Why are E=MC^2 and E=1/2MV^2 equivalent equations?

In summary, today in my physics class we learned the equation for kinetic energy; E=1/2MV^2. I found the equation strikingly similar to Einsteins famous equation E=MC^2. The only real difference is the 1/2 coefficient (Since C is just a constant for V) so i figured there should be a constant for V in the kinetic energy equation that would make the two equations yield the same result so...mc^2=1/2mv^2/c^2. Assuming a value of 1 for the mass and c^2=1/2v^2, v=423,970,560 m/s. So by substituting the constant 423,970
  • #1
physicz123
2
0
Today in my physics class we learned the equation for kinetic energy;

E = 1/2MV^2

I found the equation strikingly similar to Einsteins famous equation E=MC^2
The only real difference is the 1/2 coefficient (Since C is just a constant for V)
So i figured there should be a constant for V in the kinetic energy equation that would make the two equations yield the same result so...

mc^2 = 1/2mv^2
assume a value of 1 for the mass
c^2 = 1/2v^2
sqrt(2c^2) = v
v = 423,970,560 m/s

So by substituting the constant 423,970,560 m/s (call it 'Q') into velocity for the kinetic energy equation the two equations become equivalent.

E = mc^2 = 1/2mq^2

So basically any mass moving at the velocity 423,970,560 m/s will have the same kinetic energy as the energy contained in the mass at rest as described by E = MC^2. Which made me wonder if perhaps the reason all mass at rest has this energy is because the universe is rotating or moving at velocity Q?

Which brings me to my question..
Is there a explanation as to why E = MC^2?
 
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  • #2
If v is large enough that you have to use relativity, you can't use the classical kinetic energy formula any more. Instead, you have to use

[tex]K = mc^2 - m_0 c^2[/tex]

[tex]K = \frac{m_0 c^2}{\sqrt{1 - v^2/c^2}} - m_0 c^2[/tex]

where m is the "relativistic mass" and m0 is the "rest mass".
 
  • #3
Physics doesn't answer why questions. Physics just uses maths to model the world so that predictions can be made and experiments performed to verify those models.

While both equations you have used represent energy they are different, you can't just stick something equal to something else without justification, that is numerology not physics.
 
  • #4
cosmik debris said:
While both equations you have used represent energy they are different, you can't just stick something equal to something else without justification, that is numerology not physics.

Agreed. Get this idea out of your head that you can randomly equate 2 equations and think you have found out something about the world. It will lead you astray. What [tex]E=mc^2[/tex] really is telling you is that mass and energy are basically forms of the same entity.

For example, if I have a cup in my hand, the cup, just sitting there, has an intrinsic energy associated with it. If I were to completely annihilate this cup using nuclear reactions, the amount of energy I could get out would be the mass times the speed of light squared. In fact, that's how nuclear power works. You have nuclear material constantly undergoing fission processes. At the end of the day, you've powered a million homes for 20 years and you look at how much the nuclear material weighs and it actually has lost mass!

For actual calculations, the total energy of an object is [tex]E_{total} = mc^2 + KE + PE[/tex]. That's the actual total energy of an object. The famous [tex]E=mc^2[/tex] is telling you what the energy an object is without any kinetic energy or potential energy - in other words, the energy an object has just by the fact that it has mass. Now, since you never see nuclear reactions occurring in your intro physics courses and you are always dealing with changes of energy, the [tex]mc^2[/tex] term is just dropped because it doesn't change; that's why you see typically [tex]E_{total} = KE + PE[/tex].

Later you'll hopefully get into relativistic mechanics and you'll find that the kinetic energy is no longer just [tex]KE = .5mv^2[/tex] at velocities approaching the speed of light.
 
  • #5
physicz123 said:
Is there a explanation as to why E = MC^2?
Yes. It can be explained reasonably simply too. See this thread


AM
 
  • #6
Thanks for the quick replies. I'm just starting out in physics, still learning. :D
 

Related to Why are E=MC^2 and E=1/2MV^2 equivalent equations?

1. What does E=MC^2 mean?

E=MC^2 is a famous equation that was introduced by Albert Einstein in his theory of relativity. It states that energy (E) is equal to the mass (M) of an object multiplied by the speed of light (C) squared. This equation shows the relationship between mass and energy, and how they are interchangeable.

2. How does E=MC^2 relate to kinetic energy?

E=MC^2 is related to kinetic energy through the mass-energy equivalence principle. This principle states that any object that has mass also has an equivalent amount of energy and vice versa. In other words, the kinetic energy of an object can be calculated by using its mass and the speed of light.

3. Can E=MC^2 be applied to all forms of energy?

Yes, E=MC^2 can be applied to all forms of energy, not just kinetic energy. This equation is a fundamental principle in physics and can be used to explain various phenomena, such as nuclear reactions and the energy of electromagnetic radiation.

4. How is E=MC^2 used in practical applications?

E=MC^2 has many practical applications, such as in nuclear power plants, where it is used to calculate the energy released from nuclear reactions. It is also used in particle accelerators, where it helps scientists understand the behavior of subatomic particles.

5. Is E=MC^2 a proven equation?

Yes, E=MC^2 is a proven equation that has been tested and verified through numerous experiments and observations. It is a cornerstone of modern physics and has been used to make significant advancements in our understanding of the universe.

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