Kinetic Energy: Difference between revisions
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== See also == | == See also == | ||
[[Potential Energy]]<br> | |||
[[Rest Mass Energy]]<br> | [[Rest Mass Energy]]<br> | ||
===Further reading=== | ===Further reading=== | ||
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http://www.physicsclassroom.com/class/energy/Lesson-1/Kinetic-Energy<br> | http://www.physicsclassroom.com/class/energy/Lesson-1/Kinetic-Energy<br> | ||
https://en.wikipedia.org/wiki/Kinetic_energy<br> | https://en.wikipedia.org/wiki/Kinetic_energy<br> | ||
==References== | ==References== |
Revision as of 15:53, 13 November 2015
The energy of motion is kinetic energy. --A WORK IN PROGRESS BY JASON MORCOS--
Kinetic Energy
Objects in motion have energy associated with them. This energy of motion is called kinetic energy. Kinetic energy, often abbreviated as KE, is usually given in the standard S.I. units of Joules (J). KE is also given in units of kilo Joules (kJ). [math]\displaystyle{ 1 kJ = 1000 J }[/math]. [math]\displaystyle{ 1 J = 1 kg*(m²/s²) }[/math]. Other types of energy include Rest Mass Energy and Potential Energy.
A Mathematical Model
The relativistic equation for kinetic energy according to Einstein's Theory of Relativity is [math]\displaystyle{ KE=mc²(\frac{1}{\sqrt{1-\frac{v²}{c²}}} -1) }[/math]. However, for cases where an object's velocity is far less than the speed of light ([math]\displaystyle{ 3X10^8 m/s }[/math]), one can use the simplified kinetic energy formula: [math]\displaystyle{ KE=\frac{1}{2}mv² }[/math]. In most cases the simplified kinetic energy formula gives a result with only minimal error. However, for near light speed calculations, such as those involving subatomic particles such as electrons, protons, or photons, the relativistic equation must be used. Usually we think of the simplified kinetic energy formula as the way to calculate the kinetic energy of an average object.
By conservation of energy, energy can be converted but it cannot be created nor destroyed. Hence, in an isolated system, energy can be converted back and forth between potential and kinetic energy continuously without loss. This is an excellent visualization of energy that can be demonstrated with vpython. See A Computational Model for this demo.
A Computational Model
https://trinket.io/glowscript/87a35d5778
Examples
Simple
A ball is rolling along a frictionless surface at a constant [math]\displaystyle{ 19 m/s }[/math]. The ball has mass [math]\displaystyle{ 12 kg }[/math]. What is the kinetic energy of the ball in Joules?
Solution:
- Because the ball's velocity is far less than the speed of light, we can use the simplified kinetic energy formula.
- [math]\displaystyle{ KE=\frac{1}{2}mv² }[/math]
- [math]\displaystyle{ KE=\frac{1}{2}(12 kg)*(19 m/s)² }[/math]
- Hence, KE = [math]\displaystyle{ 2166 J }[/math] or [math]\displaystyle{ 2.166 kJ }[/math]
Middling
An electron is moving through space at a constant [math]\displaystyle{ 2.9X10^8 m/s }[/math]. The electron has mass [math]\displaystyle{ 9.1X10^-31 kg }[/math]. What is the kinetic energy of the ball in Joules?
Solution:
- Because the electron's velocity is close to the speed of light, we must use the relativistic kinetic energy formula.
- [math]\displaystyle{ KE=mc²(\frac{1}{\sqrt{1-\frac{v²}{c²}}} -1) }[/math]
- [math]\displaystyle{ KE=(9.1X10^-31 kg)(3X10^8 m/s)²(\frac{1}{\sqrt{1-\frac{(2.9X10^8 m/s)²}{(3X10^8 m/s)²}}} -1) }[/math]
- Hence, KE = [math]\displaystyle{ 2.38X10^-13 J }[/math] or [math]\displaystyle{ 2.38X10^-16 kJ }[/math]
Difficult
An proton is found to have kinetic energy [math]\displaystyle{ 2.38X10^-15 J }[/math]. The proton has mass [math]\displaystyle{ 1.67X10^-27 kg }[/math]. What velocity would the proton have to be moving at to have this kinetic energy?
Solution:
- Because we are dealing with a subatomic particle, we should probably use the relativistic kinetic energy formula as the approximate kinetic energy formula may be very inaccurate if the particle is moving at a velocity near the speed of sound.
- [math]\displaystyle{ KE=mc²(\frac{1}{\sqrt{1-\frac{v²}{c²}}} -1) }[/math]
- [math]\displaystyle{ \frac{KE}{mc²}=(\frac{1}{\sqrt{1-\frac{v²}{c²}}} -1) }[/math]
- [math]\displaystyle{ \frac{KE}{mc²} + 1=\frac{1}{\sqrt{1-\frac{v²}{c²}}} }[/math]
- [math]\displaystyle{ \frac{1}{\frac{KE}{mc²} + 1}=\sqrt{1-\frac{v²}{c²}} }[/math]
- [math]\displaystyle{ 1-\frac{v²}{c²}=(\frac{1}{\frac{KE}{mc²} + 1})² }[/math]
- [math]\displaystyle{ \frac{v²}{c²}=1 - (\frac{1}{\frac{KE}{mc²} + 1})² }[/math]
- [math]\displaystyle{ v²=c²*(1 - (\frac{1}{\frac{KE}{mc²} + 1})²) }[/math]
- [math]\displaystyle{ v=\sqrt{c²*(1 - (\frac{1}{\frac{KE}{mc²} + 1})²)} }[/math]
- [math]\displaystyle{ v=c*\sqrt{1 - (\frac{1}{\frac{KE}{mc²} + 1})²} }[/math]
- [math]\displaystyle{ v=3X10^8 m/s*\sqrt{1 - (\frac{1}{\frac{(2.38X10^-15 J)}{(1.67X10^-27 kg)(3X10^8 m/s)²} + 1})²} }[/math]
- v = [math]\displaystyle{ 2.999X10^8 m/s }[/math]. This means that the proton is moving very close to the speed of light and hence our choice to use the relativistic kinetic energy equation was a good one.
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
Potential Energy
Rest Mass Energy
Further reading
Matter and Interactions By Ruth W. Chabay, Bruce A. Sherwood - Chapter 9
External links
http://www.physicsclassroom.com/class/energy/Lesson-1/Kinetic-Energy
https://en.wikipedia.org/wiki/Kinetic_energy
References
http://www.physicsclassroom.com/class/energy/Lesson-1/Kinetic-Energy
https://en.wikipedia.org/wiki/Kinetic_energy