"Motion is impossible" claims modern Zeno

[Nathanael]

Let me set up a situation:

A piano rests on a frictionless surface. I am standing next to the piano (on a frictional surface) and I claim that the following two statements prove it is impossible for me to move the piano:

(1) ... The kinetic energy of the piano is equal to the work I've done on it.
(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

"Therefore the piano is immovable," I claim.

(1) is equivalent to the work-kinetic-energy theorem W=\Delta E
(2) is a special case (where \frac{d\vec s}{dt}=0) of the definition of work dW=\vec F\cdot d\vec s=\vec F\cdot \frac{d\vec s}{dt}dt

Please explain where and why my logic is flawed (assuming I don't know Newton's laws).

 

[A.T.]

Please explain where and why my logic is flawed

The assumption of a prerequisite relationship between two quantities which change simultaneously.

 

[Drakkith]

(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

Push it. You'll both move.

 

[Nathanael]

The assumption of a prerequisite relationship between two quantities which change simultaneously.

Sounds reasonable, but I have to say I'm not satisfied (probably because I don't entirely understand you.) Can you say more about this?

Push it. You'll both move.

Well, I probably won't move (I said my ground was frictional) but yes, it will move... but the only reason you know this is from F=ma.
But if you consider only work and energy you might come to the paradox I proposed.
I was curious if there is any way around this paradox without invoking F=ma.

 

[CWatters]

You can apply a force on the piano without doing work (the piano is stationary).

If there is no friction to counteract the force you apply the piano MUST accelerate.

Once it's moving you can doe work on it.

 

[A.T.]

Sounds reasonable, but I have to say I'm not satisfied (probably because I don't entirely understand you.) Can you say more about this?

When you have a quantitative relationship between two variables like y = x², then changing one of them implies changing the other. If you misinterpret this correlation as a two way causation, you get: A change of y requires a change of x which requires a change of y... so no change of either is possible.

 

[Nathanael]

Let me say, I do know that it will move. I do not believe my claim in the OP. (Perhaps I should have emphasized this.)

But my question is this:

Once it's moving you can doe work on it.

If I didn't do work on it then how did it get to be moving?

I understand from the F=ma perspective: I push on it therefore it accelerates.

But from an energetic perspective: it must have energy (motion) before I can give it any energy.

 

[Nathanael]

When you have a quantitative relationship between two variables like y = x², then changing one of them implies changing the other. If you misinterpret this correlation as a two way causation, you get: A change of y requires a change of x which requires a change of y... so no change of either is possible.

I feel like you're getting at the heart of it. Yet... it still doesn't resonate with me in this particular situation.
Somehow I can't get past this statement: "it must have energy (of motion) before I can give it any energy."

I will sleep on your explanation. Any different ways of explaining it are welcome. Goodnight.

 

[Drakkith]

(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

Does it have to be moving, or does it just have to move?

 

[Nathanael]

When you have a quantitative relationship between two variables like y = x², then changing one of them implies changing the other. If you misinterpret this correlation as a two way causation, you get: A change of y requires a change of x which requires a change of y... so no change of either is possible.

Sorry, I can't sleep. What you say sounds nice, but I can't understand how it is relevant:
We don't have something like y=x², we have \frac{dE}{dt}=k\sqrt{E} with E(0)=0. If E=0 then dE/dt=0 therefore the energy is in a steady state w.r.t. to time (i.e. it can't change).
(k=F\sqrt{\frac{2}{m}} not that it matters.)

Does it have to be moving, or does it just have to move?

I see what you're trying to say, but tell me if this is true or false: The only way to move is to, at some point in time, be moving.

[edited because I said "relative" instead of "relevant" ... I was tired haha]

 

[Drakkith]

I see what you're trying to say, but tell me if this is true or false: The only way to move is to, at some point in time, be moving.

At some point in time, but not every point in time.

 

[jbriggs444]

I see what you're trying to say, but tell me if this is true or false: The only way to move is to, at some point in time, be moving.

For every real number x greater than zero there is a real number y that is between x and zero. For every time when the piano is moving there is a prior time when the piano was moving.

One key to your conundrum is that you seem to want to imagine a first real number greater than zero -- a first instant when the piano is moving. But there is no such thing.

 

[russ_watters]

If I didn't do work on it then how did it get to be moving?

I understand from the F=ma perspective: I push on it therefore it accelerates.

But from an energetic perspective: it must have energy (motion) before I can give it any energy.

So what? It sounds like you are falsely equating force and energy. f=ma does not necessarily require energy.

If you apply a 1N force to a 1kg object, starting at t=0, at t=0 the object is stationary, but accelerating at 1m/s and at that time the rate of expenditure of energy is 0. There is no conflict.

 

[A.T.]

We don't have something like y=x², we have \frac{dE}{dt}=k\sqrt{E} with E(0)=0. If E=0 then dE/dt=0

This is exactly like y=x² where at x = 0 you have dy/dx = 0.

I see what you're trying to say, but tell me if this is true or false: The only way to move is to, at some point in time, be moving.

Natural language is ambiguous and another reason for apparent paradoxes.

 

[Jeff Rosenbury]

Let me set up a situation:

A piano rests on a frictionless surface. I am standing next to the piano (on a frictional surface) and I claim that the following two statements prove it is impossible for me to move the piano:

(1) ... The kinetic energy of the piano is equal to the work I've done on it.
(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

"Therefore the piano is immovable," I claim.

(1) is equivalent to the work-kinetic-energy theorem W=\Delta E
(2) is a special case (where \frac{d\vec s}{dt}=0) of the definition of work dW=\vec F\cdot d\vec s=\vec F\cdot \frac{d\vec s}{dt}dt

Please explain where and why my logic is flawed (assuming I don't know Newton's laws).

Why can't the piano have zero kinetic energy without you doing work on it?

What you seem to have proven is that you and the piano can't have a fixed position and zero momentum. Perhaps that's true instead?

 

[thefurlong]

Let me set up a situation:

A piano rests on a frictionless surface. I am standing next to the piano (on a frictional surface) and I claim that the following two statements prove it is impossible for me to move the piano:

(1) ... The kinetic energy of the piano is equal to the work I've done on it.
(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

"Therefore the piano is immovable," I claim.

(1) is equivalent to the work-kinetic-energy theorem W=\Delta E
(2) is a special case (where \frac{d\vec s}{dt}=0) of the definition of work dW=\vec F\cdot d\vec s=\vec F\cdot \frac{d\vec s}{dt}dt

Please explain where and why my logic is flawed (assuming I don't know Newton's laws).

Classically, this fails for the same reason Zeno's paradox fails. It confuses differentials with discrete steps. Think about it this way: at 0 seconds, there is still a force, f (which you are applying to the piano). Now, imagine that a differential of time, dt, has passed. Then, the piano will undergo a change in momentum, f*dt. This, after all, is the definition of force. Now that the momentum is nonzero, the instantaneous velocity is also non-zero (though infinitesimal). By definition, velocity is the rate of change of position, so now, the position will be f*dt*dt at time 2dt.

Now, if you are familiar with calculus, then to get real numbers instead of infinitesimals, you just need to integrate, and voila! The piano is now moving, and work has been done.

 

[jbriggs444]

Classically, this fails for the same reason Zeno's paradox fails. It confuses differentials with discrete steps. Think about it this way: at 0 seconds, there is still a force, f (which you are applying to the piano). Now, imagine that a differential of time, dt, has passed. Then, the piano will undergo a change in momentum, f*dt. This, after all, is the definition of force.

This imagining just pushes the problem back into infinitesimals. One is still imagining a situation where after infinitesimal time dt the piano has infinitesimal but non-zero energy. Infinitesimal work must have been done. All of which is true enough.

As you say, this confuses differentials with discrete steps. The unstated intuition is that there must be steps and that there must be a first step.

 

[Dale]

If I didn't do work on it then how did it get to be moving?

It got to be moving by accelerating, and acceleration requires non zero impulse not non zero work. The work thing is a "red herring".

 

[Tom_K]

To be honest, I don’t really understand the “problem” but:

Maybe this can be resolved by considering that the displacement of the piano does not happen instantaneously when you apply a force to it. I am thinking along the lines of a mass suspended from a spring and the mass is at the top end of travel. It is instantaneously at rest, but it is under acceleration from the force of gravity. That is, there is a phase difference between the acceleration and velocity but that certainly does not prevent the gravitational force from doing work on the mass.

I think the piano on a frictionless surface will behave in a similar fashion, with inertia causing a delay between acceleration and velocity.

 

[Jeff Rosenbury]

Sorry, I can't sleep. What you say sounds nice, but I can't understand how it is relative:
We don't have something like y=x², we have \frac{dE}{dt}=k\sqrt{E} with E(0)=0. If E=0 then dE/dt=0 therefore the energy is in a steady state w.r.t. to time (i.e. it can't change).
(k=F\sqrt{\frac{2}{m}} not that it matters.)

I agree with you Nathanael. An object with no kinetic energy cannot move -- ever. At least not under the physical model you have chosen. There are other solutions to this equation, but the trivial solution does not allow change.

In a more complete model, energy is better defined by Noether's Theorem. Plus the incompleteness theorem indicates that no particle can have zero energy. Time is not as simple as your model implies either I think. I thought these sorts of paradoxes led to the development of Quantum Mechanics?

 

[Nathanael]

This is exactly like y=x² where at x = 0 you have dy/dx = 0.

No it's not exactly like that! arhgghgharhg I'm sorry but I stand by what I said, your analogy is different and irrelevant: You have dy/dx = x =0 whereas I have dy/dx = y =0... That is importantly different! See, you're saying that I'm thinking y can't change because x is zero (which is false of course, because x is an independent variable) but I'm saying y can't change because y is zero! Isn't this importantly different?? Or have I just gone dumb...

Natural language is ambiguous and another reason for apparent paradoxes.

The paradox was is in the mathematics... dE/dt=ksqrt(E)

I agree with you Nathanael. An object with no kinetic energy cannot move -- ever. At least not under the physical model you have chosen. There are other solutions to this equation, but the trivial solution does not allow change.

Yes there are other solutions, but this is the only solution for which it starts at rest. However, I think I found the issue with this DE (see below)

One key to your conundrum is that you seem to want to imagine a first real number greater than zero -- a first instant when the piano is moving. But there is no such thing.

This is a good point, but at this point it's not even at about imagination... It's about a differential equation which is in a steady state if the energy is zero.

It got to be moving by accelerating, and acceleration requires non zero impulse not non zero work. The work thing is a "red herring".

Okay, it's a red herring... but it's a red herring which leads to a differential equation to back it up! At this point, I want to know why the math is wrong. (Anyway I think I've got it.)

If you apply a 1N force to a 1kg object, starting at t=0, at t=0 the object is stationary, but accelerating at 1m/s and at that time the rate of expenditure of energy is 0. There is no conflict.

To me there seems to be a conflict in the mathematics, let me try to emphasize what my problem is (and what I think the problem with the problem is)

From F=ma you can derive the following equation: \frac{dE}{dt}=F\sqrt{\frac{2E}{m}} ... where E=0.5mv²

This differential equation is equivalent to Newton's F=ma (just plug in E in your mind and you will see it reduces to F=ma) but they seem to be saying two very different things... This equation predicts that if E is zero it stays zero (it's in a steady state) whereas Newton's law obviously predicts that it will move. This is what the whole thread was about... (I think my verbal statement of the problem was a red herring to everyone else!)

So how can two equivalent equations make two different predictions?

My suspicion is that it has to do with dividing by zero... If you want to get from my equation to F=ma then plug in E=0.5mv² and you will get Fv=mav but then to get to Newton's F=ma you have to divide by v which is zero. Therefore, since you can't divide by zero, these equations are equivalent except at v=0.Sorry for being so stubborn, but I just felt like most replies were missing the point. A few were insightful, though. Thanks for all replies regardless.

 

[jbriggs444]

When you solve a diff

From F=ma you can derive the following equation: \frac{dE}{dt}=F\sqrt{\frac{2E}{m}} ... where E=0.5mv²

This differential equation is equivalent to Newton's F=ma (just plug in E in your mind and you will see it reduces to F=ma)

Except when v = 0. The differential equation has a singularity there. It fails to be predictive. So it is NOT equivalent to F=ma.

This equation predicts that if E is zero it stays zero (it's in a steady state).

No, it does not. It predicts that dE/dt = 0 when E=0. But that's not the same thing as predicting that E is identically zero.

 

[Nathanael]

Except when v = 0. The differential equation has a singularity there. It fails to be predictive. So it is NOT equivalent to F=ma.

I know, I pointed this out in that post. It's equivalent to f=ma except at v=0.

No, it does not. It predicts that dE/dt = 0 when E=0. But that's not the same thing as predicting that E is identically zero.

I don't understand... If E=0 (the initial condition) and dE/dt = 0 then doesn't this mean that (according to the DE) E will be forever zero?
(The DE is of course wrong at E=0 but I still don't understand your statement.)

 

[A.T.]

I'm saying y can't change because y is zero!

That makes even less sense to me.

 

[jbriggs444]

I don't understand... If E=0 (the initial condition) and dE/dt = 0 then doesn't this mean that (according to the DE) E will be forever zero?
(The DE is of course wrong at E=0 but I still don't understand your statement.)

No. E(0)=0 and E'(0)=0 does not entail that E(x) = 0 for all x.

 

[Tom_K]

I don’t believe I have ever seen the expression dE/dt = Fv before anywhere.

I see you had dW = F∙ds/dt and since W = E then dE/dt = Fv But since W = F∙s how can you treat F as a constant when differentiating W?

Since F = ma, wouldn’t you need to differentiate the ma term with respect to t also?

 

[A.T.]

If E=0 (the initial condition) and dE/dt = 0 then doesn't this mean that (according to the DE) E will be forever zero?

No, because d²E/dt² ≠ 0

Same with y=x²:
y(0) = 0
y'(0) = 0
y''(0) = 2

 

[A.T.]

I don’t believe I have ever seen the expression dE/dt = Fv before anywhere.

https://en.wikipedia.org/wiki/Power_(physics)#Mechanical_power

 

[D H]

II don't understand... If E=0 (the initial condition) and dE/dt = 0 then doesn't this mean that (according to the DE) E will be forever zero?

Not at all. If E and all of its derivatives are zero at some point and E is a smooth function, then you can say that E is identically equal to zero. Otherwise, all bets are off.

Here E=\frac 1 2 mv^2, so \frac{dE}{dt} = m \vec v \cdot \vec a and \frac{d^2E}{dt^2} = m (\vec v \cdot \dot{\vec a} + a^2). At a point where \vec v = 0, both E=0 and \frac{dE}{dt} = 0, but the second derivative is zero only if the acceleration is zero.

 

[Dale]

Okay, it's a red herring... but it's a red herring which leads to a differential equation to back it up!

The differential equation does NOT back it up. The energy is quadratic in v and therefore the derivative is 0 for small v. Your Zeno paradox (like most) is resolved by the differential equations, not confirmed.

 

[Nathanael]

Not at all. If E and all of its derivatives are zero at some point and E is a smooth function, then you can say that E is identically equal to zero. Otherwise, all bets are off.

Here E=\frac 1 2 mv^2, so \frac{dE}{dt} = m \vec v \cdot \vec a and \frac{d^2E}{dt^2} = m (\vec v \cdot \dot{\vec a} + a^2). At a point where \vec v = 0, both E=0 and \frac{dE}{dt} = 0, but the second derivative is zero only if the acceleration is zero.

Thank you. For some reason (because of my conservation with A.T.) I was thinking that y'=y implies that y''=y' which implies that all derivatives are zero. But of course it's not analogous to y'=y because of the acceleration term hidden inside of F.

 

[Dale]

This equation predicts that if E is zero it stays zero (it's in a steady state)

No. This equation does not predict that at all! $E(t)=0$ is a solution, but certainly not the only solution.

EDIT: If I did the math correctly then for a constant force:
$$E(t)=E_0 - f \sqrt{\frac{2E_0}{m}}t+\frac{f^2}{2m} t^2$$
So $E(t)=0$ corresponds to the solution where $E_0=f=0$, but there are an infinite number of other solutions also. If $E_0=0$ then $E(t)=\frac{f^2}{2m} t^2$ which is not equal to 0 except if $f=0$.

 

[A.T.]

For some reason (because of my conservation with A.T.) I was thinking that y'=y implies that y''=y'

Not sure how I gave you that idea, since my example y=x² actually disproves the above.

 

[Tom_K]

https://en.wikipedia.org/wiki/Power_(physics)#Mechanical_power

Yeah, I had one of those brain freezes going on.
dE/dt = mva is what I was trying to think of.

 

[Nathanael]

Not sure how I gave you that idea, since my example y=x² actually disproves the above.

Because I was trying to tell you y=x² is irrelevant to the example because this situation is more like y'=2y (not y'=2x) which is why I was thinking about y'=y

It's not you that made me think that, just my conversation with you.

Anyway sorry for dragging this thread out so long.

 

[darpan]

Let me set up a situation:

A piano rests on a frictionless surface. I am standing next to the piano (on a frictional surface) and I claim that the following two statements prove it is impossible for me to move the piano:

(1) ... The kinetic energy of the piano is equal to the work I've done on it.
(2) ... I can't do work on the piano unless it is moving. (But, because of (1), I can't get it moving unless I do work on it. But I can't do work on it unless it is moving... ad infinitum)

"Therefore the piano is immovable," I claim.

(1) is equivalent to the work-kinetic-energy theorem W=\Delta E
(2) is a special case (where \frac{d\vec s}{dt}=0) of the definition of work dW=\vec F\cdot d\vec s=\vec F\cdot \frac{d\vec s}{dt}dt

Please explain where and why my logic is flawed (assuming I don't know Newton's laws).

A pen is at rest on a table. I am standing next to the table. I claim that the following two statements will prove that it is impossible for me to pick up the pen:

(1) ... I will try and pick up the pen.
(2) ... I can ONLY try to pick up the pen. (But, because of (1), I will keep trying to pick up the pen. But I cannot ACTUALLY pick up the pen... ad infinitum)

"Therefore the pen can never be picked," I claim.

 

[darpan]

What if the piano rests on a frictional surface and you stand on a frictionless one.

 

[davenn]

A pen is at rest on a table. I am standing next to the table. I claim that the following two statements will prove that it is impossible for me to pick up the pen:

(1) ... I will try and pick up the pen.
(2) ... I can ONLY try to pick up the pen. (But, because of (1), I will keep trying to pick up the pen. But I cannot ACTUALLY pick up the pen... ad infinitum)

"Therefore the pen can never be picked," I claim.

But the fact, by practical demonstration, that you can pick up the pen negates your argument

 

[darpan]

But the fact, by practical demonstration, that you can pick up the pen negates your argument

Thank you Davenn. You just helped me prove, that Zeno's Paradox is practically unviable. Though its mathematical maze is an adventure to explore.

 

[votingmachine]

I think this is exactly a case parallel to Zeno's paradox. The problem is trying to define 0/0. In zeno's paradox, the infinite number of infinitely small intervals all are actually small dX's and they are still each covered in smaller and smaller dT's, allowing a constant V.

Consider that F=MA also implies that M=F/A. With no Force, and no Acceleration, we don't assume the piano has lost its mass since we last measured it.

Before a Force is applied, there is no dX. As soon as the Force is applied, there is a dX. You are asking how long does it take the piano to move no distance? dX/dT when dX=0.

The kinetic energy IS equal to the work you have done on it. The kinetic energy starts from 0, where you have done no work, also 0. You are taking the extreme of Zeno's paradox, and dividing the world into distances of 0 length. But there is no meaningful way to talk of a zero length motion.

Say that instead of the piano being at rest, it was sliding towards you and you were anchoring an ideal spring. Now it slows and slows and then STOPS, and then reverses. It only makes sense to talk of the acceleration as acting across very small distances. dx/dT/dT. The time intervals are also very small. There is no point where it is meaningful to accurately speak of what about when the dX is 0. That is a mathematical point. You need two points to have dX.

You either have something finite ... a distance between two points, no matter how small, or you have a meaningless singular point. And the problem is when you declare that the arbitrarily small intervals have to be a single point. And then the math breaks down, since it is based on differences (differentials).

I'm not sure that helps any. It is clear in my head, but seems very paradoxical as soon as you try to type it. I would recommend thinking about the spring variation, where the piano stops and reverses. How long is it at absolute rest? I think the answer is that it is at absolute rest for no time at all, even though before that moment it was in motion one direction, and then after that moment, in the other direction, it is at rest for no time at all!

 

[votingmachine]

I had another thought on this later. When you ask the amount of work for when you push during the interval from 8 inches to 9 inches, you can calculate that. But what about the work in the interval from 8 inches to 8 inches? Strictly speaking, there is no interval. I think the same problem is what you are running into when you start at 0, and define that point as an interval. You need some kind of elapsed distance or time interval.

 

[Chandra Prayaga]

Statement 2 by Nathaniel is wrong. "I can't do work on the piano unless it is moving". You can start with the piano at rest. Now push it. During the time when you are in contact with the piano, the force that you exert on the piano does work, which increases the kinetic energy from zero to some value. That is the energy point of view. You can also say that during the time you are in contact with the piano, you have delivered an impulse (force multiplied by the time of contact), which changes the momentum from zero to some value. This is the impulse-momentum point of view. You can also say, during the time that you are in contact, you are exerting a force on the piano, which, by F = ma, gives an acceleration to the piano, and over the time of contact, acceleration multiplied by time gives the final velocity. All three points of view are equivalent and give the same result.

 

[Drakkith]

Statement 2 by Nathaniel is wrong. "I can't do work on the piano unless it is moving". You can start with the piano at rest. Now push it. During the time when you are in contact with the piano, the force that you exert on the piano does work, which increases the kinetic energy from zero to some value.

Of course, but I think he was wanting an explanation that didn't involve Newton's laws or something.

 

[Chandra Prayaga]

Nathanael's statement 2 is wrong.
Start with the piano at rest. Push on it, exert a force.
Since there is no friction, you will be in contact only for a short time. You cannot keep pushing on it if there is no friction between your feet and the ground.
But during that short time:
F = ma, so there is an acceleration for a short time, and the acceleration multiplied by the tome of contact gives final velocity (view point 1)
The force multiplied by the time of contact is called the impulse, and that results in a change in momentum from zero to some final value (view point 2)
The force multiplied by the distance moved while in contact is the work done, which results in a change in kinetic energy from zero to some final value (view point 2)
All three points of view are equivalent

 

[Chandra Prayaga]

Of course, but I think he was wanting an explanation that didn't involve Newton's laws or something.

There is no such thing as an explanation that does not involve Newton's laws right? Not when you are pushing a piano.

 

[votingmachine]

Nathanael's statement 2 is wrong.
Start with the piano at rest. Push on it, exert a force.
Since there is no friction, you will be in contact only for a short time. You cannot keep pushing on it if there is no friction between your feet and the ground.
But during that short time:
F = ma, so there is an acceleration for a short time, and the acceleration multiplied by the tome of contact gives final velocity (view point 1)
The force multiplied by the time of contact is called the impulse, and that results in a change in momentum from zero to some final value (view point 2)
The force multiplied by the distance moved while in contact is the work done, which results in a change in kinetic energy from zero to some final value (view point 2)
All three points of view are equivalent

What you say is perfectly clear. I think the paradox he was building was equivalent to a force that really would not be a force ... so if you must exert a force over a distance to do work, what if you hit at the piano with a hammer which does not move beyond the starting point.

And to me the answer is that you can't hit with a hammer at point 0, and not beyond. If you did, you would not impart energy, but merely stop the hammer at the same place the piano starts. So yes, you have to actually have an interval ... even an infinitely small one to impart the force you posit. Think if you had a cam that was your force impeller. If the cam just exactly does contact the piano, there is no work. And as soon as you have more than that contact, you have an interval.

You don't have to "cross" the gap at zero. There is no gap between zero and the smallest number (as someone else pointed out). Either you exactly do not touch, or you exactly touch, or exactly do more than touch, and then do work.

The answer is the same as with Zeno. There is no gap between the most infinitely small difference and zero.

So yes, it is not in motion until work is done. And an infinitely small amount of work done in an infinitely small interval puts it into motion. I see it as asking how you cross the gap from position 0 to the first non-0 position. And there is no gap.

EDIT: I suppose the other way of saying it is to agree that there is no work until there is motion and there is no motion until there is work, but that they both begin to exist at the same infinitely small amount of time. No bit has to come before the other, but both exist as soon as the piano moves from position-zero, to the next point in space.

And as has been pointed out, that is no distance at all.

 

[Dadface]

I think the original statements should be clarified as follows:

1. The increase of kinetic energy is equal to the work done.
2. There can be no increase of kinetic energy unless work is done.

Aren't the above two statements and the original two statements different ways of saying the same thing? I can't see where there's a problem. Perhaps I'm not understanding it.

 

[A.T.]

I can't see where there's a problem. Perhaps I'm not understanding it.

See post #6.

 

[thinkandmull]

Calculus is used to say what a limit is. I struggle to understand how a limit can exist for something in space like a line. If from A to B I must always first go half a step, when will there ever be a final step. If the step can be divided in 2, its not the final step. If it cannot be divided, than I am already at my location. Again, there would be no final step.

 

[A.T.]

I must always first go half a step

The time you need to go those half steps goes towards zero.