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Vis_viva ? (Dark Energy)


Capiert

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You keep mentioning algebra vs calculus and tensors.

You are missing the inherent versatility of tensors, they provide us a means to organize all the algebraic expressions  in complex systems. Particularly in how they apply vectors and vector algebra.

This is what symmetry translations is all about, simplifying numerous vector rotations and translations. For every given coordinate.

Edited by Mordred
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14 hours ago, swansont said:

Why should there be a direct connection?

Because they are related by the same units: mass & speed (as a factor, e,g. multiplied).

m*v is common to both concepts: E(nergy) & mom.

That is so obvious (it rolls me over).

Quote

Momentum is conserved when the net external force on an object or system is zero.

Meaning no net force (F=m*a, Newton's 2nd law),

 or balanced forces (F1+F2=0, Newton's 3rd law reaction (is zero)).

 

Effectively, no acceration a=F/m ((of) Newton's 2nd law).

 

I.e. no changes of mom(entum)

 =constant mom(entum), (=conserved). (Newton's 1st law: constant speed of a mass, mom=m*v.)

Ojects in motion [tend to] stay in motion

 when NOT acted_on (=NOT accelerated).

Quote

It stems from the symmetry under coordinate transformation.

wrt distance (d).

Work's energy

WE=F*d

Quote

Energy is always conserved within a system.

?

E=KE=PE=WE=EE=...

E3#E1+E2. (See below.)

?

Quote

It stems from  symmetry under time translation.

wrt time (t).

mom=F*t.

Quote

They are not interchangeable,

?

2*m3*KE3*(v3/va3)=2*m1*KE1*(v1/va1)+2*mom1*mom2+2*m2*KE2*(v2/va2), 

m3=m1+m2.

Quote

and there are ways of defining systems such that either or both are not conserved,

?

Quote

or that both are.

?

Edited by Capiert
typo
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44 minutes ago, Capiert said:

Because they are related by the same units: mass & speed (as a factor, e,g. multiplied).

m*v is common to both concepts: E(nergy) & mom.

That is so obvious (it rolls me over).

One is a vector the other is a scalar. They are not interchangeable. Even though that have some common terms in them, there are differences. The differences matter.

Your intuition is failing you.

Quote

 

Meaning no net force (F=m*a, Newton's 2nd law),

 or balanced forces (F1+F2=0, Newton's 3rd law reaction (is zero)).

 

That's not what Newton's third law says. Action-reaction force pairs always have equal magnitudes, even when the net force is not zero. But they act on different objects, and are never both in an equation for net force.

Quote

 

Effectively, no acceration a=F/m ((of) Newton's 2nd law).

I.e. no changes of mom(entum)

 =constant mom(entum), (=conserved). (Newton's 1st law: constant speed of a mass, mom=m*v.)

Ojects in motion [tend to] stay in motion

 when NOT acted_on (=NOT accelerated).

wrt distance (d).

 

Objects colliding undergo acceleration, and yet momentum will be conserved. This comes from the proper application of the third law.

Quote

 

Work's energy

WE=F*d

?

E=KE=PE=WE=EE=...

?

wrt time (t).

mom=F*t.

 

Yes. Momentum involves force and time, while work (energy) involves force and distance, and is a dot product. That makes them different. You can have a force acting on an object that changes its momentum without doing any work on it, because the dot product is zero.

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17 minutes ago, swansont said:

One (momenetum) is a vector the other (energy) is a scalar. They are not interchangeable. Even though that have some common terms in them, there are differences.

The only difference I see is (negative) polarity (ability)

 (which I attribute to (180 degrees) angle):

 1 (=vector) has it;

 the other (=energy) does NOT.

17 minutes ago, swansont said:

The differences matter.

Differences usually (always) matter.

17 minutes ago, swansont said:

Your intuition is failing you.

How can you be (so) sure?

17 minutes ago, swansont said:

That's not what Newton's third law says.

Sorry for my sloppy style,

 I was using Newton's 3rd

 as a starting point (concept, theme)

 to explain further.

17 minutes ago, swansont said:

Action-reaction force pairs always have equal magnitudes, even when the net force is not zero.

True. No arguement there.

17 minutes ago, swansont said:

But they act on different objects,

True.

17 minutes ago, swansont said:

and are never both in an equation for net force.

?

17 minutes ago, swansont said:

Objects colliding undergo acceleration, and yet momentum will be conserved. This comes from the proper application of the third law.

?

17 minutes ago, swansont said:

Yes. Momentum involves force and time, while work (energy) involves force and distance. That makes them different.

I know they (mom & E) are different,

 & I know they are related (to each other, & (by) how (much, they are)):

KE=mom*va

mom=KE/va.

(That last 1 is rather remarkable, don't you think?)

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1 hour ago, Capiert said:

The only difference I see is (negative) polarity (ability)

 (which I attribute to (180 degrees) angle):

 1 (=vector) has it;

 the other (=energy) does NOT.

Differences usually (always) matter.

How can you be (so) sure?

Sorry for my sloppy style,

 I was using Newton's 3rd

 as a starting point (concept, theme)

 to explain further.

True. No arguement there.

True.

?

?

The reason I am sure is that you are wrong, and you have acknowledged that you don't understand my statements about Newton's 3rd law.

Quote

 

I know they (mom & E) are different,

 & I know they are related (to each other, & (by) how (much, they are)):

KE=mom*va

mom=KE/va.

(That last 1 is rather remarkable, don't you think?)

 

And yet these equations are incorrect. Partly because momentum is a vector, partly because of an assumption that has been used.

KE=mom*va as a scalar equation is only true for a constant linear acceleration. It fails in a number of other situations. e.g. an object subject to a non-constant acceleration, or an object moving in a circle at constant speed. 

 

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On ‎2017‎ ‎08‎ ‎02 at 2:03 PM, swansont said:

The reason I am sure is that you are wrong, and you have acknowledged that you don't understand my statements about Newton's 3rd law.

Which parts do you mean. I often don't (recognize=) understand your writing style (without direct examples).

(Perhaps it's too careful for my informal attitude? An example often helps to cut thru the communication gap.)

Maybe you are mentioning what I already know, in other words;

 or I've (obviously) said things wrong. ?

 

Isn't the total reacted (=reacting) force of 2 collided bodies (as a system) zero?

Although you wouldn't include the other body, I could

 (just to evaluate the system('s net total)).

Or what do you mean exactly? (Please put me on track.)

Quote

And yet these equations are incorrect. Partly because momentum is a vector,

?

mom=KE/va.  (Sorry, va. Bold type for vector.)

 

Otherwise, I'm sorry (I can't follow you), that formula is too simple to be declared incorrect.

Anybody could see its connection (if they wanted to).

Thus,

m3*E3*(v3/va3)=m1*E1*(v1/va1)+mom1*mom2+m2*E2*(v2/va2), 

m3=m1+m2.

Quote

partly because of an assumption that has been used. (i.e.)

KE=mom*va as a scalar equation is only true for a constant linear acceleration.

But, wasn't KE derived (wrt PE) from (only) linear acceleration?

How (then) can you imply otherwise.

(va is the average "linear" accelerated speed.)

I thought everyone knew KE is a "linear accelerated" energy (concept, equivalence),

 for its final resulting speed.

Quote

It fails in a number of other situations. e.g. an object subject to a non-constant acceleration,

I have never seen non_linear acceleration in the textbooks.

I guess things get too complicated for most (people)

 when non_linear (e.g. exponential). ?

Quote

or an object moving in a circle at constant speed. 

Newton's centrifugal acceleration (equivalence)

ac=(vc^2)/r,

for (tangential) circular_speed

vc=cir/T=2*Pi*r/T,

(cycle's) Period T,

& radius r.

But, circular rotation (at constant_speed)

 is non_linear acceleration

 of each x & y component.

Can we (simply) know the velocity

 of a single component?

(I.e. The displacement is obvious (with Pythagoras's r^2=y^2 + y^2), but velocity?)

Quote

 

 

Edited by Capiert
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47 minutes ago, Capiert said:

 Isn't the total reacted (=reacting) force of 2 collided bodies (as a system) zero?

Although you wouldn't include the other body, I could

 (just to evaluate the system('s net total)).

Or what do you mean exactly? (Please put me on track.)

The force on an individual particle is not zero. The force of object 1 colliding with object 2 is the the same as 2 on 1, but in the opposite direction. That's an action-reaction force pair: the force are on two different objects, while a net force is the sum of forces acting on any single object.

There is no net force on the system, so the center-of-mass does not change, but using that information alone does not help you in solving for the motion of the two colliding objects, so it's useless.

I have no idea why you think I would not include the other body.  

Quote

mom=KE/va.  (Sorry, va. Bold type for vector.)

 

Dividing by a vector is not defined. And, as I pointed out, the scalar equation is limited to one specific scenario. In every other sense it's wrong.

Quote

 

Otherwise, I'm sorry (I can't follow you), that formula is too simple to be declared incorrect.

Anybody could see its connection (if they wanted to).

 

I gave examples of how it is incorrect. An object moving in a circle does not change speed, so v = va. Its KE is not given by pvAn object subject to a non-constant acceleration will not have its average speed be (vf-vi)/2. Again, the equation will be wrong. (non-relativistic) KE is always given by 1/2 mv

Quote

But, wasn't KE derived (wrt PE) from (only) linear acceleration?

No, it wasn't. Mordred provided a link earlier. KE is the work done over some distance, but the general solution does not require any assumptions about the acceleration.  

Quote

 

How (then) can you imply otherwise.

(va is the average "linear" accelerated speed.)

I thought everyone knew KE is a "linear accelerated" energy (concept, equivalence),

 for its final resulting speed.

 

An object moving in a circle still has kinetic energy.

Quote

I have never seen non_linear acceleration in the textbooks.

When you reject calculus you limit yourself to much simpler systems, which can be solved with simple algebra. But that doesn't mean that more complicated systems don't exist, nor that that don't have solutions.

Quote

 

guess things get too complicated for most (people)

 when non_linear (e.g. exponential). ?

Newton's centrifugal acceleration (equivalence)

ac=(vc^2)/r,

for (tangential) circular_speed

 

The v in the centripetal force equation is the instantaneous linear speed. It is not the angular speed, though you can derive a formula which uses that, instead.

Quote


vc=cir/T=2*Pi*r/T,

(cycle's) Period T,

& radius r.

But, circular rotation (at constant_speed)

 is non_linear acceleration

 of each x & y component.

Can we (simply) know the velocity

 of a single component?

(I.e. The displacement is obvious (with Pythagoras's r^2=y^2 + y^2), but velocity?)

 

Someone who has studied physics would use a cylindrical or spherical coordinate system, such that the angular velocity vector is orthogonal to the radial vector.

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Stated (simply=)  linearly, that (component's velocity, e.g. vx) would (then) be .. ? (please continue).

 

Swansont: "I have no idea why you think I would not include the other body."  

 

Quote

Swansont:

Momentum is conserved when the net external force on an object or system is zero.

Capiert:

Meaning no net force (F=m*a, Newton's 2nd law),

 or balanced forces (F1+F2=0, Newton's 3rd law reaction (is zero)).

Effectively, no acceration a=F/m ((of) Newton's 2nd law).

 

Swansont:

That's not what Newton's third law says. Action-reaction force pairs always have equal magnitudes, even when the net force is not zero. But they act on different objects, and are never both in an equation for net force.

Capiert:

You've defined the net force as on a single object

because (from your experience) it's useless on a force pair system.

I did not have that restriction, because I (still) explore (whether useless or not).

But I only mentioned Newton's 3rd law

 to (further) indicate that when a single object gets accelerated (e.g. Newton's 2nd law),

 (then) it "depends" on a 2nd object, too.

We often view only half the story

 from our (exclusive) limited perspective.

I only wanted to hint,

 more was going_on (=involved) than the single particle (mass).

Edited by Capiert
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mom=KE/va.  (Sorry, va. Bold type for vector.)

 

Swansont: "Dividing by a vector is not defined."

By the look of that equation,

it sure looks (to me) like high time

that (the) defining should begin.

E.g. At least derived.

Edited by Capiert
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2 hours ago, Capiert said:

Stated (simply=)  linearly, that (component's velocity, e.g. vx) would (then) be .. ? (please continue).

 

Swansont: "I have no idea why you think I would not include the other body."  

 

Swansont:

Momentum is conserved when the net external force on an object or system is zero.

Capiert:

Meaning no net force (F=m*a, Newton's 2nd law),

 or balanced forces (F1+F2=0, Newton's 3rd law reaction (is zero)).

Effectively, no acceration a=F/m ((of) Newton's 2nd law).

Right. Momentum conservation requires that there be no net external force. That does not guarantee energy conservation for the same system.

Quote

 

Swansont:

That's not what Newton's third law says. Action-reaction force pairs always have equal magnitudes, even when the net force is not zero. But they act on different objects, and are never both in an equation for net force.

Capiert:

You've defined the net force as on a single object

because (from your experience) it's useless on a force pair system.

 

In determining the motion of an object, the only thing that matters is the net force on that object. A 3rd law force pair does not appear in a net force equation, because of the definition of net force and the statement of Newton's third law. 
"My experience" here is the proper application of physics. Useless is not strong enough of a description. It's flat-out wrong.

There's a standard question that appears in physics textbooks. How can a horse pull a cart? If forces are equal and opposite, how can anything ever move if they start at rest?

The answer lies in understanding the details I've described earlier.

https://www.lhup.edu/~dsimanek/physics/horsecart.htm

 

Quote

 

I did not have that restriction, because I (still) explore (whether useless or not).

But I only mentioned Newton's 3rd law

 to (further) indicate that when a single object gets accelerated (e.g. Newton's 2nd law),

 (then) it "depends" on a 2nd object, too.

We often view only half the story

 from our (exclusive) limited perspective.

I only wanted to hint,

 more was going_on (=involved) than the single particle (mass).

 

You don't get to "explore" anything if that involves changing the definition that everyone is using. One of the huge barriers in having a conversation with you is that you use non-standard notation and ignore definitions of terminology and math. You force everyone to learn your language, without showing the same courtesy in return.  And you do so as a visitor in our country, as it were.  It is, in that sense, very rude behavior.

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15 hours ago, swansont said:

Right. Momentum conservation requires that there be no net external force. That does not guarantee energy conservation for the same system.

In determining the motion of an object, the only thing that matters is the net force on that object. A 3rd law force pair does not appear in a net force equation, because of the definition of net force and the statement of Newton's third law. 
"My experience" here is the proper application of physics. Useless is not strong enough of a description. It's flat-out wrong.

There's a standard question that appears in physics textbooks. How can a horse pull a cart? If forces are equal and opposite, how can anything ever move if they start at rest?

That's where you & I differ.

You see force as a cause,

 & I see it (at least partly) as an effect,

an observation.

I am not completely decided

 on that as to whether force is a mechanism or not.

Quote

The answer lies in understanding the details I've described earlier.

https://www.lhup.edu/~dsimanek/physics/horsecart.htm

 

You don't get to "explore" anything if that involves changing the definition that everyone is using.

Sometimes you say rediculous things.

I can explore anything I want (without definitions, = definitionless).

Many people use physics, but not all people.

Quote

One of the huge barriers in having a conversation with you is that you use non-standard notation

I use other syntax, because yours is not adequate enough for me (& my model testing).

There are not enough non_foreign characters

 in the english alphabet.

Quote

and ignore definitions of terminology and math.

I want to test things independently (=from different perspectives), (with the ability to (test)) every step of the way.

If its foreign (for you & me) then there are better chances of (me) not falling into the same booby traps you do.

That aside, I'm not convinced of everything in (theoretical) physics

 & I would like to make (partial) comparisons

 to sort it out

 & identify (where) the problems are.

 

Quote

You force everyone to learn your language, without showing the same courtesy in return.

Sorry, (if you want whorship, you've got the wrong person)

 I'm not going to fall for that 1.

If I do everything exactly like you,

 I will fall into the snags that you do

 & never learn a different perspective (than yours)

 to solve the problems

 because I'm stuck in them.

(It's a lousey job, but somebody has got to do it.)

Nature has variation (& creation).

Quote

 And you do so as a visitor in our country, as it were.  It is, in that sense, very rude behavior.

"Take me to your leader (earthling!)"-ET.

Pun aside, I'll (try to) work on it.

Ruder was to kick the indians out,

 & now declare it (=what was stolen, is) yours. -Squatter's rights.

Edited by Capiert
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2 hours ago, Capiert said:

That's where you & I differ.

You see force as a cause,

 & I see it (at least partly) as an effect,

an observation.

I am not completely decided

 on that as to whether force is a mechanism or not.

I'm discussing Newtonian physics. I think everyone is still trying to figure out what you are discussing, because it's not Newtonian physics. I'm using the terminology, definitions and conventions of Newtonian physics. And that's the problem, because often you are not doing so in any consistent fashion.

Quote

 

Sometimes you say rediculous things.

I can explore anything I want (without definitions, = definitionless).

Many people use physics, but not all people.

 

This site has rules, so in fact you cannot "explore anything you want" while you're here.

Quote

 

I use other syntax, because yours is not adequate enough for me (& my model testing).

There are not enough non_foreign characters

 in the english alphabet.

 

And you end up with a very cluttered and confusing collection of statements as a result.

Quote

Sorry, (if you want whorship, you've got the wrong person)

In what world is courtesy considered worship?

 

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6 hours ago, Mordred said:

So your going to rewrite physics and mathematics just because your too lazy to learn.

I can't guarantee physics, they way you want, on your terms.

There is no error correction mechanism, for your faulty software which adds errors into the syntax,

 as well as the syntax changes which the new software demands.

Maybe you need better programmers to get (all) the bugs out.

I can't deal with everything.

Quote

Got it 

As for Lazy. (Buff.) My mind shuts down when too many problems exist.

I'm happy to have been able to communicate til this far.

Quote

Thanks for wasting our time.

I think your software bugs are screwing the people up.

As for got it, you now know what my pdf's va is all about.

Are there any errors?

If not then a major problem (still) needs to be solved.

8 hours ago, swansont said:

I'm discussing Newtonian physics. I think everyone is still trying to figure out what you are discussing, because it's not Newtonian physics.

"A rose by any other name smells just as sweet."

8 hours ago, swansont said:

I'm using the terminology, definitions and conventions of Newtonian physics.

That's interesting: Newton's physics; Leibnitz's (calculus) syntax. Wasn't Newton's inertia renamed momentum? What a mix. Funny that Newton didn't use it.

Quote

And that's the problem, because often you are not doing so in any consistent fashion.

This site has rules, so in fact you cannot "explore anything you want" while you're here.

And you end up with a very cluttered and confusing collection of statements as a result.

In what world is courtesy considered worship?

Heaven (on earth). ?

But I'm no fan of your mixed up conventions. (e.g. Conventional flow versus electron flow).

I don't worship & praise your syntax as the best solution, like you do.

(Your syntax conventions are more like my downfall because it goes so (=too) fast.

My brain latches (=locks_on) skipping the problem too quick.)

(I also have my preferences, like any human being does. That you want to dictate them implies a monarchial desire.)

I'll admit it is neat & fun to have a single character syntax, e.g. a minimum (syntax),

 but it doesn't help me enough to deal with (infected) computers to check the math,

 & (assist) error (reduction techniques) control.

You've got more resources than me, are richer,

 & expect everybody on the same level.

My PC constantly overheats, & stops.

The times have changed since the midages.

I'm looking for basic problems,

 but you all are convinced they do not exist (=no problems exist);

 your methods are perfect;

 & (it's) the only way to do them.

 

Btw

 You still have not answered my question:

 what is the linear x component speed

 of a rotating object.

Edited by Capiert
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I guess you didn't understand about syntax. It isn't the sites fault when you write

ke=mom+va where you intention for va is average velocity.

The correct mathematical syntax is ke=p+v_a the subscript allows one to seperate different identical variables.

This avoids the confusion I had when I first read your post as va means velocity times acceleration.

Had you properly written your mathematical expressions correctly in the first place, I would not have mentioned Dimensional analysis.

That is your error not the sites nor mine.

Swansont already informed you that there is no linear components in an angular momentum system where the conservation of angular momentum is defined.

Your formula will not work on angular systems.

The fundamental mistake you've made however is that your not paying attention to how kinetic energy is defined.

 ke= net force*displacement 

 

Edited by Mordred
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Here Pay attention to this key element in Kinetic energy.

The kinetic energy of a moving object is equal to the work required to bring it from rest to a velocity.

So average velocity is useless to us. You want to calculate how much work the bullet delivers to the wall it hits.

You need the amount of work if takes to move an object from rest to the velocity at the time of impact. 

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36 minutes ago, swansont said:

There is an error correction mechanism. It's comparing models with experimental results. And the physics passes this test.

 

Yep its taking objects and using Newtonian scales to measure the force when the object collides with a plate as one exsmple. There is a test you can perform at home.

 

In Industrial tests one example is smashing cars against a force plate. Done every single day lol. 

Edited by Mordred
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On 8/2/2017 at 4:02 AM, swansont said:

One is a vector the other is a scalar. They are not interchangeable. Even though that have some common terms in them, there are differences. The differences matter.

 

Yes. Momentum involves force and time, while work (energy) involves force and distance, and is a dot product. That makes them different. You can have a force acting on an object that changes its momentum without doing any work on it, because the dot product is zero.

lets expand on this to highlight the significants of this.

All rotations use the vector calculus of the cross product.

All linear translations use the dot product of vector calculus.

Those tensors you mentioned allow us to keep track of the vector algebra.

A nice little chart on this link is extremely useful to understand how tensors work (You will be amazed how simple it is). A dot product is two vectors on the same plane and returns a scalar quantity. A cross product is those same two vectors but includes a vector perpendicular to both. ie as per angular momentum (torque).

http://tutorial.math.lamar.edu/Classes/CalcII/CrossProduct.aspx

Please note the dot product on you 3 by 3 matrix is always on the diagonal, the cross products fill the other slots.

The site explains it well.

One lesson I've learned over the years on various forums. A large number of posters tend to reject what they don't understand. More often than not once they understand what they rejected. More often than not they then accept it.

I'm hoping once you see how useful vector calculus is including tensors this may be the case.

 

Edited by Mordred
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On 5 August 2017 at 0:48 AM, Mordred said:

I guess you didn't understand about syntax. It isn't the sites fault when you write

KE=mom*va [NOT ke=mom+va] where you intention for va is average velocity.

 

Quote

The correct mathematical syntax is

KE=p * v_a

(p & v_a are multiplied together; NOT added)

Quote

[NOT ke=p+v_a] the subscript allows one to seperate different identical variables.

This avoids the confusion I had when I first read your post as va means velocity times acceleration.

Had you properly written your mathematical expressions correctly in the first place, I would not have mentioned Dimensional analysis.

That is your error not the sites nor mine.

Swansont already informed you that there is no linear components in an angular momentum system where the conservation of angular momentum is defined.

So the answer to my question:

(Can we know the (simple=) linear velocity (components) of a rotating object?)

is no!

We do NOT have linear velocity formulas

 for rotating objects' x & y components.

Quote

Your formula will not work on angular systems.

The fundamental mistake you've made however is that your not paying attention to how kinetic energy is defined.

 ke= net force*displacement 

 

 

Edited by Capiert
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On 5 August 2017 at 1:17 AM, Mordred said:

Here Pay attention to this key element in Kinetic energy.

The kinetic energy of a moving object is equal to the work required to bring it from rest to a velocity.

So average velocity is useless to us. You want to calculate how much work the bullet delivers to the wall it hits.

You need the amount of work if takes to move an object from rest to the velocity at the time of impact. 

That means: (KE is) the amount of work(_energy)

 to (move=) accelerate a mass m,

 from initial_velocity vi=0 m/s

 to the final_velocity vf at the time of impact.

 

But that sounds like force F=m*a (instead).

I suspect the (forced, travelled) distance d=va*t

 is a clever way to involve

 the amount of time t

 into the equation,

 as d=t*va

 (for momentum mom=F*t).

Work_energy is

WE=F*d, d=t*va

WE=F*t*va, mom=F*t

WE=mom*va.

 

(Momentum is in the core

 of the (work) energy equation.)

Edited by Capiert
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On ‎2017‎ ‎08‎ ‎05 at 1:06 AM, swansont said:

There is an error correction mechanism. It's comparing models with experimental results. And the physics passes this test.

I want to agree with you there,

 but I'( woul)d have to say,

 (theoretical) physics

 has to be fit

 to the observations (=experimental results).

E.g. Dark (=unknown) energy, needs some filing (=rework, fitting).

Edited by Capiert
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On 5 August 2017 at 1:26 AM, Mordred said:

 

Yep its taking objects and using Newtonian scales to measure the force when the object collides with a plate as one exsmple. There is a test you can perform at home.

 

In Industrial tests one example is smashing cars against a force plate. Done every single day lol. 

Yes, but I suppose there are (at least) 2 different possibilities:

Elastic (e.g. bounced, having more acceleration (damage?)),

 & (absorbed) non_elastic collisions (e.g. like Gravesande's experiment, 1D).

Edited by Capiert
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2 hours ago, Capiert said:

That means: (KE is) the amount of work(_energy)

 to (move=) accelerate a mass m,

 from initial_velocity vi=0 m/s

 to the final_velocity vf at the time of impact.

 

But that sounds like force F=m*a (instead).

I

I certainly hope so as it is suppose to, the kinetic energy formula is a derivitative of f=ma. Tell me did you never notice that 1 joule of energy =1 Newton of force? Both force and energy involve the ability to perform work. Tell me did you even read the links I posted. If so why do they keep referring to Newtons three laws when discussing kinetic energy?

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