Power and Velocity

AThousandYoung
Posers and Puzzles 19 Jan '10 22:26
1. AThousandYoung
19 Jan '10 22:2612 edits
Given a Power of 1 Watt and a mass of 1 kg, what is the maximum velocity the mass can reach?

Pt = K, Pt = mv^2/2, 2Pt/m = v^2, (2Pt/m)^(1/2) = v

Thus v is proportional to t^(1/2) - asymptotic equation?

Im not sure if a finite answer exists.
2. joe shmo
Strange Egg
20 Jan '10 00:32
Originally posted by AThousandYoung
Given a Power of 1 Watt and a mass of 1 kg, what is the maximum velocity the mass can reach?

[hidden]Pt = K, Pt = mv^2/2, 2Pt/m = v^2, (2Pt/m)^(1/2) = v[/hidden]

[hidden]Thus v is proportional to t^(1/2) - asymptotic equation?[/hidden]

[hidden]Im not sure if a finite answer exists.[/hidden]
I think there is no solution.

from you equation

v = sqrt[(2P*t)/m]

This is the sqrt function, which increase indefinately. so no max value

or we can say

when the rate of change of velocity with respect to time is zero, velocity will be maximized at this time if a solution exists.

Power and mass are constant

v' = (sqrt[(2P*t)/m])'=0

dv/dt = 1/2((2P*t)/m)^(-1/2)*(2P/m)=0

= P/(m*sqrt(2P*t/m)) =0

The only way this can be true is if P=0, which it does not.

we arrive at no solutions for t

thus v has no maximum value.
3. 20 Jan '10 07:47
Originally posted by joe shmo
I think there is no solution.

from you equation

v = sqrt[(2P*t)/m]

This is the sqrt function, which increase indefinately. so no max value

or we can say

when the rate of change of velocity with respect to time is zero, velocity will be maximized at this time if a solution exists.

Power and mass are constant

v' = (sqrt[(2P*t)/m])'=0

d ...[text shortened]... hich it does not.

we arrive at no solutions for t

thus v has no maximum value.
i haven't done any work or really put much thought into this, but... isn't c (the speed of light) the limiting speed of the universe? and i believe it's known/thought that only non-massive particles can achieve this speed? though the pure newtonian physics equations can conceive of a v>c, doesn't relativity say that c will be an upper bound on v?

that being said, i have no idea what the max speed would actually be... but i think there should be a max for 1kg of mass (unless you assume that as speed increases towards c the mass is converted to energy... but then it's not really "1kg" of mass anymore is it?)

perhaps the real physics guys (yourselves included, as well as PBE6, wolfgang, themaster, etc.) have further insight into this that i lack ...? ðŸ™‚
4. AThousandYoung
20 Jan '10 07:58
Originally posted by Aetherael
i haven't done any work or really put much thought into this, but... isn't [b]c (the speed of light) the limiting speed of the universe? and i believe it's known/thought that only non-massive particles can achieve this speed? though the pure newtonian physics equations can conceive of a v>c, doesn't relativity say that c will be an upper bound on v? ...[text shortened]... ll as PBE6, wolfgang, themaster, etc.) have further insight into this that i lack ...? ðŸ™‚[/b]
I'm pretty certain this is not about relativistic effects. I got my equation from classical mechanics.
5. 20 Jan '10 08:12
Originally posted by AThousandYoung
I'm pretty certain this is not about relativistic effects. I got my equation from classical mechanics.
i was just trying to use my limited physics knowledge to create a potential upper bound on the velocity to possibly discount the "no limit for the possible velocity" notion... but anything that could approach the speed of light, for all intents and purposes (and in the spirit of the problem, i think?) would indeed be considered unlimited in terms of possible top speed.
6. joe shmo
Strange Egg
20 Jan '10 13:221 edit
Originally posted by Aetherael
i haven't done any work or really put much thought into this, but... isn't [b]c (the speed of light) the limiting speed of the universe? and i believe it's known/thought that only non-massive particles can achieve this speed? though the pure newtonian physics equations can conceive of a v>c, doesn't relativity say that c will be an upper bound on v?
ll as PBE6, wolfgang, themaster, etc.) have further insight into this that i lack ...? ðŸ™‚[/b]
Your right, with the effects of relativity c is the actual limiting velocity. I think he just wanted to see if there existed a limiting velocity that would occurr in the practical realm of velocity( actually I don't know much about physics either, you probably no more ðŸ™‚).

However, throw in some air resistance there is probably a limiting velocity.
7. wolfgang59
Mr. Wolf
20 Jan '10 16:08
Originally posted by AThousandYoung
Given a Power of 1 Watt and a mass of 1 kg, what is the maximum velocity the mass can reach?

[hidden]Pt = K, Pt = mv^2/2, 2Pt/m = v^2, (2Pt/m)^(1/2) = v[/hidden]

[hidden]Thus v is proportional to t^(1/2) - asymptotic equation?[/hidden]

[hidden]Im not sure if a finite answer exists.[/hidden]
Using Newtonian mechanics and no drag there is no limit to the velocity because no time limit is imposed. (The mass is irrelrvant)

No limit to t means that given enough time even at 1 watt you can input as much energy as you like therefore mv^2/2 is boundless therefeore v is boundless.
8. TheMaster37
Kupikupopo!
20 Jan '10 16:24
Originally posted by joe shmo
Your right, with the effects of relativity c is the actual limiting velocity. I think he just wanted to see if there existed a limiting velocity that would occurr in the practical realm of velocity( actually I don't know much about physics either, you probably no more ðŸ™‚).

However, throw in some air resistance there is probably a limiting velocity.

Here's my input.

A power of 1 Watt mean that 1J of energy is converted per second.
Since the source of the energy is not specified I'll assume it's some sort of engine.

Per second the kinetic energy 0.5*1*v^2 will increase by 1 J
Assuming v(0) = 0 we have K(t) = t, or mv^2 = 2t, or v^2 = 2t/m

This gives us a formula for the speed: v = sqrt(2t).
Therefore there is no maximm speed, aside for relativistic effects.
-------------------------------------------------

Should the mass itself be converted into energy then we have:

E = 1*c^2
We calculate the Kinetic energy at the point that the last bit of mass is converted:

K = 0,5 * 1 * v^2 = 1*c^2 ---> v = c*sqrt(2)

If somehow the energflow is capped when the mass reaches speed c we have

1*c^2 - 0,5*1*c^2 = 0,5*c^2 joules of energy left, wich corresponds to

m = E/c^2 = 0,5 kg

All this seems a bit odd, but then again a contant energyconvertion means a constant increase of kinetic energy.
9. PBE6
Bananarama
20 Jan '10 17:261 edit
Just took a quick stab at figuring this out with the assumption that the mass is relativistic and increases as the velocity approaches the speed of light according to the relationship M = m / (1 - v^2/c^2).

Starting with our expression for kinetic energy in terms of power:

Pt = 1/2Mv^2

and differentiating both sides with respect to time while allowing the mass to vary, we have:

P = 1/2(dM/dt*v^2 + 2Mv*dv/dt)

Using some implicit differentiation and a lot of algebra (which hopefully didn't get screwed up somewhere), I obtained the expression:

dv/dt = (2P/Mv)*(c^2-v^2)/(2c^2-v^2)

This gives us the acceleration of the spaceship if it has relativistic mass. If we set dv/dt = 0, we can find the stationary points for the velocity. Doing this gives us two possible solutions:

2P/Mv = 0; or (c^2-v^2)/(2c^2-v^2) = 0

The first expression has no solutions, however it's sometimes useful to think about the possibilities. P = 0 is a possible solution, but the question implies a constant power input. We know that v -> inf. doesn't work because the relativistic equations imply that v < c. One interesting possibility is M -> inf., which agrees with the assumption that the mass increases without bound as v -> c. One interpretation is that infinite energy is required to accelerate an infinite mass, therefore this point will never be reached.

The second expression has one solution, v = c. However, since v < c, this point will never be reached. Since the expression for the acceleration is always positive, this implies that the spacecraft will continually accelerate, getting closer and closer to the speed of light without ever reaching it. So I think the answer is that the velocity of the spacecraft will approach c asymptotically in the relativistic model.
10. AThousandYoung
21 Jan '10 05:35
Originally posted by PBE6
Just took a quick stab at figuring this out with the assumption that the mass is relativistic and increases as the velocity approaches the speed of light according to the relationship M = m / (1 - v^2/c^2).

Starting with our expression for kinetic energy in terms of power:

Pt = 1/2Mv^2

and differentiating both sides with respect to time while allowing ...[text shortened]... that the velocity of the spacecraft will approach c asymptotically in the relativistic model.
Does this mean that one can derive relativity from K = mv^2/2?

More likely it means that without relativity there is no asymptote.
11. AThousandYoung
21 Jan '10 05:36
Originally posted by wolfgang59
Using Newtonian mechanics and no drag there is no limit to the velocity because no time limit is imposed. (The mass is irrelrvant)

No limit to t means that given enough time even at 1 watt you can input as much energy as you like therefore mv^2/2 is boundless therefeore v is boundless.
Sounds solid to me, but I'm willing to be convinced otherwise.
12. 21 Jan '10 18:14
Originally posted by AThousandYoung
Does this mean that one can derive relativity from K = mv^2/2?.
No - because you're already assuming relativity by using the relativistic mass.
13. sonhouse
Fast and Curious
21 Jan '10 19:43
Originally posted by AThousandYoung
Sounds solid to me, but I'm willing to be convinced otherwise.
Did you work out the actual acceleration in terms of G?. I won't tell you what it is in case you want to work it out for yourself but it is surprising. 1 watt continuously applied to 1 Kg, assuming all 1 watt gets converted to acceleration. Similar to my post.
14. PBE6
Bananarama
21 Jan '10 20:03
Originally posted by sonhouse
Did you work out the actual acceleration in terms of G?. I won't tell you what it is in case you want to work it out for yourself but it is surprising. 1 watt continuously applied to 1 Kg, assuming all 1 watt gets converted to acceleration. Similar to my post.
The "actual acceleration" is a function of the velocity.

dKE/dt = P

Pt = KE = 1/2mv^2

Taking the derivative of both sides with respect to t:

P = mv*dv/dt

dv/dt = a = P/(mv)

Subbing in the values provided, we have:

a = (1)/((1)*v) = 1/v

The limit of "a" as v->0 from the right side does not exist as the function shoots off to infinity. The limit of "a" as v->inf. is 0. The acceleration, being a smooth curve, will take on all values in between at some point.
15. joe shmo
Strange Egg
21 Jan '10 20:083 edits
Originally posted by sonhouse
Did you work out the actual acceleration in terms of G?. I won't tell you what it is in case you want to work it out for yourself but it is surprising. 1 watt continuously applied to 1 Kg, assuming all 1 watt gets converted to acceleration. Similar to my post.
how many times must we go over this???? YOU NEED TO STIPULATE A TIME......:'(

I don't know what your calculating, but its not the acceleration of the body in question given the parameters.

have a look at this

in this problem

acceleration = sqrt(2*power/(mass*time))

you have stipulated

Power = 1watt
mass = 1kg
time is required to find acceleration otherwise it has an infinite number of solutions.

do you have a graphing calculator? If you do graph the equation that PBE6 derrived for you
y=acceleratio;x=velocity

y = 1/x

What you are telling us (with your solution) is that this Graph of y = 1/x, is not a curve, but a horizontal line at some positive y value.

So graph the equation and see for yourself that it is indeed a curve, and NOT A HORIZONTAL LINE(meaning acceleration is NOT CONSTANT)