# Force from power at rest = infinity?

1. Apr 17, 2014

### Pharrahnox

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?

2. Apr 17, 2014

### UltrafastPED

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. Apr 17, 2014

### Staff: Mentor

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. Apr 17, 2014

### peterparker

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. Apr 17, 2014

### dauto

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. Apr 17, 2014

### Staff: Mentor

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. Apr 17, 2014

### dauto

What's the force between two electrons in the limit their distance goes to zero?

8. Apr 17, 2014

### my2cts

F=dP/dv and will be finite at v=0.

9. Apr 17, 2014

### Pharrahnox

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 realise that P/0 doesn't actually = $\infty$, but for a velocity of 0.000.....1, the value would be "tending towards infinity".

10. Apr 18, 2014

### my2cts

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. Apr 18, 2014

### Staff: Mentor

A finite expression due to quantum mechanics, but the interaction depends on the way the electrons approach each other.

12. Apr 18, 2014

### mikeph

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. Apr 18, 2014

### dauto

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.

Last edited: Apr 18, 2014
14. Apr 18, 2014

### Staff: Mentor

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.

Last edited: Apr 18, 2014
15. Apr 18, 2014

### my2cts

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. Apr 18, 2014

### jbriggs444

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. Apr 18, 2014

### my2cts

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. Apr 18, 2014

### dauto

Thank you very much. That's exactly the point I'm making. You said it better than I could do.

19. Apr 18, 2014

### jbriggs444

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.

Last edited: Apr 18, 2014
20. Apr 18, 2014

### A.T.

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.