Power?

[NTuft]

On 6/14/2022 at 3:28 AM, swansont said:

Who is saying it's a pseudovector? I don't see that in your book excerpt. It's a vector, following the protocol of a cross product.

I may have misunderstood: the link to the un-vetted reference manual qualifies angular momentum or velocity as a pseudovector, but torque is a time derivative of the angular momentum or velocity??? Likely 2nd deriv of momentum and 1st deriv of velocity?

Quote

By definition, torque τ = r × F. Therefore, torque on a particle is equal to the first derivative of its angular momentum with respect to time.

Handwiki.org: Physics: Torque

 

 

I like to pretend the quartz crystal in my watch has an AC Josephson junction running it, but, it has stopped, and who am I kidding. It is almost Dada's day in the U.S.; happy belated St. Joseph's Day to those in Europe.

 

The question at hand is, "Power?", but capiert has gone AWOL. We have to extensively formulate all of classical physics first, and then teach capiert calculus, joigus; so no rest 'til then. What is the math in the background of your profile?!? Is /(P)/ for Power? Please define equations here.

 

 

U or V for voltage... 

Power is also  , torque x angular velocity . M L T 

 

 

22 hours ago, Markus Hanke said:

I think it’s also important to mention here that we don’t actually live in a Newtonian universe, except as an approximation in the low-energy, low-velocity regime. What this means is that Noether’s theorem actually applies to an action within at least a 4D Minkowski spacetime (ignoring gravity for now) - in which case the conserved quantity associated with time-translation invariance is not just ‘energy’, but the full stress-energy-momentum tensor. This should be obvious, since ‘energy’ on its own depends on the observer, and is thus conserved only within a given frame.

If you could explain the stress-energy-momentum tensor to @icarus2 you can save him from most recent "imaginary" gravity madness, but I think you did try in his gravitational self-energy thread. All these power units seem to include  , I don't know if we're walking down to Planck units for gravity but they're there, too. To get to 4D Minkowski space we've appended our "imaginary" friend in the Lorentz transformations to the time component, making the time dimension complex compared to the spatial. The ubiquity of the "imaginary" here, and in the Fourier transformations (from complex valued functions) we're doing in astrophysics when measuring vibration have our friend icarus2 confounded on the physical groundedness of the "imaginary" time, or otherwise complex (e.g. it's used in the quantum momentum operator -- is that because momentum has to be gained in time, or why?) values that are possible (so far off-topic here we may need a moderator). Thanks from me for mentioning s-e-m tensor, and to other mentions of worthwhile topics on the matter at hand.

 

 

@exchemist, I take one swig from the jug of methylene blue as punishment for losing the point. But i maintain energy directs work.

Edited June 15, 2022 by NTuft
Henri Poincaré may have appended the imaginary unit to complexify the time dimension. I mentioned Einstein-Cartan to icarus2 but I think there is negative Gauss curvature and he is saying it's flat; optical measures support bending light in support of GR.

[exchemist]

7 hours ago, NTuft said:

I may have misunderstood: the link to the un-vetted reference manual qualifies angular momentum or velocity as a pseudovector, but torque is a time derivative of the angular momentum or velocity??? Likely 2nd deriv of momentum and 1st deriv of velocity?

 

 

I like to pretend the quartz crystal in my watch has an AC Josephson junction running it, but, it has stopped, and who am I kidding. It is almost Dada's day in the U.S.; happy belated St. Joseph's Day to those in Europe.

 

The question at hand is, "Power?", but capiert has gone AWOL. We have to extensively formulate all of classical physics first, and then teach capiert calculus, joigus; so no rest 'til then. What is the math in the background of your profile?!? Is /(P)/ for Power? Please define equations here.

 

 

U or V for voltage... 

Power is also  , torque x angular velocity . M L T 

 

 

If you could explain the stress-energy-momentum tensor to @icarus2 you can save him from most recent "imaginary" gravity madness, but I think you did try in his gravitational self-energy thread. All these power units seem to include  , I don't know if we're walking down to Planck units for gravity but they're there, too. To get to 4D Minkowski space we've appended our "imaginary" friend in the Lorentz transformations to the time component, making the time dimension complex compared to the spatial. The ubiquity of the "imaginary" here, and in the Fourier transformations (from complex valued functions) we're doing in astrophysics when measuring vibration have our friend icarus2 confounded on the physical groundedness of the "imaginary" time, or otherwise complex (e.g. it's used in the quantum momentum operator -- is that because momentum has to be gained in time, or why?) values that are possible (so far off-topic here we may need a moderator). Thanks from me for mentioning s-e-m tensor, and to other mentions of worthwhile topics on the matter at hand.

 

 

@exchemist, I take one swig from the jug of methylene blue as punishment for losing the point. But i maintain energy directs work.

No, work is a form of energy. A force, acting through a distance, F x d, = W. Mechanical work is one form of energy. For example if you lift a mass you exert a force to overcome gravity and do work on the object, thereby increasing its gravitational potential energy.

To say energy “directs” work makes no sense.

[NTuft]

8 hours ago, exchemist said:

No, work is a form of energy. A force, acting through a distance, F x d, = W. Mechanical work is one form of energy. For example if you lift a mass you exert a force to overcome [ed: i.e. against the direction of the other force involved,] gravity and do work on the object, thereby increasing its gravitational potential energy.

To say energy “directs” work makes no sense.

"In general, modern physics holds the view that the universe - and systems therein - spontaneously drives toward a state of lower energy, if possible."
When an enzyme is directing alterations, do you conceive the energy involved as a nebulous statement on the system? I would argue that the work is very specifically directed and also can be localized in space, and while directed by net energetic concerns of the system the specificity of the alterations and requisite structural demands again speak towards directionality buried in the conventions.

[exchemist]

41 minutes ago, NTuft said:

"In general, modern physics holds the view that the universe - and systems therein - spontaneously drives toward a state of lower energy, if possible."
When an enzyme is directing alterations, do you conceive the energy involved as a nebulous statement on the system? I would argue that the work is very specifically directed and also can be localized in space, and while directed by net energetic concerns of the system the specificity of the alterations and requisite structural demands again speak towards directionality buried in the conventions.

There is no mechanical work done by an enzyme. And while systems tend to change towards lower energy states, that is not in general a spatial direction. You are using terms in too vague a way for a scientific conversation to be possible.

[NTuft]

3 hours ago, exchemist said:

There is no mechanical work done by an enzyme. And while systems tend to change towards lower energy states, that is not in general a spatial direction. You are using terms in too vague a way for a scientific conversation to be possible.

Judging by displacement wrought by the chemical reaction mechanism, it is doing work by applying force over time: "In physics, work is the energy transferred to or from an object via the application of force along a displacement." 

See biophysical chemistry, e.g.: "A ribosome is a biological machine that utilizes protein dynamics on nanoscales to translate RNA into proteins"

Edited June 16, 2022 by NTuft
tex equation

[exchemist]

36 minutes ago, NTuft said:

Judging by displacement wrought by the chemical reaction mechanism, it is doing work by applying force over time: "In physics, work is the energy transferred to or from an object via the application of force along a displacement." 

See biophysical chemistry, e.g.: "A ribosome is a biological machine that utilizes protein dynamics on nanoscales to translate RNA into proteins"

That’s rather ridiculous. An enzyme is a catalyst for a reaction. The energy change that brings the reaction about is to be found in the reactants and products. And this change does not take the form of  mechanical work, but changes in chemical potential energy, in chemical bonds.

[NTuft]

4 hours ago, exchemist said:

That’s rather ridiculous. An enzyme is a catalyst for a reaction. The energy change that brings the reaction about is to be found in the reactants and products. And this change does not take the form of  mechanical work, but changes in chemical potential energy, in chemical bonds.

My chemicals and enzymes and much more nuts and Boltz, mann, than yourzz... 
Your perspective, knowledge and understanding is appreciated. I think I can understand what you're saying about Energy as descriptive merely system-wide? I don't think we're too off base, as I think an enzyme kinetics graph is literally plotting  against "reaction progress". Also plotted is E + S ... E + P or some such business.

[NTuft]

Quote

 

"Power is related to intensity at a radius ; the power emitted by a source can be written as:

 "

"In physics, intensity of radiant energy is the power transferred per unit area, where the area is measured on the plane perpendicular to the direction of propagation of the energy. In the SI system, it has units watts per square metre (W/m2), or kg⋅s−3 in base units. Intensity is used most frequently with waves such as acoustic waves (sound) or electromagnetic waves such as light or radio waves, in which case the average power transfer over one period of the wave is used. Intensity can be applied to other circumstances where energy is transferred. For example, one could calculate the intensity of the kinetic energy carried by drops of water from a garden sprinkler.

The word "intensity" as used here is not synonymous with "strength", "amplitude", "magnitude", or "level", as it sometimes is in colloquial speech.

Intensity can be found by taking the energy density (energy per unit volume) at a point in space and multiplying it by the velocity at which the energy is moving. The resulting vector has the units of power divided by area (i.e., surface power density)."

 

from handwiki reference.

 

quoting/paraphrasing here:

"Transitions between harmonic-oscillator states can occur only between adjacent levels -- a selection rule.

We shall show... that if a harmonic oscillator is irradiated with electromagnetic radiation propagating in the z direction, then the probability that the oscillator makes a transition from state  to ' is proportional to the square of the integral

where  is the z component of the dipole moment. This type of integral, which occurs frequently in spectroscopy, is called a dipole transition moment.
...
the selection rule for vibrational transitions under the harmonic-oscillator approximation... normal modes... infrared active... infrared inactive."¹

 

 

¹McQuarrie, Donald A. (Donald Allan). Quantum Chemisty. 2nd ed., University Science Books, 2008. pgs. 237-238

²Attached pgs. 138-139 from ¹ on Newton's equations (calculus), gradient operator, level curves, equipotential lines, Fick's law of diffusion, Fourier's law of heat flow, mechanical potential energy, electrostatic potential. I think this argues is exchemist's favor; those level curves look like they direct the work; I don't understand energy well or walls properly, , or 

[exchemist]

11 hours ago, NTuft said:

My chemicals and enzymes and much more nuts and Boltz, mann, than yourzz... 
Your perspective, knowledge and understanding is appreciated. I think I can understand what you're saying about Energy as descriptive merely system-wide? I don't think we're too off base, as I think an enzyme kinetics graph is literally plotting  against "reaction progress". Also plotted is E + S ... E + P or some such business.

Agree about the ZZZzzzz..........😆

[Capiert]

Yuk!
What a messterpiece.
I'm trying to simplify (to algebra)
 so (even) a 5 year old can understand (Trump mentality, noted (by reporters) for its advantages),
 but it explodes & gets scattered
 into so much complexity (some interesting though),
 (really avoiding the question, with (fake) substitutes)
 (unfortunately demonstrating how little was understood, in some cases).

When the cat'( i)s away
 the mouse will play.

To get back on track.

 

On 6/3/2022 at 2:50 AM, joigus said:

What's the speculation here?

 

Capiert:
The (basic) question was
 whether the (Jame's Watt's)
 Power_equation's (forced) speed
 is a speed_difference v=v_f-v_i (No?, but most commonly claimed?)
 or the average_speed v_a=d/t (Yes? New).

The algebra seems to confirm (the later, & itsself),

(Unfortunately opening the question
 why the calculus did NOT do the job (right,
 for that question),
 which I DON'T really want to discuss (calculus yet).
Maybe for later.)

(I guess) I was a little sloppy
 because gravity's weight force
 was intended only as an example of force;
 NOT all forces in general
 as Swansont put it.

On 6/2/2022 at 12:28 PM, swansont said:

On 5/31/2022 at 7:07 PM, Capiert said:

Power

 P=F*d/t, va=d/t

 is the Force (e.g. Weight Wt=m*g)
 F=m*g
 multiplied
 by (e.g. the height h=) distance d
 per time t.

Here is an example of using a specific equation and trying to apply it in general.

Power applies to more than the situation of an object falling under gravity, so one cannot make a general claim that force is weight, since there are other forces. If you use weight, then the equation will only apply to a falling object, and also only if you can assume that this is a constant force.

Should (better be)

Power

 P=F*d/t

 is the Force F=m*a,
 multiplied
 by the distance d
 per time t.

The average_speed
 va=d/t
 is the distance d
 per time t

E.g.
Let the (linear) acceleration a=g be gravitational free_fall's acceleration,
& the force F=Wt be the Weight Wt=m*g,
& the distance d=h height (fallen).

P=Wt*h/t.

On 5/31/2022 at 7:07 PM, Capiert said:

KE=m*(v_f²-v_i²)/2.

On 5/31/2022 at 7:36 PM, Bufofrog said:

It looks like you are saying an object moving at a constant velocity has zero KE,

I did NOT say that (an object moving at a constant velocity has zero KE);
 you did.

On 5/31/2022 at 7:36 PM, Bufofrog said:

since v_{f =} v_i   

If initial & final speeds are the same
 then the KE (between them)
 is zero.
E.g. For 2 objects=bodies.

E.g. For 1 object=body
 the (same, single) mass
 has NOT be accelerated (e.g. faster)
 from its initial speed;
 or
 e.g. it has NOT been accelerated
 from its
 final speed.

Take your pick.

On 5/31/2022 at 7:36 PM, Bufofrog said:

That's a problem. 

I (still) do NOT see your problem.

E.g. Same speeds
 (for same mass(es))
 means NO kinetic_energy(_difference) KE,
 but KE is a difference (KE_d=KE_f-KEi)
 of (the 2) KE's;
 although I think
 Swansont will want to correct me there.

On 6/1/2022 at 1:33 AM, Bufofrog said:

The KE equation you wrote would mean that an object moving at a constant speed would have zero KE. 

No it does NOT.

The equation uses 2 (different) speeds
 even if 1 of them (speeds)
 is allowed to be zero.

A difference of KE's
 means that you are subtracting 2 KE's.
So your components are KE's.

E.g.
KE_d=m*(v_f²/2)-m*(v_i²/2).
KE_f=m*(v_f²/2).
KE_i=m*(v_i²/2).

v_f=v_i+v
v_i=v_f-v
v=v_d=v_f-v_i.

On 6/1/2022 at 1:33 AM, Bufofrog said:

That is incorrect, hence the equation is incorrect.

Please show me the error (if so).

As far as I know
 that math (syntax) works (for me).
Maybe you have a task, example
 where I can use it?

My concept (syntax)
 of KE (=KE_d=KE_f-KE_i)
 is (a (KE) difference,
 &) similar to yours,
 but (differs in that it)
 is (also) extrapolated thru
 to (both) the initial_KE KE_i=KE_f-KE
 & the final_KE KE_f=KE_i+KE.

(But Swansont commented=identified
 that is delta_KE.)

 What I'm saying
 is, the initial_speed vi
 is [often]
 invisible
 (for same speed objects);
 but NOT zero!

Even if objects=bodies
 seem static;
 they are (really) moving!

Same speed objects
 can NOT accelerate
 each other
 (because they do NOT collide
 with each other).

Again
PS: Very interesting.

It looks like you did NOT get it (=the (2) perspective(s)).
How can anything moving,
 NOT have KE?

 If v_f = v_i (then that) means you can NOT see the(ir) motion.
 (..when compared).

The 2 objects (seem to) stand still,
 (perhaps) with a constant (separation) distance.

E.g. 2 billiard balls
 on a (billiard) table
 (although the world=Earth is turning).

Our reference(_speed)
 vi
 could be the Earth's rotation speed
 e.g. ruffly ~1000 [m/s] eastwards.

The (2) balls are NOT going
 to do anything
 (e.g. (they are NOT going to) collide)
 with each other
 because they stay still
 (wrt the billiard table,
 on Earth);
 but (both (balls)) are rotating
 with the Earth_speed
 vi~1000 [m/s] eastwards
 (i.e. wrt the Earth)!

Their KE would NOT become obvious
 until they collide
 with something else
 moving
 at a completely different speed.

E.g. a cue hitting a ball.

((Please) allow me to exaggerate.)

E.g. If the cue travelled ~-1000 [m/s] e.g. westwards
 (to hit the ball)
 to compensate against the Earth's rotational speed
 (so that the cue would seem like zero speed
 wrt the Earth's center).

KE, (e.g. the speed_(energy)_change)
 can only be [acquired=] "received
 from", or else "transferred
 to" another mass('s motion).

 

 

Edited July 18, 2022 by Capiert
The following text was automatically (=NOT intentionally!) crossed out,  I CAN'T remove the crossout. So I deleted.

[Bufofrog]

25 minutes ago, Capiert said:

Please show me the error (if so).

Sure, your equation is:

KE=m*(v_f²-v_i²)/2.

Let's plug in some numbers into your equation.  Assume a mass of 10 kg is moving at 10 m/s.

KE = 10 kg ( 100 m^2/s^2 - 100 m^2/s^2)/2 = 0.  That is the wrong answer.

The correct answer is 500 J.

Edited July 18, 2022 by Bufofrog

[Capiert]

Why is (your) initial_speed v_i (=v_ref) NOT zero?
You have said the 10 kg mass is "moving" at (v_f=v_i+v=0+v=) 10 m/s,
 so what is the reference
 that you used?

[swansont]

41 minutes ago, Capiert said:

Capiert:
The (basic) question was
 whether the (Jame's Watt's)
 Power_equation's (forced) speed
 is a speed_difference v=v_f-v_i (No?, but most commonly claimed?)
 or the average_speed v_a=d/t (Yes? New).

...

Should (better be)

Power

 P=F*d/t

Your assertion of Watt's equation is questionable, and this last equation is not quite correct

There's a difference between instantaneous power and average power

 

Instantaneous power is given by P = dW/dt, which can be rewritten as F.ds/dt = F.v

Force dot product with velocity is the instantaneous power. It's not a velocity difference. It's the velocity at the moment you are calculating the power.

 

5 minutes ago, Capiert said:

Why is (your) initial_speed v_i (=v_ref) NOT zero?
You have said the 10 kg mass is "moving" at (v_f=v_i+v=0+v=) 10 m/s,
 so what is the reference
 that you used?

Because they are saying the speed is constant. Your equation says the KE is zero, which is wrong. As I said before, your equation is for the change in KE, not for KE

[Capiert]

Continued, (from my previous)

That speed_change
 is the acceleration
 e.g. to or from
 a collision,
 but once that collision (duration e.g. t<1 , stops=) finishes
 the(n that) acceleration stops;
 so that friction decelerates the ball (further)
 to a (slow, much longer duration, e.g. t>1)
 stop.

 But,
 an Earthquake
 (e.g. an abrupt change
 in the Earth_plates' (rotational) speed)
 could shake
 the (table &) balls
 (in)to motion.

Done.

How would you describe (your) KE?
 (..without problems).

 That was mine (=my description).

Identical speeds
 does NOT mean
 zero KE.

 It only means
 zero KE
 between
 the 2 objects
 that have the same speed.

Constant speed
 does NOT mean
 zero KE.

 In fact any speed
 (helps)
 indicate a KE.

(E.g. If you (also) know the mass(es), too.)

But any (observed) speed(_difference) v
 is a (speed_)difference v,
 i.e. that uses a reference (initial_)speed v_i
 (as basis),
 to observe the (final_)speed v_f=v_i+v.

(But)
 we do NOT see the (final_)speed v_f
 because it has the (initial_)speed v_i
 in it,
 which we exclude.

E.g. The Earth's_rotation_speed
 v_i~1000 [m/s] eastwards
 remains invisible
 (to us)
 because everything (static) around us
 is (also) moving
 at (approx.) that same speed.

How can we solve
 your problem? 

P.S.
When Earthquakes happen
 1 of the problems
 is the ((tectonic) plates') rotation
 slows down
 & then continues
 to speed up;
 repeatedly.

That kind of (kinetic) Energy
 can bring down skyscrapers (crashing),
 (it's so powerful!)

(m*PE=m*KE)

 m*m*a*d=m*m*(v_f²-v_i²)
 only equates
 the accelerated distance
 with the speed change
 of a mass;
 NOTHING more. 

a*d=((v_f²)-(v_i²)).

Edited July 18, 2022 by Capiert
That text was automatically (=NOT intentionally!) crossed out,  I CAN'T remove the crossout. Please read (if you can).

[swansont]

1 minute ago, Capiert said:

Continued, (from my previous)

That speed_change
 is the acceleration
 e.g. to or from
 a collision,

What if there is no acceleration?

 

[Capiert]

51 minutes ago, swansont said:

Your assertion of Watt's equation is questionable, and this last equation is not quite correct

There's a difference between instantaneous power and average power

Undoubtably average_power.

I was NOT aware
 that James Watt used instantaneous power.

He used how fast
 a (180 lb (force) ~82 kg (mass)) "horse"
 walking (continuously) a circle,
 using yards (=3 feet=0.9144 m)
 & seconds (time),
 could wheel up water
 from a well
 against (the horse's) gravity's weight (806 N).

~737 Watt (rounded to 740 W for 75 kg).
746 W would have needed 75.59 kg (~167 lb) at 1m/s.

I question that math
 was any great deed
 of calculus.

Watt determined that a horse could turn a mill wheel 144 times in an hour (or 2.4 times a minute).^[6] The wheel was 12 feet (3.7 m) in radius; therefore, the horse travelled 2.4 × 2π × 12 feet in one minute. Watt judged that the horse could pull with a force of 180 pounds-force (800 N). So:

Watt defined and calculated the horsepower as 32,572 ft⋅lbf/min, which was rounded to an even 33,000 ft⋅lbf/min.^[7]

Watt determined that a pony could lift an average 220 lbf (0.98 kN) 100 ft (30 m) per minute over a four-hour working shift.^[8] Watt then judged a horse was 50% more powerful than a pony and thus arrived at the 33,000 ft⋅lbf/min figure.

https://en.wikipedia.org/wiki/Horsepower

50 minutes ago, swansont said:

What if there is no acceleration?

Then the "speed" should be constant.

But the real (more probing) question
 is from where
 or what "speed" (reference)
 are you using
 to measure the "speed"
 you are observing.

(Obviously, your own speed
is the reference, there.)
 

Edited July 18, 2022 by Capiert

[swansont]

1 hour ago, Capiert said:

Undoubtably average_power.

I was NOT aware
 that James Watt used instantaneous power.

He used how fast
 a (180 lb (force) ~82 kg (mass)) "horse"
 walking (continuously) a circle,
 using yards (=3 feet=0.9144 m)
 & seconds (time),
 could wheel up water
 from a well
 against (the horse's) gravity's weight (806 N).

~737 Watt (rounded to 740 W for 75 kg).
746 W would have needed 75.59 kg (~167 lb) at 1m/s.

I question that math
 was any great deed
 of calculus.

Watt determined that a horse could turn a mill wheel 144 times in an hour (or 2.4 times a minute).^[6] The wheel was 12 feet (3.7 m) in radius; therefore, the horse travelled 2.4 × 2π × 12 feet in one minute. Watt judged that the horse could pull with a force of 180 pounds-force (800 N). So:

Watt defined and calculated the horsepower as 32,572 ft⋅lbf/min, which was rounded to an even 33,000 ft⋅lbf/min.^[7]

Watt determined that a pony could lift an average 220 lbf (0.98 kN) 100 ft (30 m) per minute over a four-hour working shift.^[8] Watt then judged a horse was 50% more powerful than a pony and thus arrived at the 33,000 ft⋅lbf/min figure.

https://en.wikipedia.org/wiki/Horsepower

If the speed is constant, then why all the commotion about average speed? It's constant! There's no reason to worry about initial and final speed, since it doesn't change.

 

1 hour ago, Capiert said:

Then the "speed" should be constant.

But the real (more probing) question
 is from where
 or what "speed" (reference)
 are you using
 to measure the "speed"
 you are observing.

(Obviously, your own speed
is the reference, there.)
 

Speed is always relative to the frame of reference to which it is measured. As is kinetic energy.

What's your point?

[Bufofrog]

2 hours ago, Capiert said:

Why is (your) initial_speed v_i (=v_ref) NOT zero?

Because we are talking about a constant velocity.

2 hours ago, Capiert said:

You have said the 10 kg mass is "moving" at (v_f=v_i+v=0+v=) 10 m/s,
 so what is the reference
 that you used?

That would be the reference frame that measures the objects velocity at 10 m/s.

[Capiert]

17 hours ago, swansont said:

If the speed is constant, then why all the commotion about average speed? It's constant!

Constant speed is NOT how the universe is working.
& I am looking for a more general, (but) simpler description.
Constant speed is the exception.
 (E.g. "Special" Relativity.)
Interaction is the (higher, more dominating) rule.
(Actually more basic=fundamental.)

& there is NOTHING
 which says
 acceleration must always be linear.

I get the idea "average" speed
 is an eye_sore for you
 because you prefer
 your instantaneous speed, instead.
But that (instantaneous speed)
is theoretical,
 & I suspect sometimes misleading.
(I DON'T want to be lulled
(or else tempted)
into a false security.
A(n unexpected surprise)
 jack_in_the_box
 (e.g. delayed) time bomb
 is NOT my idea of fun
 (for accuracy
 in extrapolation).)

Calculus works "perfectly" (=flawlessly, NO errors)
 for "linear" acceleration
 because it is based on an (exact) "average";
 (e.g. 2 triangles making a rectangle
 using a diagonal (line))
 but I have my doubts
 for NON_linear (line) examples
 & irregular curves.

E.g.
Sigma_summing
 & (versus) the integrals (integrating, integration)
 should give
 the same (=identical)
 answers;
 but DON'T always. 

I would prefer
 to develop
 something
 more simple(?)
 & experimental,
 (e.g. averages),
 based on algebra, instead.

But I also notice(d)
 that algebra has its limits
 (& errors) too.

(So I can NOT trust algebra completely, either.)

But I have NOT had enough chance
 to pinpoint
 exactly what those flaws=weaknesses are.
(It happens so rarely, like a flipped coin standing on its edge.)
I suspect it(s (math) flaws, or errors) has
 to do with (math) syntax compromises
 when dealing with negative exponents
 for totally=completely (symmetrically) reversible (e.g. recoverable?) math.
E.g. A "negative" exponent
 should mean "rooting"
 as the reverse (operation)
 of exponentiation;
 (instead of a divided exponentiation).
(Not to mention the meaning of the exponent zero, then.)
& then there is the theme of sequence (history).

(Math is a language
 & its sentences
 are equations
 (the phrases formulas).)

But that'( i)s (all) too speculative
 right now.
Too early to say (anything useful).

--

You are content with
 that you have learned
 some (of th(os)e complexity's) math rules,
 & use them.

I am NOT.

I am more interested
 in finding the reasons
 why some (math) things (fail)
 do NOT work sometimes (as we (would) wish);
 & (then) try to fix=repair them.

Experimenting
 with equations of motion
 & (so_called) mass
 are a good place=way
 to start,
 at seeing & discovering
 how this universe works.
Geometry as well.
(E.g. The "basis". Real proof!)

17 hours ago, swansont said:

There's no reason to worry about initial and final speed, since it doesn't change.

NOT for me.

I CAN'T follow you there,
 because "getting" (e.g. interacting)
 from initial_speed
 to final_speed
 (when they (speeds) do NOT equal)
 is acceleration.
Making them (2 speeds) equal
 is an exception,
 because (that is why) they have different names
 so that they do NOT have to be the same.
The (2 speeds)
 "can" be quite different.

I'm aimed (=intended) at collisions, or jerks!

(E.g. Unifying) the changes. 

Maybe you can tell me what I "should" be saying
 (or have NOT said)
 in my syntax description
 for initial (versus final) speeds?

Because what I say
 (just) does NOT seem
 to land
 to you (all)
 as intended.

17 hours ago, swansont said:

Speed is always relative to the frame of reference to which it is measured. As is kinetic energy.

What's your point?

Energy is (all) about (linear) acceleration (i.e. gravity).

That was the reason
 for the vis_viva (living force, ~2*KE)
 controversy;
 & why Gravesande "abandoned" Descartes & Newton's momentum.

(& that is the reason why we live with that mess NOW.
i.e. Dark Energy ERRORS.)

The speed is NOT constant for acceleration(s),
 such as (for, in) collisions
 where m*E is transferred.
 KE=PE (pronounced "keep", versus PE=KE pronounce "peek")
 (e.g. Conservation of Energy COE)
 might work (well)
 for (accelerating) a single (moving) mass
 that stays the same (sized mass);
 but (it, KE=PE)
 does NOT (always) work well
 when that mass interacts
 (& is accelerated)
 by other various sized masses
 (&) of different speeds.

For them (collisions, interactions),
 m*E works better,
 because it is more general, universal,
 uniting conservation of momentum COM,
 & Energy COE
 together
 into 1 formula.
 

On 6/2/2022 at 12:28 PM, swansont said:

And mass-energy is famously not a conserved quantity.

I guess I am an early bird.

(DON'T forget the worm.)

On 6/2/2022 at 12:28 PM, swansont said:

∆KE = KE_f-KE_i

i.e. KE_f-KE_i is the change in kinetic energy, not the kinetic energy. This is a very important distinction.

What then is KE?
Forced distance?

On 6/2/2022 at 12:28 PM, swansont said:

Here is an example of using a specific equation and trying to apply it in general.

Power applies to more than the situation of an object falling under gravity, so one cannot make a general claim that force is weight, since there are other forces. If you use weight, then the equation will only apply to a falling object, and also only if you can assume that this is a constant force.

I guess you mean
 I can NOT rely on
 gravity's force
 as constant.
So its a poor example.
?

20 hours ago, Bufofrog said:

22 hours ago, Capiert said:

Why is (your) initial_speed v_i (=v_ref) NOT zero?

Because we are talking about a constant velocity.

22 hours ago, Capiert said:

You have said the 10 kg mass is "moving" at (v_f=v_i+v=0+v=) 10 m/s,
 so what is the reference
 that you used?

That would be the reference frame that measures the objects velocity at 10 m/s.

That seems like an indirect answer
 by trying to be (more) general.

21 hours ago, Capiert said:

But the real (more probing) question
 is from where
 or what "speed" (reference)
 are you using
 to measure the "speed"
 you are observing.

(Obviously, your own speed
is the reference, there.)

E.g. The Earth,
 ground or surface
 that you are standing on.
Isn't a "reference frame"
 rather fictive imaginary.

I'm talking
about a real (physical) world.

 

[swansont]

1 hour ago, Capiert said:

Constant speed is NOT how the universe is working.

But it's how the example worked that you gave. When you solve a problem, you use the appropriate physics for the problem, so if the speed is constant, you can use an equation for constant speed. And there are lots of problems where speed is constant, or that's a reasonable approximation of a situation.

 

1 hour ago, Capiert said:

 

& I am looking for a more general, (but) simpler description.

And yet you use average speed, which makes assumptions as well.

 

1 hour ago, Capiert said:

 

& there is NOTHING
 which says
 acceleration must always be linear.

Indeed.

 

1 hour ago, Capiert said:

I get the idea "average" speed
 is an eye_sore for you
 because you prefer
 your instantaneous speed, instead.
But that (instantaneous speed)
is theoretical,

Nothing theoretical about it. It's the value of the speed at a particular time.

 

1 hour ago, Capiert said:

Calculus works "perfectly" (=flawlessly, NO errors)
 for "linear" acceleration
 because it is based on an (exact) "average";
 (e.g. 2 triangles making a rectangle
 using a diagonal (line))
 but I have my doubts
 for NON_linear (line) examples
 & irregular curves.

So this is based on a mistrust of math, and as a result you use math that's less appropriate. 

 

1 hour ago, Capiert said:

 

You are content with
 that you have learned
 some (of th(os)e complexity's) math rules,
 & use them.

I am NOT.

I am more interested
 in finding the reasons
 why some (math) things (fail)
 do NOT work sometimes (as we (would) wish);
 & (then) try to fix=repair them.

If you can show that math is inconsistent that's a purely mathematical issue and has nothing to do with physics.

 

1 hour ago, Capiert said:

 

I CAN'T follow you there,
 because "getting" (e.g. interacting)
 from initial_speed
 to final_speed
 (when they (speeds) do NOT equal)
 is acceleration.

In the example given they were the same, because speed was constant.

Changing the parameters of someone else's example and then complaining about a problem that arises is not an argument made in good faith.

1 hour ago, Capiert said:

 

Making them (2 speeds) equal
 is an exception,
 because (that is why) they have different names
 so that they do NOT have to be the same.
The (2 speeds)
 "can" be quite different.

They can be, but the point was that if they aren't your equation quite obviously fails. It's wrong. But you're ignoring that. Nobody else is fooled by the distraction.

1 hour ago, Capiert said:

I'm aimed (=intended) at collisions, or jerks!

Really? You only mentioned collisions in passing until now.

 

1 hour ago, Capiert said:

 

The speed is NOT constant for acceleration(s),
 such as (for, in) collisions
 where m*E is transferred.
 KE=PE (pronounced "keep", versus PE=KE pronounce "peek")

No, really, that's not how it's pronounced. They are pronounced "Kinetic Energy" and "Potential Energy" and they are sometimes equal to each other.

1 hour ago, Capiert said:

 (e.g. Conservation of Energy COE)
 might work (well)
 for (accelerating) a single (moving) mass
 that stays the same (sized mass);
 but (it, KE=PE)
 does NOT (always) work well
 when that mass interacts
 (& is accelerated)
 by other various sized masses
 (&) of different speeds.

For them (collisions, interactions),
 m*E works better,
 because it is more general, universal,
 uniting conservation of momentum COM,
 & Energy COE
 together
 into 1 formula.
 

Those who study and understand physics know the limitations; KE is not a conserved quantity in collisions except in a special case (elastic collision) because there are other possible forms of energy (e.g. thermal, sound, deformation)

 

 

[Bufofrog]

On 7/18/2022 at 12:47 PM, Bufofrog said:

That would be the reference frame that measures the objects velocity at 10 m/s.

 

11 hours ago, Capiert said:

That seems like an indirect answer

Nope that is as direct as I can get.  Any reference frame that measures the objects speed at 10 m/s meets the requirement for the measurement of  the KE.

[Capiert]

I suspect the confusion arises
 (between us)
 from the difference
 between my syntax
 versus your syntax.

I have extended the syntax description details
 beyond yours.

Your "the" (what?) KE,
 is my "final"_KE KE_f.

My KE_d is the (KE_)difference
 of 2 ((perhaps) different) Kinetic Energies.
In this case (final_KE minus initial_KE) KE_f-KE_i.
(We can thank Swansont for helping us point out that 1.)

I (would) find it (quite) peculiar
 if you would deny
 that the difference
 between 2 (different) KE's
 is NOT a KE (itself),
 because of:
 (1.) conservation of Energy COE:
 i.e. energy can NEITHER be created;
 NOR destroyed,
 & so the parts must add (up, together);
 &
(2.) per definition_name
 kinetic_energy
 is (=means) the energy of "motion".
There speed &/or acceleration apply.
Delta_KE=KE_d must be a KE
 (even if it is NOT your "the" KE,
 whatever "the" is suppose to be.
Disclaimer: I am only trying to be thorough,
 with the(=my, algebra_)syntax.)
 

If a mass object
 maintains a constant_speed

 (for that brief_time that "you" measure (it);

 instead of (versus, compared
 to) following its complete history
 as to how it (ever) acquired its (constant_)speed
 in the 1st place=originally),

 then, its initial_KE KE_i
 (which is NOT ZERO, (&) which you seem to want to ignore
 (maybe) just to makes things difficult(?),
 but which I have already taken into account (of))
 will be equal
 to its final_KE KE_f
 & (but, thus)
 its KE_difference KE_d=KE_f-KE_i
 will be (exactly) zero.

(The analogy
 is taken directly
 from my speeds(_syntax)
 for the speed_difference
 v_d=v_f-v_i.

That is NOTHING new!

(Except for (now) adding _d subscripts
 (for "difference"),
 where they were NOT before.)

Swansont should know
 that for linear acceleration
 the average_speed, is
 v_a=v_i+v_d/2
 &
 his final_(instantaineous)_speed (equivalent), is
 v_f=v_a+v_d/2.

E.g.
v_f=v_i+v_d

On 7/19/2022 at 4:18 PM, swansont said:

So this is based on a mistrust of math, and as a result you use math that's less appropriate. 

I have NOT yet recognized
 that that algebra will NOT do.
Meaning it (=the algebra) is still valid
 (as far as I can see).

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

I'm aimed (=intended) at collisions, or jerks!

Really?

Yes!

On 7/19/2022 at 4:18 PM, swansont said:

You only mentioned collisions in passing until now.

Jerks are also accelerations,
 (similar to collision accelerations).
Whether they (accelerations) are linear(?), I doubt it(!).

On 7/19/2022 at 4:18 PM, swansont said:

Those who study and understand physics know the limitations; KE is not a conserved quantity in collisions except in a special case (elastic collision) because there are other possible forms of energy (e.g. thermal, sound, deformation)

Unfortunately there I suspect you assume too much for what is actually happening
 instead of measuring.

Am I right?

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

& I am looking for a more general, (but) simpler description.

And yet you use average speed, which makes assumptions as well.

Quite right.

(NO bout a_doubt_it.)

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

 KE=PE (pronounced "keep", versus PE=KE pronounce "peek")

No, really, that's not how it's pronounced. They are pronounced "Kinetic Energy" and "Potential Energy" and they are sometimes equal to each other.

"keep" & "peek" are (only) how I have nicknamed the formulas (acronymns if you will)
 as an easy reminder.
NOT intended to offend anyone('s physics, indeed).

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

But that (instantaneous speed)
is theoretical,

Nothing theoretical about it. It's the value of the speed at a particular time.

How then do you "measure" it (=instananeous_speed) exactly,
 especially for collisions?
Relying on math (e.g. calculus) sounds pretty theoretical to me.

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

Constant speed is NOT how the universe is working.

But it's how the example worked that you gave. When you solve a problem, you use the appropriate physics for the problem, so if the speed is constant, you can use an equation for constant speed. And there are lots of problems where speed is constant, or that's a reasonable approximation of a situation.

Please recognize (=acknowledge) my initial_KE Ki syntax into the problem.
That'( i)s what it is there for,
 i.e. (it was designed)
 for such problems
 with (initial) constant_speed.

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

I am more interested
 in finding the reasons
 why some (math) things (fail)
 do NOT work sometimes (as we (would) wish);
 & (then) try to fix=repair them.

If you can show that math is inconsistent that's a purely mathematical issue and has nothing to do with physics.

If physics is using such math,
 then DON'T expect it (=physics)
 to hold itself together.

I would (rather) say, then it (=that purely mathematical issue) should NOT have to do with physics
 (but unfortunately (does &) is messing it (=physics) up).

Universities live & breathe calculus. (They love it!)
I choke on it.

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

 

I CAN'T follow you there,
 because "getting" (e.g. interacting)
 from initial_speed
 to final_speed
 (when they (speeds) do NOT equal)
 is acceleration.

In the example given they were the same, because speed was constant.

Changing the parameters of someone else's example and then complaining about a problem that arises is not an argument made in good faith.

Sorry (I was so slow), I must 1st grasp the problem
 before I can understand
 (what is happening)
 in order to tackle it (later).
Stating what (examples) I know
 helps get me nearer
 e.g. Later then excluded.
Stated otherwise:
 I had (wrongly) assumed
 you could follow me or my syntax,
 thus I could NOT understand
 where a problem was, if any.

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

 

Making them (2 speeds) equal
 is an exception,
 because (that is why) they have different names
 so that they do NOT have to be the same.
The (2 speeds)
 "can" be quite different.

They can be, but the point was that if they aren't your equation quite obviously fails. It's wrong. But you're ignoring that. Nobody else is fooled by the distraction.

Again, (when using my KE_i, which I had thought you were already aware of)
 my 02:40 comment
 described what else my equation could also do.
If you see that (extra) as ignorance,
 well then I'm sorry
 (for the misunderstanding).

(But) I still do NOT see that my equation has failed.

Simply take the (=my) KE_i
 because it equals the (=my) KE_f
 (which has (=is) your KE).

The KE_d has (=is) NOTHING!

That has obviously succeeded,
 don't you think?

To summarize the syntaxes:

Mine:   Yours:
KE_d      delta_KE
KE_f      "the" KE
KE_i        ?

I hope I have answered that sufficiently
 to your satisfaction(s).

Edited August 4, 2022 by Capiert

[swansont]

5 hours ago, Capiert said:

I suspect the confusion arises
 (between us)
 from the difference
 between my syntax
 versus your syntax.

Almost certainly. The thing is, people have been doing physics for quite some time, so the syntax is well-established. 

What you're doing is similar to showing up in a foreign country and expecting them to speak your language

5 hours ago, Capiert said:

I have extended the syntax description details
 beyond yours.

No, not really. 

Quote

 

Your "the" (what?) KE,
 is my "final"_KE KEf.

My KEd is the (KE_)difference
 of 2 ((perhaps) different) Kinetic Energies.
In this case (final_KE minus initial_KE) KEf-KEi.

 

KE_f being final kinetic energy is fine, if you mean it the same as everyone else: the KE at the end of the example you're analyzing

For difference we use ∆, so ∆KE = KE_f - KE_i 

So what you are doing is not "extending" the syntax, you are introducing new syntax where it already exists, which is confusing. 

Quote

I (would) find it (quite) peculiar
 if you would deny
 that the difference
 between 2 (different) KE's
 is NOT a KE (itself),
 

∆KE is not a KE that any one particle has, so it's a value for energy, but it's no longer describing the energy of a particle, so calling it a KE isn't correct.

Example: Two 1 kg objects moving at 1 m/s collide head-on and stick together, coming to rest. Their change in KE is -1 Joule, but at the end of the example nothing is moving, so saying ∆KE is a kinetic energy is incorrect. There is nothing in that example that has -1 J of kinetic energy. A kinetic energy can't be negative.

Quote

because of:
 (1.) conservation of Energy COE:
 i.e. energy can NEITHER be created;
 NOR destroyed,
 & so the parts must add (up, together);
 

Yes, energy is conserved, but kinetic is only one form of energy. KE itself is not a conserved quantity. See the above example of a completely inelastic collision.

Quote

(2.) per definition_name
 kinetic_energy
 is (=means) the energy of "motion".
There speed &/or acceleration apply.
Delta_KE=KEd must be a KE
 (even if it is NOT your "the" KE,
 whatever "the" is suppose to be.
Disclaimer: I am only trying to be thorough,
 with the(=my, algebra_)syntax.)

As I have shown, ∆KE is not a value associated with any one particle, or even anything having motion. It is a useful value to know in may problems, but to be useful it must be properly labeled, so one can do a proper accounting of the energy present.

5 hours ago, Capiert said:

Universities live & breathe calculus. (They love it!)
I choke on it.

Well, that's your problem. Calculus works regardless of your understanding or dislike of it.

5 hours ago, Capiert said:

Simply take the (=my) KE_i
 because it equals the (=my) KE_f
 (which has (=is) your KE).

The KE_d has (=is) NOTHING!

That has obviously succeeded,
 don't you think?

To summarize the syntaxes:

Mine:   Yours:
KE_d      delta_KE
KE_f      "the" KE
KE_i        ?

I hope I have answered that sufficiently
 to your satisfaction(s).

I agree that ∆KE = 0. That's the problem. You had said KE=m*(v_f²-v_i²)/2, rather than saying this was ∆KE

IOW, you were claiming that some object's KE was described by the equation. And for an object moving at some constant speed, it's kinetic energy is decidedly NOT zero. For an object not starting from rest, this does not give the object's KE. Which makes your equation wrong.  

[Bufofrog]

On 5/31/2022 at 1:07 PM, Capiert said:

 

Deleted since I cannot do arithmetic, this morning.

Edited August 4, 2022 by Bufofrog

[Capiert]

(Our) Confusion arises
 from different syntax:
 mine; versus yours.

15 hours ago, swansont said:

Almost certainly. The thing is, people have been doing physics for quite some time, so the syntax is well-established. 

What you're doing is similar to showing up in a foreign country

I agree.

15 hours ago, swansont said:

and expecting them to speak your language.

& they do speak English.

15 hours ago, swansont said:

  16 hours ago, Capiert said:

I have extended the syntax description details
 beyond yours.

15 hours ago, swansont said:

No, not really. 

KE_f being final kinetic energy is fine, if you mean it the same as everyone else: the KE at the end of the example you're analyzing

Yes, I think we can assume so.
(& It's nice to hear it is fine.)

15 hours ago, swansont said:

For difference we use ∆, so ∆KE = KE_f - KE_i 

So what you are doing is not "extending" the syntax, you are introducing new syntax where it already exists, which is confusing. 

You still have NOT commented on KE_i.

15 hours ago, swansont said:

∆KE is not a

(final_)

15 hours ago, swansont said:

KE

KE_f

15 hours ago, swansont said:

that any one particle has, so it's a value for energy,

What kind of Energy?
It's NOT chemical, thermal, NOR potential etc.
So what KIND of energy is that?

& also,
 only a particle
 could have had that energy
 when it was by=with the particle.

If a particle had lost that KE_d
 then that particle has decelerated
 & is now moving slower,
 than it was (before the loss).
But it was lost to another particle
 (forced to accelerate,
 E.g. Newton's 2nd Law).

KE_d can only be had by a particle (or object) e.g. mass.
The KE_d formula can discuss (either): the same mass m;
 or (else) it can discuss 2 different masses.
But (the) K(inetic )E(nergy) d(ifference)
 can NOT be transferred
 without a mass (as mediator, or transporter).
So what you said, e.g.
 that KE_d does NOT belong to a particle,
 (simply) does NOT (seem to) make sense
 (to me).
KE_d can only be transferred by (a) mass, (instead).
But:
The bigger, more important, question
(to ask)
 is: to which mass does KE_d belong to;
 & when? (before versus after, a(n elastic) collision).
E.g. Which mass lost the -KE_d?
 & which mass gained the +KE_d?

NO mass?
Then NO energy!
You can NOT describe (e.g. formulate) energy
 without mass.
(& charge is intimately
 related to the inverse "mass",
 as in charge to mass ratio.)

15 hours ago, swansont said:

but it's no longer describing the

motion_

15 hours ago, swansont said:

energy of a particle, so calling it a

 final_

15 hours ago, swansont said:

KE isn't correct.

I think we can fictively equate KE_d
 to various other masses & speeds
 just to get equivalents
 (or equivalence).
(But that'( i)s modelling,
 e.g. NOT physical.)

I find your next example (description)
 (non_elastic collision)
 rather interesting.

15 hours ago, swansont said:

Example: Two 1 kg objects

Let (their masses, be)
 m₁=1 [kg], &
 m₂=1 [kg].

15 hours ago, swansont said:

moving at 1 m/s collide head-on 

Let their initial_speeds, be
 v_i1=1 [m/s], & =j⁰*1 [m/s], j⁰=1 (0°)
 v_i2=-1 [m/s]=j²*1 [m/s], j²=-1 (180°)

 with initial_KE's
 KE_i1=m₁*(v_i1²-0²)/2=1 [kg]*((1 [m/s])²-0)/2=0.5 [kg*m²/s²]=j⁰*0.5 [J] (0°)
 KE_i2=m₁*(v_i2²-0²)/2=1 [kg]*((-1 [m/s])²-0)/2=1 [kg]*((j²*1 [m/s])²-0)/2=j⁴*0.5 [kg*m²/s²]=j⁴*0.5 [J] (360°)

15 hours ago, swansont said:

and stick together, coming to rest.

Their final_speeds, are
 v_f1= 0 [m/s], &
 v_f2= 0 [m/s]

 with final_KE's
 KE_f1=m₁*(v_f1²-0²)/2=1 [kg]*((0 [m/s])²-0)/2=0.5 [kg*m²/s²]=j⁰*0 [J] (0°)
 KE_f2=m₁*(v_f1²-0²)/2=1 [kg]*((0 [m/s])²-0)/2=0.5 [kg*m²/s²]=j⁰*0 [J] (0°)

with KE_differences
 KE_d1=m₁*(v_f1²-v_i1²)/2=1 [kg]*((0 [m/s])²-(1 [m/s])²)/2=-0.5 [kg*m²/s²]=j²*0.5 [J] (180°)
 KE_d2=m₁*(v_f2²-v_i2²)/2=1 [kg]*((0 [m/s])²-(-1 [m/s])²)/2=1 [kg]*(-(j²*1 [m/s])²)/2=-j⁴*0.5 [kg*m²/s²]=j⁶*0.5 [J] (540°).

Both (Energy) losses (together) are -1 [J],
 each -0.5 [J].

15 hours ago, swansont said:

Their change in KE is -1 Joule,

KE_d1=KE_f1-KE_i1=-0.5 [J]
KE_d2=KE_f2-KE_i2=(-)³*0.5 [J]
KE_d3=KE_d1+KE_d2=-0.5 [J]+(-)³*0.5 [J], ~-1 [J].

15 hours ago, swansont said:

but at the end of the example nothing is moving,

How then can you claim
 that their (kinetic) Energy
 was NOT destroyed (=annihilated)?

Are you claiming out of (good) "belief"?
(E.g. Scientific religion.
 E.g. Hoping the theory
 will justify every "odd" detail
 in the end?)
A reasonable person
 could NOT claim that,
 (when) seeing the facts.

(Seeing is believing.)
(You must (=most probably, surely) have good faith, (brother)!)

15 hours ago, swansont said:

so saying ∆KE is a

 kind (=type)
 of (moving (=motion) energy)

15 hours ago, swansont said:

kinetic energy is incorrect.
There is nothing in that example that has -1 J of kinetic energy.

Although we can loose energy
 (by using the minus sign symbol "-"),
 ((but) into NOTHING!).

15 hours ago, swansont said:

A kinetic energy can't be negative.

Because the math
(e.g. (final_speed) "squaring")
 does NOT allow negative( value)s.

15 hours ago, swansont said:

A kinetic energy can't be negative.

Quote

because of:
 (1.) conservation of Energy COE:
 i.e. energy can NEITHER be created;
 NOR destroyed,
 & so the parts must add (up, together);

15 hours ago, swansont said:

Yes, energy is conserved, but kinetic is only one form of energy. KE itself is not a conserved quantity. See the above example of a completely inelastic collision.

I have (seen the inelastic example above), & it looks convincing
 (& puzzling, at the same time).
It does NOT fit, like indefinite integrals (do).

15 hours ago, swansont said:

As I have shown, ∆KE is not a value associated with any one particle, or even anything having motion. It is a useful value to know in ma(n)y problems, but to be useful it must be properly labeled, so one can do a proper accounting of the energy present.

I DON'T want to argue with you there
 (not a value associated with any one particle, or even anything having motion.)
But I suspect a closer examination (in detail)
 will allow associating to particles,
 & of motion.

15 hours ago, swansont said:

I choke on (calculus).

15 hours ago, swansont said:

Well, that's your problem. Calculus works regardless of your understanding or dislike of it.

3 indefinite integrals begin to stand (=argue) against (a) calculus working regardlessly.
Serious mathematicians know calculus does NOT "always" work right.
they have been successful
 at isolating those few quirks;
 but they CAN'T quite
 put their thumb on why
 to eliminate them completely.
They simply set up warnings, DON'T do.
With those Jack in the boxes,
 I'm curious what else
 (for problems)
 could be found
 awaiting us
 for: in the future.

Btw. What is half of infinity?
Calculus works with infinitesimal limits,
 hoping to make the problem so small
 so it disappears.
The problem is the proportioning
 should stay the same,
 instead of approximating
 that it is gone.

15 hours ago, swansont said:

I agree that ∆KE = 0. That's the problem. You had said KE_d=m*(v_f²-v_i²)/2, rather than saying this was ∆KE

Yes, with your help
 we have identified a problem
 that could be cleared.
Thank you.

That'( i)s the way we learned it
 in school.
& looking at it (now)
 it seems
 like a (more/most) general description:
 dealing on a macroscopic level/perspective;
 rather than a microscopic infinitesimal view.

15 hours ago, swansont said:

IOW, you were claiming that some object's KE (=KE_f) was described by the equation.

I'm sorry, but I have to disagree with you there.
That is how you interpreted my text's syntax.

You assumed I was talking about KE_f;
 but I was always talking about KE_d, instead.

I DIDN'T use the _d subscript
 as is typical for many variables
 in Physics.
E.g. (duration) time t=t_f-t_i
Eg. distance d=d_f-d_i
E.g. speed v=v_f-v_i.

When you measure a distance
 you rarely say it is a distance_"difference"
 of e.g. 100 [km] from 1 big city to another.
You just say the distance is ..;
 & forget about the fact
 that it is a difference, too=also.
Why the (complexity &) inconsistency
 in your syntax?
Answer: Because normal people DON'T talk that way.
You guys have to turn everything into a unique exception
 rather than talk like normal human beings.
I mean "sometimes" there are advantages
 to the (few extra) details;
 but NOT always.

15 hours ago, swansont said:

And for an object moving at some constant speed, it's kinetic energy is decidedly NOT zero. For an object not starting from rest, this does not give the object's KE (=KE_f). Which makes your equation wrong.  

The equation is KE_d (NOT your (agreed upon) KE_f);
 & as KE_d=m*(v_f²-v_i²)/2
 that equation is NOT wrong
 but instead absolutely correct.

∆KE=KE_d

It should have been obvious
 (excluding typo errors),
 that
 KE_f#m*(v_f²-v_i²)/2
 could NOT possibly be (correct)
 as KE_f
 because it was missing (the term) -m*(v_i²)/2;
 instead of being exclusively KE_f~m*(v_f²)/2
 using v_f²;
 without -m*(v_i²)/2 losses.

That is,
 unless you have found some(thing else) new, breath_taking news
 that we should (all) be cautious (about), for.

Btw.
I must comment
 that I was rather enthusiast
 at Bufofrog's original comment,
 (but unfortunately wrongly)
 assuming he might have found an interesting (new) problem
 that we could work on,
 e.g. he was thinking outside of the box.

But instead,
 it only turned out to be (boring) syntax misunderstandings.

But at least we got that cleared too.

Thanks gang (=team)!

Edited August 5, 2022 by Capiert