Does E=mc2 Work with Different Metric Systems?

[Jarfi]

Ok so I am confused with the face the different metric systems and sizes use different numbers for the same size f.ex. 1000 meters can be 1 km or 100 mph can be 135 km, And the mass can be told in pounds, kg, grams, and more. And energy in calories. So it seems to me that you can get different outcomes when calculating with e=mc2 if you use different systems, its like it doesn't seem right, they use the international metric system, so e is than actually just a number that is (300000)2 times the mass but what if you use a system from usa with (180000)2 times the mass, than you would get a totally different outcome, I don't know if you understand me, it's just that I don't understand how it can work out with man made numbers.

 

[Phyisab****]

Pick some quantities and calculate E in English units. Convert your quantities to SI units. Calculate E again. Post your results.

 

[SteamKing]

In any physical formula, you must use consistent units. If you have mass in kilograms and velocity in miles per hour, the calculated E in E = mc^2 will be nonsense.

In SI, to calculate E measured in joules, m must be measured in kilograms and c must be converted to meters per second.

In imperial units, to calculate E measured in foot-pounds, m must be in slugs and c must be converted to feet per second.

 

[Janus]

Ok so I am confused with the face the different metric systems and sizes use different numbers for the same size f.ex. 1000 meters can be 1 km or 100 mph can be 135 km, And the mass can be told in pounds, kg, grams, and more. And energy in calories. So it seems to me that you can get different outcomes when calculating with e=mc2 if you use different systems, its like it doesn't seem right, they use the international metric system, so e is than actually just a number that is (300000)2 times the mass but what if you use a system from usa with (180000)2 times the mass, than you would get a totally different outcome, I don't know if you understand me, it's just that I don't understand how it can work out with man made numbers.

As long as you stay within the same system, it doesn't matter, your answer will come out the same.

Example:

1kg mass
c=300,000,000 m/sec

gives you 9 x 10^16 joules.

converting to fps

1 kg = ~2.2 lbs
c= 990,000,000 ft/sec

gives you ~2.16 x 10^18 ft-poundals

If you convert joules to ft-poundals you will find there are 23.7 ft-poundals to a joule and 9 x10^16 joules = 2.16 x 10^18 ft-poundals. (More or less, I rounded out a bit which means you won't get an exact answer.

The trick is that when you do the calculation, all the units have to be in the same system

You use grams with centimeters, Kg with meters, etc. You can't use kg and ft together, for example.

 

[Jarfi]

In any physical formula, you must use consistent units. If you have mass in kilograms and velocity in miles per hour, the calculated E in E = mc^2 will be nonsense.

In SI, to calculate E measured in joules, m must be measured in kilograms and c must be converted to meters per second.

In imperial units, to calculate E measured in foot-pounds, m must be in slugs and c must be converted to feet per second.

This is what has been confusing me! Why does kmph
fit with kg and not pounds or ounces, if i where to create my own measuring system, how would i coordinate speed and energy to make it work out. Why do some measuring units work together but others dont? Why is it better to use foot with pounds than with kiloes?

 

[bp_psy]

This is what has been confusing me! Why does kmph
fit with kg and not pounds or ounces, if i where to create my own measuring system, how would i coordinate speed and energy to make it work out. Why do some measuring units work together but others dont? Why is it better to use foot with pounds than with kiloes?

Units in a system work because there exist a well defined conventions of what the derived units are. You could of course make a system that uses basic units form other systems but then you have to define other derived units. An example J=Kg*m^2/s^2 but Kg*ft^2/s^2 =X which is still a valid unit of energy, X=0.093J. Not much of a point of doing this though.

 

[Monsterleg]

This is what has been confusing me! Why does kmph
fit with kg and not pounds or ounces, if i where to create my own measuring system, how would i coordinate speed and energy to make it work out. Why do some measuring units work together but others dont? Why is it better to use foot with pounds than with kiloes?

All the parts of your equation have to equal the unit for energy your using, for example Joules (J):
J= kg(m/s)²
In order to get the energy in Joules you have to have all of those components in your mathematical equation.
So if:
E= J and J= kg(m/s)²

Then:
E= kg(m/s)²
-------------------
If:
E= kg(m/s)² and E= Mc²

Then:
kg(m/s)²= Mc²
--------------------
If:
kg(m/s)²= Mc² and M=kg

Then:
kg(m/s)²= (kg)c²
--------------------
If:
kg(m/s)²= (kg)c² and c= (m/s)

Then:
kg(m/s)²= kg(m/s)²

That's why you have to use kg for m and such. You still can solve an equation using other units but we don't have simple words for those units like Joules you'd have to write out the unit.

 

[D H]

As long as you stay within the same system, it doesn't matter, your answer will come out the same.

I beg to differ.

As long as you stay within a consistent set of units, it doesn't matter.

SI comprises a consistent set of units (so long as you steer clear of archaic units such as calories). The standard English system, with mass expressed in pounds mass, force in pounds force, energy in BTUs is anything but consistent.

A consistent set of units has

• 1 unit of energy = 1 unit of length * 1 unit of force
• 1 unit of force = 1 unit of mass * 1 unit of acceleration
• 1 unit of acceleration = 1 unit of velocity / 1 unit of time
• 1 unit of velocity = 1 unit of length / 1 unit of time

An inconsistent set of units requires the use of values other than one in the above. For example, when force is expressed in pounds force, mass is expressed in pounds mass, and acceleration is expressed in ft/sec² you get 1 unit of force = 1 unit of mass * 32.17404855643 units of acceleration.
There's a very good reason scientists prefer to work with consistent units. To steal a phrase from Steve Jobs, "It just works".

 

[jetwaterluffy]

Energy(J)=Force(N)*Distance(m)=
distance(m)*(Δmomentum(kgms^-1)/time(s))=Energy(kgm²s^-2)Energy=Mass(kg)*(speed_light(ms^-1))²=Energy(kgm²s^-2)


See those units at the end of both? They are the same units. Therefore... the equation works.