Force from power at rest = infinity?

In summary: Force can be expressed as F = P/v where P = power (watts) and v = velocity (metres per second)So what happens when you use a given amount of power to accelerate an object from rest? This would be given by:a = P/(mv)And since v = 0 => a = P/(m*0)This would be division by 0, so is this correct that from rest the acceleration would be infinite?If so, if the acceleration is infinite, it would only be for an infintesimal time period. Would that make the results a little easier to work with?What would be the
  • #1
Pharrahnox
106
0
Force can be expressed as F = P/v where P = power (watts) and v = velocity (metres per second)

So what happens when you use a given amount of power to accelerate an object from rest? This would be given by:
a = P/(mv)

And since v = 0 => a = P/(m*0)

This would be division by 0, so is this correct that from rest the acceleration would be infinite?

If so, if the acceleration is infinite, it would only be for an infintesimal time period. Would that make the results a little easier to work with?
What would be the change in velocity for an object that accelerates with an infinite magnitude for an infintesimal time?
 
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  • #2
From the expression P = F x v, where the velocity and the force are in the same direction, you are suggesting:

1. If we know the power
2. and we know the speed,
3. then we can determine the magnitude of the force: F = P/v.

This is OK.

Then you suggest that we use the case where v=zero, and expect to obtain something meaningful.

If you are standing still, the force isn't doing any work ... and the power will be zero.

So I ask: Why do you want to divide by zero?

Your assumption is that the power expended and the applied force are remaining constant while you are stopping. Do you have a physically meaningful situation where this could happen?
 
  • #3
Pharrahnox said:
So what happens when you use a given amount of power to accelerate an object from rest? This would be given by:
a = P/(mv)

And since v = 0 => a = P/(m*0)

This would be division by 0, so is this correct that from rest the acceleration would be infinite?

If so, if the acceleration is infinite

The problem here is that you cannot instantaneously apply a given amount of power to an object at rest; the power starts at zero and ramps up along with the velocity. Thus, the pathological equation you get from ##a=\frac{P}{mv}## is not ##\frac{P}{0}=\infty##; you get a 0/0 which is indeterminate - it's not infinite but we don't know what it is.

To actually solve the physics of what happens when the force is initially applied, you have to consider things like the compressibility of the object, the details of the mechanism that's applying the force and your power source. It gets to be rather complicated so we usually don't worry about exactly what happens at time zero when ##a=\frac{P}{mv}## behaves badly.
 
  • #4
A Division by zero isn't an infinite but an undefined result.
It's because If the divisor approaches zero from the negative side, the result tends to minus infinity.
 
  • #5
Everybody is beating up the division by zero but the problem as stated by the OP can be solved easily. It is possible to ingrate a function that diverges at one point and still find a finite result.
 
  • #6
Pharrahnox said:
Force can be expressed as F = P/v where P = power (watts) and v = velocity (metres per second)

So what happens when you use a given amount of power to accelerate an object from rest? This would be given by:
a = P/(mv)

And since v = 0 => a = P/(m*0)

This would be division by 0, so is this correct that from rest the acceleration would be infinite?
Yes, which is why no real mechanism can provide a given (non-zero) amount of power to an object at rest.

A real mechanism can provide a given (non-zero) force, and that force delivers 0 power at 0 velocity.
 
  • #7
DaleSpam said:
Yes, which is why no real mechanism can provide a given (non-zero) amount of power to an object at rest.

A real mechanism can provide a given (non-zero) force, and that force delivers 0 power at 0 velocity.

What's the force between two electrons in the limit their distance goes to zero?
 
  • #8
F=dP/dv and will be finite at v=0.
 
  • #9
Ah, so the error was the assumption that just sticking power into something would just magically make it do something.

I guess it makes sense, now, as you can't directly propel something with just power, you need the power to be creating a force of some sort. And you would then have force anyway, to determine the acceleration, and not power.

And yes, I do realize that P/0 doesn't actually = [itex]\infty[/itex], but for a velocity of 0.000...1, the value would be "tending towards infinity".
 
  • #10
The mistake is that F#P/v, especially at v=0. F=dP/dv.
Also, while v=0 the energy of the particle is constant and the power delivered to it is zero.
 
  • #11
dauto said:
What's the force between two electrons in the limit their distance goes to zero?
A finite expression due to quantum mechanics, but the interaction depends on the way the electrons approach each other.
 
  • #12
Seems a lot simpler to think about this in terms of work as force times distance. W=Fx.

Then power is P=dW/dt=F(dx/dt)+x(dF/dt). If the object is at rest at t=0 then both terms equal zero and P=0.

The step from 0=Fv to F=infinity is simply a mathematical error.
 
  • #13
mfb said:
A finite expression due to quantum mechanics, but the interaction depends on the way the electrons approach each other.

Who said anything about quantum mechanics? My point in this thread is not that an infinite force is possible. Rather, I'm simply saying that the constant power problem proposed in the OP is mathematically sound and easily solvable as stated. It cannot be dismissed just because of the divergent force at the initial instant. When confronted with infinities our reaction should not be to run to the hills. Now, whether or not the model is realizable in the real world is a different question. All models are in one way or another simplified versions of the real world and that one is no exception, but real insight can be obtained from solving it so one should solve it.
 
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  • #14
dauto said:
My point in this thread is not that a infinite force is possible. Rather, I'm simply saying that the constant power problem proposed in the OP is mathematically sound and easily solvable as stated. It cannot be dismissed just because of the divergent force at the initial instant. When confronted with infinities our reaction should not be to run to the hills. Now, whether or not the model is realizable in the real world is a different question. All models are in one way or another simplified versions of the real world and that one is no exception, but real insight can be obtained from solving it so one should solve it.
That's all fine, but it appears to me the OP wanted to know about reality.

Also, yes, all models involve simplifications and error, but this one breaks down completely for the specified case. At the same time, ironically, it correctly predicts failure of machines being described, for that case.
 
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  • #15
dauto said:
the constant power problem proposed in the OP is mathematically sound

The problem is not mathematically sound. Dividing by zero never is. Please read my earlier posts. If v=0 then P=0 and these instantaneous values do not determine the force.
 
  • #16
my2cts said:
The problem is not mathematically sound. Dividing by zero never is. Please read my earlier posts. If v=0 then P=0 and these instantaneous values do not determine the force.

The point being made (as I understand it) is that the motion of the object is determinable, even though the second derivitive of its position at t=0 may be non-existent (infinite).

One can get velocity as a function of time by using kinetic energy. From the given input power KE(t) = Pt. But KE(t) = 1/2 mv2. So v(t) = sqrt(2Pt/m).

Integrate that with respect to time and one can get x(t) without ever needing acceleration or force. dauto's point was made more generally about improper integrals and functions that are undefined at a single point as I recall. The acceleration or force or even power precisely at t=0, x=0, v=0 is irrelevant to determining the position of the object at all future times.
 
  • #17
jbriggs444 said:
The point being made (as I understand it) is that the motion of the object is determinable, even though the second derivitive of its position at t=0 may be non-existent (infinite).

One can get velocity as a function of time by using kinetic energy. From the given input power KE(t) = Pt. But KE(t) = 1/2 mv2. So v(t) = sqrt(2Pt/m).

Integrate that with respect to time and one can get x(t) without ever needing acceleration or force. dauto's point was made more generally about improper integrals and functions that are undefined at a single point as I recall. The acceleration or force or even power precisely at t=0, x=0, v=0 is irrelevant to determining the position of the object at all future times.

KE=Pt is only correct if P has been constant since t=0 and zero before that. The general expression is KE=Integral Pdt.
 
  • #18
jbriggs444 said:
The point being made (as I understand it) is that the motion of the object is determinable, even though the second derivitive of its position at t=0 may be non-existent (infinite).

One can get velocity as a function of time by using kinetic energy. From the given input power KE(t) = Pt. But KE(t) = 1/2 mv2. So v(t) = sqrt(2Pt/m).

Integrate that with respect to time and one can get x(t) without ever needing acceleration or force. dauto's point was made more generally about improper integrals and functions that are undefined at a single point as I recall. The acceleration or force or even power precisely at t=0, x=0, v=0 is irrelevant to determining the position of the object at all future times.

Thank you very much. That's exactly the point I'm making. You said it better than I could do.
 
  • #19
my2cts said:
KE=Pt is only correct if P has been constant since t=0 and zero before that. The general expression is KE=Integral Pdt.

The relevant integral can be improper. That, in turn, allows for the possibility that P(0) is not defined.

If we are given that KE=0 at t=0 then it is the (improper) definite integral from 0 to t that matters. If we are only interested in the motion for t>0 then the values for P(t) for t<0 simply do not enter in.

If one insists on extending positive constant power into the past, we would admittedly have the problem of an imaginary KE at t<0. That would not seem to be very physical.

Edit: The point about power being possibly undefined at t=0 is irrelevant. The trajectory you get if you crank the numbers will have a well defined P at t=0 even if the force or acceleration at t=0 are undefined. It will be a one-sided limit, but that's OK as long as we are only considering a trajectory from the starting point forward and not from the starting point going into the past.
 
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  • #20
Pharrahnox said:
Ah, so the error was the assumption that just sticking power into something would just magically make it do something.
The problem might be the idea, that power is something "that you stick into something". Power is just a numerical value, that some observer computes based on the velocities in his reference frame. If the velocity is zero in his frame, then so is power and he cannot use F = P/v to determine F, which still can have some finite value.
 
  • #21
jbriggs444 said:
The point being made (as I understand it) is that the motion of the object is determinable, even though the second derivitive of its position at t=0 may be non-existent (infinite).
Mathematically sure, but no real force can do that.

I don't know what you and dauto are going on about. Even if you can mathematically work around the infinite force, and even if other quantities do not diverge, the fact remains that an infinite force is not physical. The objections based on that fact, which point out that F is infinite or undefined at v=0 for a finite P, are entirely correct objections.
 
  • #23
my2cts said:
F=dP/dv and will be finite at v=0.
No, P=F.v and F will be infinite if P is finite and v is 0.
 
  • #24
DaleSpam said:
Mathematically sure, but no real force can do that.

I don't know what you and dauto are going on about. Even if you can mathematically work around the infinite force, and even if other quantities do not diverge, the fact remains that an infinite force is not physical. The objections based on that fact, which point out that F is infinite or undefined at v=0 for a finite P, are entirely correct objections.

An objection that F is infinite or undefined at zero is mathematically correct. But that is not a physical objection.
 
  • #25
DaleSpam said:
No, P=F.v and F will be infinite if P is finite and v is 0.

That is not the only definition of P.

It is valid to define P as the derivitive of KE with respect to time. That formulation does not suffer from the problem of an undefined force multiplied by a zero velocity.
 
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  • #26
DaleSpam said:
No, P=F.v and F will be infinite if P is finite and v is 0.

I am confident that your approach is in disagreement with classical mechanics.
 
  • #27
jbriggs444 said:
That is not the only definition of P.

It is valid to define P as the derivitive of KE with respect to time. That formulation does not suffer from the problem of an undefined force multiplied by a zero velocity.
That is indeed a valid definition, but it doesn't change anything. If you start with some fixed power under that definition and consider what happens for an object at rest you still find that it undergoes an infinite acceleration and therefore an infinite force.
 
  • #28
my2cts said:
I am confident that your approach is in disagreement with classical mechanics.
Then provide a reference for you formula: f=dP/dv
 
  • #29
DaleSpam said:
That is indeed a valid definition, but it doesn't change anything. If you start with some fixed power under that definition and consider what happens for an object at rest you still find that it undergoes an infinite acceleration and therefore an infinite force.

Yes, certainly. An infinite acceleration and a corresponding infinite force applied over a zero time interval. That makes it almost certainly non-physical and almost certainly non-measurable.
 

1. What is the concept of "force from power at rest = infinity"?

The concept of "force from power at rest = infinity" is a theoretical concept in physics that suggests that an object at rest has an infinite amount of potential energy, which can be converted into force when the object is set in motion.

2. How does "force from power at rest = infinity" relate to Newton's laws of motion?

According to Newton's laws of motion, an object at rest will remain at rest unless acted upon by an external force. In the concept of "force from power at rest = infinity", an infinite amount of potential energy is considered to be the external force that can cause an object at rest to move.

3. Is "force from power at rest = infinity" a proven theory?

No, "force from power at rest = infinity" is not a proven theory. It is a theoretical concept that has not been empirically tested or confirmed by scientific experiments.

4. Can the concept of "force from power at rest = infinity" be applied to all objects at rest?

The concept of "force from power at rest = infinity" is often used in theoretical physics to understand the behavior of objects at rest in extreme scenarios, such as in black holes or the Big Bang. It may not be applicable to all objects at rest in everyday scenarios.

5. How does the infinite potential energy in "force from power at rest = infinity" affect the behavior of an object?

The infinite potential energy in "force from power at rest = infinity" suggests that an object at rest has an infinite amount of force that can be applied to it. This can potentially lead to extreme acceleration and unpredictable behavior of the object when it is set in motion.

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