPS.

== See also ==

*[[Linear Momentum]]
*[[Newton's Second Law: the Momentum Principle]]
*[[Impulse and Momentum]]
*[[Mass]]
*[[Vectors]]

===External Links===

*http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html

==References==
*https://opentextbc.ca/physicstestbook2/chapter/relativistic-momentum/

[[Category:Momentum]]

Relativistic Momentum

2026-10-30T05:35:51Z

Lheiman07:

This page gives the relativistic definition of linear momentum and compares it to the traditional definition of linear momentum.

==The Main Idea==

The [[Linear Momentum]] of an object is traditionally defined as <math>\vec{p} = m \vec{v}</math>. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to [[Newton's Second Law: the Momentum Principle]], and in certain situations, [[Conservation of Momentum|the momentum of a system is conserved]]. However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:

<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>

where <math>\vec{p}</math> is the momentum of the particle, <math>m</math> is mass, <math>\vec{v}</math> is the velocity of the particle, <math>v</math> is the magnitude of the velocity (the speed of the particle), and <math>c</math> is the speed of light (about <math> 3 * 10^8 </math> m/s).

The factor <math>\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} </math> appears in many other relativistic formulas, so it has been given a name (the Lorentz factor) and a symbol (<math>\gamma</math>). This turns the relativistic momentum formula into

<math> \vec{p} = \gamma m \vec{v} </math>.

The Lorentz factor is a ratio and therefore has no units.

Note: sometimes, particularly in older textbooks, the factor <math>\gamma * m</math> is considered the "relativistic mass" of the particle. This would reduce the relativistic momentum formula to <math> \vec{p} = m_{rel} * \vec{v} </math>. However, this way of thinking is no longer favored, since the mass of a moving particle cannot be measured independently from its momentum, and there is therefore no evidence to suggest that a particle's mass increases as its speed approaches the speed of light. It is recommended that you use the formula <math> \vec{p} = \gamma m \vec{v} </math>, keeping <math>\gamma</math> and <math>m</math> separate and using the particle's rest mass (traditional notion of mass) for m.

===A Mathematical Model===

Let us mathematically analyze the equations

<math> \vec{p} = \gamma m \vec{v} </math>

and

<math>\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} </math>.

One observation worth noting is that in the limit where the speed of the particle <math>v</math> is much smaller than the speed of light <math>c</math>, the fraction <math>\frac{v^2}{c^2}</math> is approximately 0 and so <math>\gamma</math> is approximately 1. This explains why the traditional equation <math>\vec{p} = m \vec{v}</math> works very well for objects moving at everyday speeds; the relativistic definition and the traditional definition agree when <math>\gamma</math> is 1. The relativistic definition should be used if <math>v > \frac{1}{10} c</math>; for slower particles, the error caused by using the traditional formula is negligible.

Another observation worth noting is that in the limit where the speed of the particle <math>v</math> approaches the speed of light <math>c</math>, the fraction <math>\frac{v^2}{c^2}</math> approaches 1 and so <math>\gamma</math> approaches infinity. This has far-reaching implications. An object of non-zero mass traveling at the speed of light would have infinite momentum according to the relativistic definition, so an infinite impulse must be applied to a particle at rest to change its speed to the speed of light. As real forces are always finite in both magnitude and duration, no particle of nonzero mass can ever achieve the speed of light. Let us analyze what happens if a constant force is applied to a particle at rest for a long time. The particle's momentum will increase linearly over time as a result of the force, as described by Newton's second law. At first, this means its speed increases linearly over time as well. However, as the particle enters the relativistic regime, <math>\gamma</math> begins to increase as well, meaning that the particle's speed doesn't have to increase as much to sustain the steady increase in momentum. Finally, when the particle's speed is nearly the speed of light, even the tiniest increases in <math>v</math> cause <math>\gamma</math> to skyrocket, meaning that the particle hardly speeds up while the particle's momentum continues to steadily increase. The speed of the particle would asymptotically approach the speed of light over time if the force continues to be applied.

===Relativistic Kinetic Energy and Total Energy===

In classical mechanics, the kinetic energy of an object is defined as

:<math>K = \frac{1}{2}mv^2</math>

However, this equation becomes inaccurate when an object moves at speeds comparable to the speed of light. Similar to momentum, kinetic energy must be modified to account for relativistic effects.

The relativistic definition of total energy is

:<math>E = \gamma mc^2</math>

where <math>\gamma</math> is the Lorentz factor, <math>m</math> is the rest mass of the particle, and <math>c</math> is the speed of light.

An important distinction is that this equation represents the TOTAL energy of a particle, not just its kinetic energy. Even when a particle is completely at rest, it possesses energy due to its mass. This is called rest energy and is described by Einstein's famous equation:

:<math>E_0 = mc^2</math>

The relativistic kinetic energy is the difference between the total energy and the rest energy:

:<math>K = E - E_0</math>

Substituting the equations for total energy and rest energy gives

:<math>K = \gamma mc^2 - mc^2</math>

Factoring out <math>mc^2</math> gives the relativistic kinetic energy equation:

:<math>\boxed{K = (\gamma - 1)mc^2}</math>

This equation demonstrates that an object's kinetic energy increases as its velocity increases and the Lorentz factor becomes larger.

'''Relationship Between Kinetic Energy and Total Energy'''

The total energy of a relativistic particle can be expressed as the sum of its rest energy and kinetic energy:

:<math>E = K + E_0</math>

:<math>E = (\gamma - 1)mc^2 + mc^2</math>

Simplifying:

:<math>\boxed{E = \gamma mc^2}</math>

This relationship is important because it demonstrates that the total energy of a particle consists of both its rest energy and its kinetic energy.

'''What Happens at Different Velocities?'''

There are two important limiting cases to consider:

* '''When an object is at rest:''' Its velocity is zero, meaning <math>\gamma = 1</math>. Therefore, its kinetic energy is zero, but it still possesses rest energy equal to <math>mc^2</math>.

* '''When an object approaches the speed of light:''' The Lorentz factor increases dramatically. Consequently, the kinetic energy and total energy increase without bound as the velocity approaches the speed of light. This is another reason why an object with nonzero rest mass cannot reach the speed of light.

'''Connection to Classical Kinetic Energy'''

At everyday speeds, where <math>v \ll c</math>, the relativistic kinetic energy equation approaches the classical kinetic energy equation.

In this limit, the Lorentz factor can be approximated as

:<math>\gamma \approx 1 + \frac{v^2}{2c^2}</math>

Substituting this approximation into the relativistic kinetic energy equation gives

:<math>K = (\gamma - 1)mc^2</math>

:<math>K \approx \left(1 + \frac{v^2}{2c^2} - 1\right)mc^2</math>

Simplifying:

:<math>K \approx \frac{1}{2}mv^2</math>

This demonstrates that classical kinetic energy is simply an approximation of relativistic kinetic energy when an object's speed is much smaller than the speed of light.

==Examples==

===Simple===

Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a speed of <math> .97c </math> 
 What is the magnitude of the momentum of the proton?

Solution:

:<math> \frac{v}{c} = \frac{.97c}{c}</math> <math> = .97 </math>
:<math> \gamma = \frac{1}{\sqrt{1-(.97)^2}} = 4.1135 </math>
::Plug values in
:<math> p = \gamma * m*v = (4.1135)*(1.7*10^{-27})*(.97*(3*10^8)) = 2.035 * 10^{-18} </math> kg*m/s

===Middling===

Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a velocity <math> <1 * 10^7 , 2 * 10^7 , 3 * 10^7> </math> m/s. 
 What is the momentum of the proton?

Solution:

:<math> \left\vert \overrightarrow{v} \right\vert = \sqrt{(1*10^7)^2 + (2*10^7)^2 +(3*10^7)^2} </math> m/s <math> = 3.7 * 10^7 </math> m/s.
:<math> \frac{\left\vert \overrightarrow{v} \right\vert}{c} = \frac{3.7 * 10^7 m/s}{3 * 10^8 m/s}</math> <math> = .12 </math>
:<math> \gamma = \frac{1}{\sqrt{1-(.12)^2}} = 1.007 </math>
:<math> \overrightarrow{p} = \gamma * m * \overrightarrow{v} </math>
::Plug values in
:<math> \overrightarrow{p} = (1.007)*(1.7*10^{-27})*<1 * 10^7 , 2 * 10^7 , 3 * 10^7> = <1.7 * 10^{-20}, 3.4 * 10^{-20}, 5.1 *10^{-20}> </math> kg*m/s

===Relativistic Kinetic Energy and Total Energy===

In classical mechanics, the kinetic energy of an object is defined as

:<math>K = \frac{1}{2}mv^2</math>

However, this equation becomes inaccurate when an object moves at speeds comparable to the speed of light. Similar to momentum, kinetic energy must be modified to account for relativistic effects.

The relativistic definition of total energy is

:<math>E = \gamma mc^2</math>

where <math>\gamma</math> is the Lorentz factor, <math>m</math> is the rest mass of the particle, and <math>c</math> is the speed of light.

An important distinction is that this equation represents the TOTAL energy of a particle, not just its kinetic energy. Even when a particle is completely at rest, it possesses energy due to its mass. This is called rest energy and is described by Einstein's famous equation:

:<math>E_0 = mc^2</math>

The relativistic kinetic energy is the difference between the total energy and the rest energy:

:<math>K = E - E_0</math>

Substituting the equations for total energy and rest energy gives

:<math>K = \gamma mc^2 - mc^2</math>

Factoring out <math>mc^2</math> gives the relativistic kinetic energy equation:

:<math>\boxed{K = (\gamma - 1)mc^2}</math>

This equation demonstrates that an object's kinetic energy increases as its velocity increases and the Lorentz factor becomes larger.

'''Relationship Between Kinetic Energy and Total Energy'''

The total energy of a relativistic particle can be expressed as the sum of its rest energy and kinetic energy:

:<math>E = K + E_0</math>

:<math>E = (\gamma - 1)mc^2 + mc^2</math>

Simplifying:

:<math>\boxed{E = \gamma mc^2}</math>

This relationship is important because it demonstrates that the total energy of a particle consists of both its rest energy and its kinetic energy.

'''What Happens at Different Velocities?'''

There are two important limiting cases to consider:

* '''When an object is at rest:''' Its velocity is zero, meaning <math>\gamma = 1</math>. Therefore, its kinetic energy is zero, but it still possesses rest energy equal to <math>mc^2</math>.

* '''When an object approaches the speed of light:''' The Lorentz factor increases dramatically. Consequently, the kinetic energy and total energy increase without bound as the velocity approaches the speed of light. This is another reason why an object with nonzero rest mass cannot reach the speed of light.

'''Connection to Classical Kinetic Energy'''

At everyday speeds, where <math>v \ll c</math>, the relativistic kinetic energy equation approaches the classical kinetic energy equation.

In this limit, the Lorentz factor can be approximated as

:<math>\gamma \approx 1 + \frac{v^2}{2c^2}</math>

Substituting this approximation into the relativistic kinetic energy equation gives

:<math>K = (\gamma - 1)mc^2</math>

:<math>K \approx \left(1 + \frac{v^2}{2c^2} - 1\right)mc^2</math>

Simplifying:

:<math>K \approx \frac{1}{2}mv^2</math>

This demonstrates that classical kinetic energy is simply an approximation of relativistic kinetic energy when an object's speed is much smaller than the speed of light.

===Difficult===

Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving at a speed of <math>.95c</math>. A force of magnitude <math>2 * 10^{-17}</math>N acts on it for .5s in the direction of the proton's velocity, causing it to speed up. What is its new speed at the end of the .5s time interval? Use <math>3*10^8</math> for <math>c</math>.

Solution:

Let us first find the initial momentum of the particle:

<math> \vec{p} = \gamma m \vec{v} </math>

<math> \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-.95^2}} = 3.203 </math>

<math> \vec{p} = 3.203 * 1.7 * 10^{-27} * (.95 * 3 * 10^8) = 1.55 * 10^{-18} </math> kg*m/s

Now let us find the momentum at the end of the impulse using the [[Impulse and Momentum|impulse-momentum theorem]].

<math> \vec{p}_f = \vec{p}_i + \vec{J} </math>

Since the impulse is in the same direction as the velocity, this vector addition becomes scalar (regular) addition.

<math> \vec{p}_f = 1.55 * 10^{-18} + .5 * 2 * 10^{-17} = 1.155 * 10^{-17} </math> kg*m/s

Now let us solve for the final speed using

<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>

<math> 1.155 * 10^{-17} = \frac{1}{\sqrt{1-\frac{v^2}{(3 * 10^8)^2}}} * 1.7 * 10^{-27} * v </math>

It is impossible to algebraically solve for <math>v</math>, but using an equation solver yields

<math>v = 2.997 * 10^8 </math> m/s.

This demonstrates that when a particle is moving at nearly the speed of light, even a large impulse cannot cause it to meet or exceed the speed of light. An impulse of <math>10^{-17}</math> N*s may not seem very large, but it increased the particle's momentum by approximately a factor of 10.

==Connectedness==

Relativistic momentum has applications in quantum physics, particle physics, and astrophysics. In these branches of physics, particles routinely travel at relativistic speeds, and often, their momenta are directly measurable by the particles' interactions with matter. In industry, relativistic momentum is mostly used by space flight institutions.

==History==

In 1905, [[Albert Einstein]] published his Special Theory of Relativity. This theory set the "speed limit" for the universe at the speed of light and redefined many properties such as momentum and energy for fast-moving objects. As particles' speeds approach the speed of light, the laws of physics change even for concepts as fundamental as space and time. After this discovery, physicists were better able to understand the motion of fast-moving particles. Einstein published the more comprehensive General Theory of Relativity in 1915, which allowed differently accelerating frames of reference to be reconciled. Some aspects of relativity have since been used in a variety of technologies, including GPS.

== See also ==

*[[Linear Momentum]]
*[[Newton's Second Law: the Momentum Principle]]
*[[Impulse and Momentum]]
*[[Mass]]
*[[Vectors]]

===External Links===

*http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html

==References==
*https://opentextbc.ca/physicstestbook2/chapter/relativistic-momentum/

[[Category:Momentum]]

Relativistic Momentum

2026-10-30T05:34:42Z

Lheiman07:

This page gives the relativistic definition of linear momentum and compares it to the traditional definition of linear momentum.

==The Main Idea==

The [[Linear Momentum]] of an object is traditionally defined as <math>\vec{p} = m \vec{v}</math>. Momentum is a useful quantity because many other concepts relate to it. For example, forces cause the momentum of a system to change according to [[Newton's Second Law: the Momentum Principle]], and in certain situations, [[Conservation of Momentum|the momentum of a system is conserved]]. However, for systems containing objects moving at speeds comparable to the speed of light, both Newton's second law and conservation of momentum appear to be violated. As it turns out, if the definition of the momentum of a particle is adjusted to accommodate high-speed objects, these relations hold true after all. The adjusted definition of momentum is called the relativistic definition and defines momentum as follows:

<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>

where <math>\vec{p}</math> is the momentum of the particle, <math>m</math> is mass, <math>\vec{v}</math> is the velocity of the particle, <math>v</math> is the magnitude of the velocity (the speed of the particle), and <math>c</math> is the speed of light (about <math> 3 * 10^8 </math> m/s).

The factor <math>\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} </math> appears in many other relativistic formulas, so it has been given a name (the Lorentz factor) and a symbol (<math>\gamma</math>). This turns the relativistic momentum formula into

<math> \vec{p} = \gamma m \vec{v} </math>.

The Lorentz factor is a ratio and therefore has no units.

Note: sometimes, particularly in older textbooks, the factor <math>\gamma * m</math> is considered the "relativistic mass" of the particle. This would reduce the relativistic momentum formula to <math> \vec{p} = m_{rel} * \vec{v} </math>. However, this way of thinking is no longer favored, since the mass of a moving particle cannot be measured independently from its momentum, and there is therefore no evidence to suggest that a particle's mass increases as its speed approaches the speed of light. It is recommended that you use the formula <math> \vec{p} = \gamma m \vec{v} </math>, keeping <math>\gamma</math> and <math>m</math> separate and using the particle's rest mass (traditional notion of mass) for m.

===A Mathematical Model===

Let us mathematically analyze the equations

<math> \vec{p} = \gamma m \vec{v} </math>

and

<math>\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} </math>.

One observation worth noting is that in the limit where the speed of the particle <math>v</math> is much smaller than the speed of light <math>c</math>, the fraction <math>\frac{v^2}{c^2}</math> is approximately 0 and so <math>\gamma</math> is approximately 1. This explains why the traditional equation <math>\vec{p} = m \vec{v}</math> works very well for objects moving at everyday speeds; the relativistic definition and the traditional definition agree when <math>\gamma</math> is 1. The relativistic definition should be used if <math>v > \frac{1}{10} c</math>; for slower particles, the error caused by using the traditional formula is negligible.

Another observation worth noting is that in the limit where the speed of the particle <math>v</math> approaches the speed of light <math>c</math>, the fraction <math>\frac{v^2}{c^2}</math> approaches 1 and so <math>\gamma</math> approaches infinity. This has far-reaching implications. An object of non-zero mass traveling at the speed of light would have infinite momentum according to the relativistic definition, so an infinite impulse must be applied to a particle at rest to change its speed to the speed of light. As real forces are always finite in both magnitude and duration, no particle of nonzero mass can ever achieve the speed of light. Let us analyze what happens if a constant force is applied to a particle at rest for a long time. The particle's momentum will increase linearly over time as a result of the force, as described by Newton's second law. At first, this means its speed increases linearly over time as well. However, as the particle enters the relativistic regime, <math>\gamma</math> begins to increase as well, meaning that the particle's speed doesn't have to increase as much to sustain the steady increase in momentum. Finally, when the particle's speed is nearly the speed of light, even the tiniest increases in <math>v</math> cause <math>\gamma</math> to skyrocket, meaning that the particle hardly speeds up while the particle's momentum continues to steadily increase. The speed of the particle would asymptotically approach the speed of light over time if the force continues to be applied.

===Relativistic Kinetic Energy and Total Energy===

In classical mechanics, the kinetic energy of an object is defined as

:<math>K = \frac{1}{2}mv^2</math>

However, this equation becomes inaccurate when an object moves at speeds comparable to the speed of light. Similar to momentum, kinetic energy must be modified to account for relativistic effects.

The relativistic definition of total energy is

:<math>E = \gamma mc^2</math>

where <math>\gamma</math> is the Lorentz factor, <math>m</math> is the rest mass of the particle, and <math>c</math> is the speed of light.

An important distinction is that this equation represents the TOTAL energy of a particle, not just its kinetic energy. Even when a particle is completely at rest, it possesses energy due to its mass. This is called rest energy and is described by Einstein's famous equation:

:<math>E_0 = mc^2</math>

The relativistic kinetic energy is the difference between the total energy and the rest energy:

:<math>K = E - E_0</math>

Substituting the equations for total energy and rest energy gives

:<math>K = \gamma mc^2 - mc^2</math>

Factoring out <math>mc^2</math> gives the relativistic kinetic energy equation:

:<math>\boxed{K = (\gamma - 1)mc^2}</math>

This equation demonstrates that an object's kinetic energy increases as its velocity increases and the Lorentz factor becomes larger.

'''Relationship Between Kinetic Energy and Total Energy'''

The total energy of a relativistic particle can be expressed as the sum of its rest energy and kinetic energy:

:<math>E = K + E_0</math>

:<math>E = (\gamma - 1)mc^2 + mc^2</math>

Simplifying:

:<math>\boxed{E = \gamma mc^2}</math>

This relationship is important because it demonstrates that the total energy of a particle consists of both its rest energy and its kinetic energy.

'''What Happens at Different Velocities?'''

There are two important limiting cases to consider:

* '''When an object is at rest:''' Its velocity is zero, meaning <math>\gamma = 1</math>. Therefore, its kinetic energy is zero, but it still possesses rest energy equal to <math>mc^2</math>.

* '''When an object approaches the speed of light:''' The Lorentz factor increases dramatically. Consequently, the kinetic energy and total energy increase without bound as the velocity approaches the speed of light. This is another reason why an object with nonzero rest mass cannot reach the speed of light.

'''Connection to Classical Kinetic Energy'''

At everyday speeds, where <math>v \ll c</math>, the relativistic kinetic energy equation approaches the classical kinetic energy equation.

In this limit, the Lorentz factor can be approximated as

:<math>\gamma \approx 1 + \frac{v^2}{2c^2}</math>

Substituting this approximation into the relativistic kinetic energy equation gives

:<math>K = (\gamma - 1)mc^2</math>

:<math>K \approx \left(1 + \frac{v^2}{2c^2} - 1\right)mc^2</math>

Simplifying:

:<math>K \approx \frac{1}{2}mv^2</math>

This demonstrates that classical kinetic energy is simply an approximation of relativistic kinetic energy when an object's speed is much smaller than the speed of light.

==Examples==

===Simple===

Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a speed of <math> .97c </math> 
 What is the magnitude of the momentum of the proton?

Solution:

:<math> \frac{v}{c} = \frac{.97c}{c}</math> <math> = .97 </math>
:<math> \gamma = \frac{1}{\sqrt{1-(.97)^2}} = 4.1135 </math>
::Plug values in
:<math> p = \gamma * m*v = (4.1135)*(1.7*10^{-27})*(.97*(3*10^8)) = 2.035 * 10^{-18} </math> kg*m/s

===Middling===

Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving with a velocity <math> <1 * 10^7 , 2 * 10^7 , 3 * 10^7> </math> m/s. 
 What is the momentum of the proton?

Solution:

:<math> \left\vert \overrightarrow{v} \right\vert = \sqrt{(1*10^7)^2 + (2*10^7)^2 +(3*10^7)^2} </math> m/s <math> = 3.7 * 10^7 </math> m/s.
:<math> \frac{\left\vert \overrightarrow{v} \right\vert}{c} = \frac{3.7 * 10^7 m/s}{3 * 10^8 m/s}</math> <math> = .12 </math>
:<math> \gamma = \frac{1}{\sqrt{1-(.12)^2}} = 1.007 </math>
:<math> \overrightarrow{p} = \gamma * m * \overrightarrow{v} </math>
::Plug values in
:<math> \overrightarrow{p} = (1.007)*(1.7*10^{-27})*<1 * 10^7 , 2 * 10^7 , 3 * 10^7> = <1.7 * 10^{-20}, 3.4 * 10^{-20}, 5.1 *10^{-20}> </math> kg*m/s

===Difficult===

Suppose that a proton (mass = <math> 1.7 * 10^{-27} </math> kg) is moving at a speed of <math>.95c</math>. A force of magnitude <math>2 * 10^{-17}</math>N acts on it for .5s in the direction of the proton's velocity, causing it to speed up. What is its new speed at the end of the .5s time interval? Use <math>3*10^8</math> for <math>c</math>.

Solution:

Let us first find the initial momentum of the particle:

<math> \vec{p} = \gamma m \vec{v} </math>

<math> \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-.95^2}} = 3.203 </math>

<math> \vec{p} = 3.203 * 1.7 * 10^{-27} * (.95 * 3 * 10^8) = 1.55 * 10^{-18} </math> kg*m/s

Now let us find the momentum at the end of the impulse using the [[Impulse and Momentum|impulse-momentum theorem]].

<math> \vec{p}_f = \vec{p}_i + \vec{J} </math>

Since the impulse is in the same direction as the velocity, this vector addition becomes scalar (regular) addition.

<math> \vec{p}_f = 1.55 * 10^{-18} + .5 * 2 * 10^{-17} = 1.155 * 10^{-17} </math> kg*m/s

Now let us solve for the final speed using

<math> \vec{p} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} * m \vec{v} </math>

<math> 1.155 * 10^{-17} = \frac{1}{\sqrt{1-\frac{v^2}{(3 * 10^8)^2}}} * 1.7 * 10^{-27} * v </math>

It is impossible to algebraically solve for <math>v</math>, but using an equation solver yields

<math>v = 2.997 * 10^8 </math> m/s.

This demonstrates that when a particle is moving at nearly the speed of light, even a large impulse cannot cause it to meet or exceed the speed of light. An impulse of <math>10^{-17}</math> N*s may not seem very large, but it increased the particle's momentum by approximately a factor of 10.

==Connectedness==

Relativistic momentum has applications in quantum physics, particle physics, and astrophysics. In these branches of physics, particles routinely travel at relativistic speeds, and often, their momenta are directly measurable by the particles' interactions with matter. In industry, relativistic momentum is mostly used by space flight institutions.

==History==

In 1905, [[Albert Einstein]] published his Special Theory of Relativity. This theory set the "speed limit" for the universe at the speed of light and redefined many properties such as momentum and energy for fast-moving objects. As particles' speeds approach the speed of light, the laws of physics change even for concepts as fundamental as space and time. After this discovery, physicists were better able to understand the motion of fast-moving particles. Einstein published the more comprehensive General Theory of Relativity in 1915, which allowed differently accelerating frames of reference to be reconciled. Some aspects of relativity have since been used in a variety of technologies, including GPS.

== See also ==

*[[Linear Momentum]]
*[[Newton's Second Law: the Momentum Principle]]
*[[Impulse and Momentum]]
*[[Mass]]
*[[Vectors]]

===External Links===

*http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html

==References==
*https://opentextbc.ca/physicstestbook2/chapter/relativistic-momentum/

[[Category:Momentum]]

Power

2026-10-30T05:22:57Z

Lheiman07: /* Difficult */

Claimed by Aeze3

This topic covers Power.

==The Main Idea==

Power is the rate of doing work or the amount of energy consumed over an interval of time.

===A Mathematical Model===

When a force is applied over a distance in a unit of time, power is calculated by

[[File:power(1).jpg |border|right]]

:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>

where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.

Power can also be defined as the rate of change of energy transfer:

:<math>P = \frac{\Delta E}{\Delta t}</math>

When there is no heat transfer (Q = 0), the change in energy is equal to the work done. Therefore,

:<math>P = \frac{\Delta E}{\Delta t} = \frac{W}{\Delta t}</math>

Since work is the dot product of force and displacement, we can further expand the equation:

:<math>P = \frac{\vec{F}\cdot\Delta\vec{r}}{\Delta t}</math>

Rearranging gives:

:<math>P = \vec{F}\cdot\frac{\Delta\vec{r}}{\Delta t}</math>

It then follows that instantaneous power is

:<math>power = \vec{F}\cdot\vec{v}</math>

where '''v''' is velocity.

The SI unit for power is watts (J/s). The units can also be expressed in terms of fundamental SI units:

:<math>Watt = \frac{J}{s} = \frac{N\cdot m}{s} = \frac{kg\cdot m^2}{s^3}</math>

Note that '''W''' can represent either watts or work, depending on the context.

===A Computational Model===

How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A certain motor is capable of doing 3000 J of work in 12 s
What is the power output of this motor?

:<math>power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts </math>

===Middling===

Here are questions dealing with human power. '''(a)''' If you follow a diet of 2000 food calories per day (2000 kC), what is your average rate of energy consumption in watts (power input)? (A food or “large” calorie is a unit of energy equal to 4.2 J; a regular or “small” calorie is equal to 4200 J.) '''(b)''' How many days of a diet of 2000 large calories are equivalent to the gravitational energy change from sea level to the top of Mount Everest, 8848 m above sea level? Assume your weight is 58 kg. (The body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)

'''(a)''' <math>power = \frac{W}{\Delta t} = \frac{2000 kC}{day} \cdot \frac{4200 J}{1 kC} \cdot \frac{1 day}{24 h} \cdot \frac{1 h}{3600 s} = 97.2 watts </math>

'''(b)''' <math> {\frac{97.2 J}{s}} \cdot \frac{3600 s}{1 h} \cdot \frac{24 h}{1 day} = 8398080 J/day </math>

:<math> \Delta U_g = mg\Delta y = (58 kg)(9.8 m/s^2)(8848 m - 0 m) = 5029203.2 J </math>

:<math> \Delta t = \frac{W}{power} = \frac{5029203.2 J}{8398080 J/day} = 0.599 days </math>

===Difficult===

A 1200 kg elevator is lifted vertically from rest through a height of 45 m in 15 seconds. At the end of the lift, the elevator is moving upward at a speed of 6 m/s. Assuming the motor operates at a constant power output, calculate the average power output required to lift the elevator. Ignore friction and air resistance.

'''Step 1:''' Identify the work done by the motor.

The motor must provide enough energy to increase both the gravitational potential energy and the kinetic energy of the elevator.

:<math>W = \Delta U_g + \Delta K</math>

'''Step 2:''' Calculate the change in gravitational potential energy.

:<math>\Delta U_g = mg\Delta y</math>

:<math>= (1200 \ kg)(9.8 \ m/s^2)(45 \ m)</math>

:<math>= 529200 \ J</math>

'''Step 3:''' Calculate the change in kinetic energy.

Since the elevator starts from rest, its initial velocity is zero.

:<math>\Delta K = \frac{1}{2}m(v_f^2-v_i^2)</math>

:<math>= \frac{1}{2}(1200 \ kg)(6^2-0^2)</math>

:<math>= 21600 \ J</math>

'''Step 4:''' Calculate the total work performed by the motor.

:<math>W = \Delta U_g + \Delta K</math>

:<math>= 529200 \ J + 21600 \ J</math>

:<math>= 550800 \ J</math>

'''Step 5:''' Calculate the average power output.

Using the relationship between work, time, and power:

:<math>power = \frac{W}{\Delta t}</math>

:<math>= \frac{550800 \ J}{15 \ s}</math>

:<math>= 36720 \ watts</math>

'''Final Answer:'''

:<math>\boxed{power = 36720 \ W = 36.72 \ kW}</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

*[[Work]]
*[[Gravitational Potential Energy]]

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

"General Mechanics/Work and Power." - Wikibooks, Open Books for an Open World. Web. [https://en.wikibooks.org/wiki/General_Mechanics/Work_and_Power General Mechanics/Work and Power]

==References==

Chabay, Ruth W.; Sherwood, Bruce A. Matter and Interactions, 4th Edition: 1-2. Wiley.

[[Category:Which Category did you place this in?]]

Power

2026-10-30T05:22:09Z

Lheiman07:

Claimed by Aeze3

This topic covers Power.

==The Main Idea==

Power is the rate of doing work or the amount of energy consumed over an interval of time.

===A Mathematical Model===

When a force is applied over a distance in a unit of time, power is calculated by

[[File:power(1).jpg |border|right]]

:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>

where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.

Power can also be defined as the rate of change of energy transfer:

:<math>P = \frac{\Delta E}{\Delta t}</math>

When there is no heat transfer (Q = 0), the change in energy is equal to the work done. Therefore,

:<math>P = \frac{\Delta E}{\Delta t} = \frac{W}{\Delta t}</math>

Since work is the dot product of force and displacement, we can further expand the equation:

:<math>P = \frac{\vec{F}\cdot\Delta\vec{r}}{\Delta t}</math>

Rearranging gives:

:<math>P = \vec{F}\cdot\frac{\Delta\vec{r}}{\Delta t}</math>

It then follows that instantaneous power is

:<math>power = \vec{F}\cdot\vec{v}</math>

where '''v''' is velocity.

The SI unit for power is watts (J/s). The units can also be expressed in terms of fundamental SI units:

:<math>Watt = \frac{J}{s} = \frac{N\cdot m}{s} = \frac{kg\cdot m^2}{s^3}</math>

Note that '''W''' can represent either watts or work, depending on the context.

===A Computational Model===

How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A certain motor is capable of doing 3000 J of work in 12 s
What is the power output of this motor?

:<math>power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts </math>

===Middling===

Here are questions dealing with human power. '''(a)''' If you follow a diet of 2000 food calories per day (2000 kC), what is your average rate of energy consumption in watts (power input)? (A food or “large” calorie is a unit of energy equal to 4.2 J; a regular or “small” calorie is equal to 4200 J.) '''(b)''' How many days of a diet of 2000 large calories are equivalent to the gravitational energy change from sea level to the top of Mount Everest, 8848 m above sea level? Assume your weight is 58 kg. (The body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)

'''(a)''' <math>power = \frac{W}{\Delta t} = \frac{2000 kC}{day} \cdot \frac{4200 J}{1 kC} \cdot \frac{1 day}{24 h} \cdot \frac{1 h}{3600 s} = 97.2 watts </math>

'''(b)''' <math> {\frac{97.2 J}{s}} \cdot \frac{3600 s}{1 h} \cdot \frac{24 h}{1 day} = 8398080 J/day </math>

:<math> \Delta U_g = mg\Delta y = (58 kg)(9.8 m/s^2)(8848 m - 0 m) = 5029203.2 J </math>

:<math> \Delta t = \frac{W}{power} = \frac{5029203.2 J}{8398080 J/day} = 0.599 days </math>

===Difficult===

A 1200 kg elevator is lifted vertically from rest through a height of 45 m in 15 seconds. At the end of the lift, the elevator is moving upward at a speed of 6 m/s. Assuming the motor operates at a constant power output, calculate the average power output required to lift the elevator. Ignore friction and air resistance.

‘’‘Step 1:’’’ Identify the work done by the motor.

The motor must provide enough energy to increase both the gravitational potential energy and the kinetic energy of the elevator.

W = \Delta U_g + \Delta K

‘’‘Step 2:’’’ Calculate the change in gravitational potential energy.

\Delta U_g = mg\Delta y

= (1200 kg)(9.8 m/s^2)(45 m)

= 529200 J

‘’‘Step 3:’’’ Calculate the change in kinetic energy.

Since the elevator starts from rest, its initial velocity is zero.

\Delta K = \frac{1}{2}m(v_f^2-v_i^2)

= \frac{1}{2}(1200 kg)(6^2-0^2)

= 21600 J

‘’‘Step 4:’’’ Calculate the total work performed by the motor.

W = \Delta U_g + \Delta K

= 529200 J + 21600 J

= 550800 J

‘’‘Step 5:’’’ Calculate the average power output.

Using the relationship between work, time, and power:

power = \frac{W}{\Delta t}

= \frac{550800 J}{15 s}

= 36720 watts

‘’‘Final Answer:’’’

\boxed{power = 36,720 \text{ W} = 36.72 \text{ kW}}

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

*[[Work]]
*[[Gravitational Potential Energy]]

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===

Internet resources on this topic

"General Mechanics/Work and Power." - Wikibooks, Open Books for an Open World. Web. [https://en.wikibooks.org/wiki/General_Mechanics/Work_and_Power General Mechanics/Work and Power]

==References==

Chabay, Ruth W.; Sherwood, Bruce A. Matter and Interactions, 4th Edition: 1-2. Wiley.

[[Category:Which Category did you place this in?]]

Power

2026-10-30T03:08:29Z

Lheiman07: /* A Mathematical Model */

Claimed by Aeze3

This topic covers Power.

==The Main Idea==

Power is the rate of doing work or the amount of energy consumed over an interval of time.

===A Mathematical Model===

When a force is applied over a distance in a unit of time, power is calculated by

[[File:power(1).jpg |border|right]]

:<math>power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t}</math>

where '''F''' is force, '''Δr''' is displacement, '''Δt''' is the duration of time and '''W''' is work.

Power can also be defined as the rate of change of energy transfer:

:<math>P = \frac{\Delta E}{\Delta t}</math>

When there is no heat transfer (Q = 0), the change in energy is equal to the work done. Therefore,

:<math>P = \frac{\Delta E}{\Delta t} = \frac{W}{\Delta t}</math>

Since work is the dot product of force and displacement, we can further expand the equation:

:<math>P = \frac{\vec{F}\cdot\Delta\vec{r}}{\Delta t}</math>

Rearranging gives:

:<math>P = \vec{F}\cdot\frac{\Delta\vec{r}}{\Delta t}</math>

It then follows that instantaneous power is

:<math>power = \vec{F}\cdot\vec{v}</math>

where '''v''' is velocity.

The SI unit for power is watts (J/s). The units can also be expressed in terms of fundamental SI units:

:<math>Watt = \frac{J}{s} = \frac{N\cdot m}{s} = \frac{kg\cdot m^2}{s^3}</math>

Note that '''W''' can represent either watts or work, depending on the context.

===A Computational Model===

How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===

A certain motor is capable of doing 3000 J of work in 12 s
What is the power output of this motor?

:<math>power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts </math>

===Middling===

Here are questions dealing with human power. '''(a)''' If you follow a diet of 2000 food calories per day (2000 kC), what is your average rate of energy consumption in watts (power input)? (A food or “large” calorie is a unit of energy equal to 4.2 J; a regular or “small” calorie is equal to 4200 J.) '''(b)''' How many days of a diet of 2000 large calories are equivalent to the gravitational energy change from sea level to the top of Mount Everest, 8848 m above sea level? Assume your weight is 58 kg. (The body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)

'''(a)''' <math>power = \frac{W}{\Delta t} = \frac{2000 kC}{day} \cdot \frac{4200 J}{1 kC} \cdot \frac{1 day}{24 h} \cdot \frac{1 h}{3600 s} = 97.2 watts </math>

'''(b)''' <math> {\frac{97.2 J}{s}} \cdot \frac{3600 s}{1 h} \cdot \frac{24 h}{1 day} = 8398080 J/day </math>

:<math> \Delta U_g = mg\Delta y = (58 kg)(9.8 m/s^2)(8848 m - 0 m) = 5029203.2 J </math>

:<math> \Delta t = \frac{W}{power} = \frac{5029203.2 J}{8398080 J/day} = 0.599 days </math>

===Difficult===

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== See also ==

*[[Work]]
*[[Gravitational Potential Energy]]

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===Further reading===

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===External links===

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"General Mechanics/Work and Power." - Wikibooks, Open Books for an Open World. Web. [https://en.wikibooks.org/wiki/General_Mechanics/Work_and_Power General Mechanics/Work and Power]

==References==

Chabay, Ruth W.; Sherwood, Bruce A. Matter and Interactions, 4th Edition: 1-2. Wiley.

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