Why is the Velocity equal to joules per kilogram?

[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.

 

[Chestermiller]

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

$$F = m\frac{dv}{dt}$$
Multiplying both sides by velocity:
$$Fv = mv\frac{dv}{dt}$$or equivalently:$$F\frac{ds}{dt} = mv\frac{dv}{dt}$$where s is the distance over which the force is applied. Multiplying both sides by dt yields:
$$Fds = mvdv$$Integrating yields:
$$Fs=\frac{1}{2}mv^2$$
Notice that the left hand side of this equation is the force times the displacement over which it is applied. This in expressed in Joules. Notice that the right hand side of the equation is the increase in kinetic energy, equal to the number of Joules of work done by the force. So kinetic energy Joules are the same as force times distance Joules.

 

[kuruman]

To add my two bits at this point, there is another dimension for confusion concerning Joules. If you break down the dimensions of energy units, you get$$[J]=\frac{[M][L]^2}{[T]^2} $$ but so do you when you break down dimensions of torque. Two completely different beasts, same dimensions.

 

[JR Jonsson]

The area of a rektangel is for ex.
2 m • 4 m = 8 m^2

The area of a triangel is for ex.
2 m • 4 m /2 = 4 m^2

Make a diagram for velosity and time in the case of free fall and you get a triangel.
The area of this triangel is for the first second
1 s • 10/2 m/s = 5 m
( gravity 10 m/s^2, or 10 m/s every second of free fall, friction neglected)

The kinetik energy for 10 m/s and potential energy for 5 m

For two seconds we get;
2 s • 20/2 m/s = 20 m
And speed 2 s • 10 m/s^2 = 20 m/s

If you can follow this you see that the speed has doubled while the fall is four times longer.

That is why the 2 is needed between kinetik and potential energy, for bodies as well as for fluids.

Abaut the torque;
It really helps if you think of torque as Nm/rad

If you turn a wheel then you get a work done;
Nm/rad • rad = Nm = J

For power;
Nm/rad • rad/s = W

PS. An angel is not considered to have a dimension by the expertise, but it helps me to think of radius as a special case of length; m/rad DS

 

[Ricardo Reyes]

I just want to add my logic for the units.

J= Kg*m2*s2, so J/Kg is the same as Kg*m2*s2/kg Got it? In this case, the kilogram unit cancels out, and we're left with m2/s2 and if we take the square root of that we have the velocity units (m/s).
Hope that this helps anyone

 

[kuruman]

I just want to add my logic for the units.

J= Kg*m2*s2, so J/Kg is the same as Kg*m2*s2/kg Got it? In this case, the kilogram unit cancels out, and we're left with m2/s2 and if we take the square root of that we have the velocity units (m/s).
Hope that this helps anyone

That's good, but this thread is more than 4 years old.