Why is the Velocity equal to joules per kilogram?

[grandpa2390]

Homework Statement

Calculating the rms speed of an N2 molecule. The units I end up with are J/kg and this is equivalent to Velocity. my question is why?

Homework Equations

v = J / kg
KE = .5mv^2

The Attempt at a Solution

I tried to derive it myself using the kinetic energy formula. What I end up with is that $v=\sqrt{\frac{2*J}{kg}}$

The units are right, J/kg, but it isn't a direct conversion like the solution on Chegg says. it looks like to convert, my 517 J/kg doesn't equal 517 m/s. It would seem I would have to take the square root of 1034

 

[Chestermiller]

J/ kg has units of velocity square.

 

[fresh_42]

I'm not sure that I understand what you mean.
$1J = \frac{kg\cdot m^2}{s^2}$ so Joule per kilogram is $\frac{m^2}{s^2}$ which is the unity of the squared velocity: $[v^2] = \frac{m^2}{s^2}$
The reason is, that all these units base on the metric SI system, i.e. $m\, , \,kg\, , \,s\, , \,A\,$.

 

[grandpa2390]

@Chestermiller @fresh_42

you are absolutely right. the $V_{rms} = \sqrt{\frac{3kT}{m}}$ is the square root of J/kg or v^2. But what about the 2? I guess that is already factored into the formula? What I am seeing now, based on your responses, is that velocity is equal to $v=\sqrt{2*v^2}$ it is just not working out in my mind. I am missing something.

the kinetic energy formula is $KE={\frac{1}{2}*m*v^2}$ or in units $J=\frac{1}{2}*kg*\frac{m^2}{s^2}$
solved for $v$ $v=\sqrt{\frac{2*KE}{m}}$ the units being $\frac{m}{s} = \sqrt{\frac{2*J}{kg}}$
and you are saying that $\frac{J}{kg} = \frac{m^2}{s^2}$ or $\frac{J}{kg} = v^2$
so that would be $\frac{m}{s} = \sqrt{2*\frac{m^2}{s^2}}$ or $v = \sqrt{2*v^2}$

that 2 is throwing me off. what am I missing here?

 

[TSny]

the kinetic energy formula is $KE={\frac{1}{2}*m*v^2}$ or in units $J=\frac{1}{2}*kg*\frac{m^2}{s^2}$

The factor of $\frac{1}{2}$ in the KE expression has no dimensions. So, $\frac{1}{2}$ doesn't contribute any units.

Another example: The SI units for an area are m². The formula for the area of a circle is $A = \pi r^2$. When analyzing the units on the right side of $A$, you would not say that the units for the area are $\pi$m².

 

[fresh_42]

@Chestermiller @fresh_42

you are absolutely right. the $V_{rms} = \sqrt{\frac{3kT}{m}}$ is the square root of J/kg or v^2. But what about the 2? I guess that is already factored into the formula? What I am seeing now, based on your responses, is that velocity is equal to $v=\sqrt{2*v^2}$ it is just not working out in my mind. I am missing something.

the kinetic energy formula is $KE={\frac{1}{2}*m*v^2}$ or in units $J=\frac{1}{2}*kg*\frac{m^2}{s^2}$
solved for $v$ $v=\sqrt{\frac{2*KE}{m}}$ the units being $\frac{m}{s} = \sqrt{\frac{2*J}{kg}}$
and you are saying that $\frac{J}{kg} = \frac{m^2}{s^2}$ or $\frac{J}{kg} = v^2$
so that would be $\frac{m}{s} = \sqrt{2*\frac{m^2}{s^2}}$ or $v = \sqrt{2*v^2}$

that 2 is throwing me off. what am I missing here?

You shouldn't mix notations of physical quantities like $v$ with units. $\frac{J}{kg} = v^2$ isn't a good way to write it. That's why I wrote $[v^2] = \frac{J}{kg}$ with brackets to indicate "unit of". The two doesn't play a role for the units. If we have three yards in each of three downs, we get $3 \cdot (3 \cdot yd) = (3 \cdot 3) \cdot yd = 9 \, yd$ in total. So the unit of the drive $dr$ was $[dr] = yd$ and it doesn't matter, that it was the result of $3 \cdot 3\, yd$ or $0\,yd +0\,yd +9\,yd\;$.

 

[Chestermiller]

@Chestermiller @fresh_42

you are absolutely right. the $V_{rms} = \sqrt{\frac{3kT}{m}}$ is the square root of J/kg or v^2. But what about the 2? I guess that is already factored into the formula? What I am seeing now, based on your responses, is that velocity is equal to $v=\sqrt{2*v^2}$ it is just not working out in my mind. I am missing something.

the kinetic energy formula is $KE={\frac{1}{2}*m*v^2}$ or in units $J=\frac{1}{2}*kg*\frac{m^2}{s^2}$
solved for $v$ $v=\sqrt{\frac{2*KE}{m}}$ the units being $\frac{m}{s} = \sqrt{\frac{2*J}{kg}}$
and you are saying that $\frac{J}{kg} = \frac{m^2}{s^2}$ or $\frac{J}{kg} = v^2$
so that would be $\frac{m}{s} = \sqrt{2*\frac{m^2}{s^2}}$ or $v = \sqrt{2*v^2}$

that 2 is throwing me off. what am I missing here?

What you are missing is that just because something has units of velocity doesn't mean that it is velocity.

 

[grandpa2390]

I'm not arguing, I'm just trying to understand a well-accepted conversion. I know it is right, I just really want to understand why.

J/ kg has units of velocity square.

What about the 1/2. What happened to it during conversion.

You shouldn't mix notations of physical quantities like $v$ with units. $\frac{J}{kg} = v^2$ isn't a good way to write it. That's why I wrote $[v^2] = \frac{J}{kg}$ with brackets to indicate "unit of". The two doesn't play a role for the units. If we have three yards in each of three downs, we get $3 \cdot (3 \cdot yd) = (3 \cdot 3) \cdot yd = 9 \, yd$ in total. So the unit of the drive $dr$ was $[dr] = yd$ and it doesn't matter, that it was the result of $3 \cdot 3\, yd$ or $0\,yd +0\,yd +9\,yd\;$.

Right but if I want to convert 9 yd to downs, I don't just say 9 yards equals 9 downs, I have to divide by that 3.
this is the way I am thinking of it. 1 yd = 3 x 1ft,3*1*ft the 3 doesn't play a role for the units. except that it does. I am not seeing why 1 yd must equal 3 ft, but with a joule we can ignore the two. I'm just not seeing it. Perhaps it is something that needs to be explained with pencil and paper.
If the two doesn't matter, why is it present in the kinetic energy formula. It doesn't have a unit so it doesn't change the units, but according to that formula, Joules equal $\frac{1}{2}*kg*v^2$, not $kg*v^2$

What you are missing is that just because something has units of velocity doesn't mean that it is velocity.

I don't know what that means. It doesn't have a direction, but why does that dump the sqrt(2).

at this point, the only explanation that seems to ease my mind is the the root mean squared speed formula as that two factored in already. that it would be $v=\sqrt{\frac{2*1.5*k*T}{m}}$ that the 2 has already been factored in.
That in any other conversion from Joules and kg to speed (or velocity), I would need that 2.

 

[grandpa2390]

it's ok if you all want to give up on me. I'm sorry if I am a pain. You can go ahead and delete the thread. I'm thinking maybe I should just ask my professor. Maybe he can get through to me. It's a lot easier when you can talk in person and the other person can yell at you and hammer it in. ;)

I mean it is not really "important". can we say that anytime I am presented with Joules/kg, I can just take the square root and get the speed? That it's always a

edit: lol a $\sqrt{9}:3$ conversion.

 

[fresh_42]

A factor in front of a unit doesn't affect the unit, only the quantity. $cm$ and $m$ both measure the physical quantity length $L$. If this length is achieved by a velocity $V$ during a time $T$, say $V=10\frac{cm}{h}$ and $T=\frac{1}{2}\,h$, probably a snail, then the snail moved $L=V \cdot T = 10\frac{cm}{h} \cdot \frac{1}{2}\,h = 5 \, cm$ during this half hour. If we do the same calculation in the units meter and seconds, we get $L = 10 \cdot 10^{-2} \cdot m \cdot \frac{1}{3600\,s} \cdot \frac{1}{2} \cdot 3600 \,s = 5 \cdot 10^{-2} \,m = 0.05 \, m = 5\, cm$. Now the only difference between $0.05 \, m$ and $5\, cm$ is whether the $\frac{1}{100}$ is swallowed by the $5$ making it $0.05$ or by the unit, making it centi.

You can always handle units as "times something", i.e. a factor. This is the reason why we can add $1\,m + 2\,m = 1 \cdot m + 2 \cdot m = (1+2) \cdot m = 3\, m$ and can not add $1\,m + 2$. And it is the reason, why we can multiply both: $1\,m \cdot 2\,m= 2\,m^2$ and $1\,m \cdot 2 = 2\, m$.
Therefore any conversion factors can be treated as such, as factors, and you can likewise multiply the amount of quantity or change the unit, but you must not do both for this would square the factor because it would be applied twice. But it's only available once.

 

[Chestermiller]

it's ok if you all want to give up on me. I'm sorry if I am a pain. You can go ahead and delete the thread. I'm thinking maybe I should just ask my professor. Maybe he can get through to me. It's a lot easier when you can talk in person and the other person can yell at you and hammer it in. ;)

I mean it is not really "important". can we say that anytime I am presented with Joules/kg, I can just take the square root and get the speed? That it's always a

edit: lol a $\sqrt{9}:3$ conversion.

According to your rationale, if s is distance and a is acceleration, then s/a is time squared:$$\frac{s}{a}=t^2$$and $$t=\sqrt{\frac{s}{a}}$$But wait! The SUVAT equations tell us that $$s=\frac{1}{2}at^2$$and$$t=\sqrt{\frac{2s}{a}}$$So,$$t=\sqrt{2t^2}$$There's that pesky little 2 again. How do you explain this?

 

[gneill]

the $V_{rms} = \sqrt{\frac{3kT}{m}}$ is the square root of J/kg or v^2. But what about the 2? I guess that is already factored into the formula?

Yes, it's already factored into the formula. When you use the formula $v = \sqrt{\frac{2 \cdot KE}{m}}$ the constant "2" arrived via the derivation of $KE = \frac{1}{2}m v^2$ from the work-energy theorem, Newton's second law, and basic kinematics (you can find the derivation easily enough via a google search).

The expression for Vrms took a different route in its derivation, involving statistical mechanics. If you take the KE to be an average KE for a particle in an ensemble of particles of mass m then if you equate the two velocity formulae:

$\sqrt{\frac{3 k T}{m}} = \sqrt{\frac{2 KE}{m}}$

so

$3 k T = 2 KE$

$\frac{3}{2}k T = KE$

The left hand side of the last expression should look familiar from chemistry classes (or statistical mechanics)

 

[grandpa2390]

According to your rationale, if s is distance and a is acceleration, then s/a is time squared:$$\frac{s}{a}=t^2$$and $$t=\sqrt{\frac{s}{a}}$$But wait! The SUVAT equations tell us that $$s=\frac{1}{2}at^2$$and$$t=\sqrt{\frac{2s}{a}}$$So,$$t=\sqrt{2t^2}$$There's that pesky little 2 again. How do you explain this?

I don't explain it. That's what I am asking. I am not arguing that the 2 belongs. Obviously it doesn't because it is a well accepted conversion.

I am asking why. How do you explain the pesky little 2

 

[grandpa2390]

Yes, it's already factored into the formula. When you use the formula $v = \sqrt{\frac{2 \cdot KE}{m}}$ the constant "2" arrived via the derivation of $KE = \frac{1}{2}m v^2$ from the work-energy theorem, Newton's second law, and basic kinematics (you can find the derivation easily enough via a google search).

The expression for Vrms took a different route in its derivation, involving statistical mechanics. If you take the KE to be an average KE for a particle in an ensemble of particles of mass m then if you equate the two velocity formulae:

$\sqrt{\frac{3 k T}{m}} = \sqrt{\frac{2 KE}{m}}$

$3 k T = 2 KE$

$\frac{3}{2}k T = KE$

The left hand side of the last expression should look familiar from chemistry classes (or statistical mechanics)

this is exactly the issue that I am having. Thankyou.$\frac{3}{2}k T = KE$

and if I want to convert that kinetic energy to v, I have to multiply it by 2, divide by the mass, and take the square root.

$\sqrt{\frac{2*\frac{3}{2}kT}{m}}$

 

[Chestermiller]

I don't explain it. That's what I am asking. I am not arguing that the 2 belongs. Obviously it doesn't because it is a well accepted conversion.

I am asking why. How do you explain the pesky little 2

The 2 obviously does belong.

 

[grandpa2390]

The 2 obviously does belong.

... so then how does $t=\sqrt{2}t$

 

[Chestermiller]

... so then how does $t=\sqrt{2}t$

It doesn't. That's the result of your derivation, not the SUVAT result.

 

[grandpa2390]

A factor in front of a unit doesn't affect the unit, only the quantity. $cm$ and $m$ both measure the physical quantity length $L$. If this length is achieved by a velocity $V$ during a time $T$, say $V=10\frac{cm}{h}$ and $T=\frac{1}{2}\,h$, probably a snail, then the snail moved $L=V \cdot T = 10\frac{cm}{h} \cdot \frac{1}{2}\,h = 5 \, cm$ during this half hour. If we do the same calculation in the units meter and seconds, we get $L = 10 \cdot 10^{-2} \cdot m \cdot \frac{1}{3600\,s} \cdot \frac{1}{2} \cdot 3600 \,s = 5 \cdot 10^{-2} \,m = 0.05 \, m = 5\, cm$. Now the only difference between $0.05 \, m$ and $5\, cm$ is whether the $\frac{1}{100}$ is swallowed by the $5$ making it $0.05$ or by the unit, making it centi.

You can always handle units as "times something", i.e. a factor. This is the reason why we can add $1\,m + 2\,m = 1 \cdot m + 2 \cdot m = (1+2) \cdot m = 3\, m$ and can not add $1\,m + 2$. And it is the reason, why we can multiply both: $1\,m \cdot 2\,m= 2\,m^2$ and $1\,m \cdot 2 = 2\, m$.
Therefore any conversion factors can be treated as such, as factors, and you can likewise multiply the amount of quantity or change the unit, but you must not do both for this would square the factor because it would be applied twice. But it's only available once.

I don't think you understand the question I am asking. I already know that a constant doesn't change the units.

 

[grandpa2390]

It doesn't. That's the result of your derivation, not the SUVAT result.

so then the sqrt(2) doesn't belong. I'm going to backtrack a bit for a second

 

[grandpa2390]

so then the sqrt(2) doesn't belong. I'm going to backtrack a bit for a second

I think @gnell has answered my question completely. that kinetic energy is equal to $KE = \frac{3}{2}kT$ it looks like my problem is solved.

 

[Chestermiller]

so then the sqrt(2) doesn't belong. I'm going to backtrack a bit for a second

The 2 belongs in the SUVAT equations and the equation for the kinetic energy. The equations that you wrote are the ones that are incorrect.

 

[grandpa2390]

hold up

 

[grandpa2390]

Yes, it's already factored into the formula. When you use the formula $v = \sqrt{\frac{2 \cdot KE}{m}}$ the constant "2" arrived via the derivation of $KE = \frac{1}{2}m v^2$ from the work-energy theorem, Newton's second law, and basic kinematics (you can find the derivation easily enough via a google search).

The expression for Vrms took a different route in its derivation, involving statistical mechanics. If you take the KE to be an average KE for a particle in an ensemble of particles of mass m then if you equate the two velocity formulae:

$\sqrt{\frac{3 k T}{m}} = \sqrt{\frac{2 KE}{m}}$

so

$3 k T = 2 KE$
b
$\frac{3}{2}k T = KE$

The left hand side of the last expression should look familiar from chemistry classes (or statistical mechanics)

That formula is familiar to me now. This is all I needed. Is for someone to say "gramps, you Dummy! 3kT equal 2*KE not KE" and so it fits nicely in the $v=\sqrt{\frac{2KE}{kg}}$

that brings me back to units.
so there is a difference between a Kinetic Energy Joule (1/2 Mass x Velocity squared) and a Joule (Mass x Velocity squared)

 

[grandpa2390]

so there is a difference between a Kinetic Energy Joule (1/2 Mass x Velocity squared) and a Joule (Mass x Velocity squared)

it seems clear that a joule in potential energy is different from a Joule of KE. etc.

 

[Chestermiller]

so there is a difference between a Kinetic Energy Joule (1/2 Mass x Velocity squared) and a Joule (Mass x Velocity squared)

Nope. A Joule is a unit of energy (any kind).

 

[fresh_42]

There is a difference between the kinetic energy $E=\frac{1}{2}mv^2$ and the energy $E=mc^2$. It's the energy of a different thing, not a different unit.

 

[grandpa2390]

Nope. A Joule is a unit of energy (any kind).

I remember we had to be able to derive the Kinetic Energy (via integration). So I know how to do it. I guess I just wasn't really thinking about what I was doing at the time because I was coming from a different angle. And so, now I am confused when it is appearing to me that a joule is defined two different ways.

 

[Chestermiller]

I remember we had to be able to derive the Kinetic Energy (via integration). So I know how to do it. I guess I just wasn't really thinking about what I was doing at the time because I was coming from a different angle. And so, now I am confused when it is appearing to me that a joule is defined two different ways.

I can only suggest you go back and review the derivation of the kinetic energy. Then you will see specifically where the 2 comes from.

 

[grandpa2390]

I can only suggest you go back and review the derivation of the kinetic energy. Then you will see specifically where the 2 comes from.

it comes from the integration of momentum right? v becomes 1/2 v ^2
I got that, I'm just trying to connect the pieces.

At this point, I guess, the question has evolved into. What is a Joule. Besides being a measurement of energy. Because if the solution for understanding the KE Joule is by integration of Momentum. It seems like there is no connection between the formula for kinetic energy and a joule except for the units and the word joule. and that there would be a conversion of 1/2. of some sort. which only makes sense if they are not the same joule.

What is a Joule. Why can we have 1/2 mv2 and mv2 and they both be joules. Not half joules and whole joules, just joules.

 

[fresh_42]

... when it is appearing to me that a joule is defined two different ways.

Many: $$1J=1Nm=1VAs=1CV=1Ws=1\frac{kg\cdot m^2}{s^2}=\frac{1}{4,1868} cal=10^{7}erg$$ This doesn't mean that something measured in Watt seconds (output of a power plant) is directly comparable to something measured in Newton meter (torque of the motor in your car) or by kilogram times speed squared (kinetic energy in crash) or by calories (food) or by erg (temperature transport).

 

[Chestermiller]

it comes from the integration of momentum right? v becomes 1/2 v ^2
I got that, I'm just trying to connect the pieces.

At this point, I guess, the question has evolved into. What is a Joule. Besides being a measurement of energy. Because if the solution for understanding the KE Joule is by integration of Momentum. It seems like there is no connection between the formula for kinetic energy and a joule except for the units and the word joule. and that there would be a conversion of 1/2. of some sort. which only makes sense if they are not the same joule.

What is a Joule. Why can we have 1/2 mv2 and mv2 and they both be joules. Not half joules and whole joules, just joules.

I Googled the definition of a Joule: One joule is defined as the amount of energy exerted when a force of one Newton is applied over a displacement of one meter. Do you see the connection to kinetic energy now?

 

[grandpa2390]

Many: $$1J=1Nm=1VAs=1CV=1Ws=1\frac{kg\cdot m^2}{s^2}=10^{7}erg$$ This doesn't mean that something measured in Watt seconds (output of a power plant) is directly comparable to something measured in Newton meter (torque of the motor in your car) or by kilogram times speed squared (kinetic energy in crash) or by calories (food) or by erg (temperature transport).

so when we are plugging in joules into a formula. we should be mindful of what kind of energy we are measuring. It sounds like a nobrainer. you couldn't really plug the formula for potential energy into the speed root mean squared formula... it just wouldn't make sense (to me at least. Though at this point I could believe anything lol)

 

[grandpa2390]

I Googl
ed the definition of a Joule: One joule is defined as the amount of energy exerted when a force of one Newton is applied over a displacement of one meter. Do you see the connection to kinetic energy now?

except for the 1/2. yeah. f=ma m*a*d = mv^2

I see how kinetic energy appears to be half a joule. it throws me off that the only way to get that 1/2 is by the integration. it doesn't make sense if you try to derive it by manipulating formulas.

edit: so far it doesn't make sense

 

[gneill]

That formula is familiar to me now. This is all I needed. Is for someone to say "gramps, you Dummy! 3kT equal 2*KE not KE" and so it fits nicely in the $v=\sqrt{\frac{2KE}{kg}}$

that brings me back to units.
so there is a difference between a Kinetic Energy Joule (1/2 Mass x Velocity squared) and a Joule (Mass x Velocity squared)

Numeric constants are never involved when you are doing unit (or dimensional) analysis. Drop all numeric constants immediately as they are unitless.

Don't marry units to a particular equation that happens to involve quantities with units. The constants that relate the variables in different contexts may be different. Note that KE is not a unit. We define units of energy in terms of more fundamental units, so for example the Joule is a compound unit:

$J = \frac{kg~m^2}{s^2} \Longrightarrow \frac{[M] [L]^2}{[T]^2}$

It's okay to ask about how to compare energy calculated or expressed in different forms, particularly when we know that they are related through conservation in a given system. The formulas are associated with the system under study and tell you how energy moves or manifests in different forms in a given system. The constants related to the particular system come into play to tell you how the "trade" is managed.

For example, consider KE and PE in a system where mass moves in a gravitational field. The two are related through the conservation of energy and both have units of energy, but they are calculated via different formulas using different constants. That does not mean that the units of energy are different for each, or that " J = 2J" to paraphrase your earlier post. In fact we know that KE and PE can be traded in the system and the constants used to relate the quantities depend upon the system under study. The amount of energy represented by 1 Joule is a fixed quantity that does not vary, but will show up as different measures of some observables in a given system (speed versus height, for example).

 

[fresh_42]

so when we are plugging in joules into a formula.

We do not plug in Joule. We plug in measurements of certain physical quantities and calculate an energy, and hopefully Joule come out of the equation.

we should be mindful of what kind of energy we are measuring.

You don't need to. You can write the output of a power plant in calories. You will probably only face problems to be understood. Or you can ask the local car seller about the torque in Watt seconds, but Newton meter will be more likely to get a quick response. The energy is what we compare, the unit is with what we compare it to. And the energy is calculated or measured, the unit a choice of communication to tell the result of the calculation or measurement. The formulas hold for any unit, as long as it announces energy.