# Power and Velocity II

AThousandYoung
Posers and Puzzles 28 Jan '10 23:01
1. AThousandYoung
28 Jan '10 23:012 edits
You are in a spaceship which is not accelerating. It has a rocket engine which supplies a power = P to accelerate the spaceship when it is on.

You light the rocket for time = T, making the ship accelerate. Then you turn off the engine. Now the ship has velocity = V relative to the first inertial frame.

You light the rocket for time = T, making the ship accelerate. Then you turn off the engine.

The ship does not accelerate to velocity = 2V

Why?

Energy is proportional to square of velocity

This implies that not all inertial reference frames are equal

This implies Newtonian relativity is wrong; there is a preferred frame of reference

The real puzzle for this thread is how do I reconcile this?
2. wolfgang59
Mr. Wolf
28 Jan '10 23:27
Originally posted by AThousandYoung
You are in a spaceship which is not accelerating. It has a rocket engine which supplies a power = P to accelerate the spaceship when it is on.

You light the rocket for time = T, making the ship accelerate. Then you turn off the engine. Now the ship has velocity = V relative to the first inertial frame.

You light the rocket for time = T, makin ...[text shortened]... ference[/hidden]

[hidden]The real puzzle for this thread is how do I reconcile this?[/hidden]
I think we have explored this enough! The problem I think is one of language and intuition.

Energy is defined as a force over a distance.
Power is therefore a force over distance per unit time.

As we increase velocity we cover a GREATER DISTANCE per unit time.

therefore
per unit POWER we exert a SMALLER FORCE over a GREATER DISTANCE

As velocity increases the distance per unit time increases and therefore the force is smaller.
3. PBE6
Bananarama
28 Jan '10 23:34
Originally posted by AThousandYoung
You are in a spaceship which is not accelerating. It has a rocket engine which supplies a power = P to accelerate the spaceship when it is on.

You light the rocket for time = T, making the ship accelerate. Then you turn off the engine. Now the ship has velocity = V relative to the first inertial frame.

You light the rocket for time = T, makin ...[text shortened]... ference[/hidden]

[hidden]The real puzzle for this thread is how do I reconcile this?[/hidden]
It's all relative:

http://en.wikipedia.org/wiki/Kinetic_energy

"The kinetic energy of an object is...defined as the work needed to accelerate a body of a given mass from rest to its current velocity... The kinetic energy of a single object is completely frame-dependent (relative)."
4. AThousandYoung
29 Jan '10 04:54
Originally posted by PBE6
It's all relative:

http://en.wikipedia.org/wiki/Kinetic_energy

"The kinetic energy of an object is...defined as the work needed to accelerate a body of a given mass from rest to its current velocity... The kinetic energy of a single object is completely frame-dependent (relative)."
So when I switched the frame of reference the kinetic energy "disappeared" with respect to that frame. OK.
5. AThousandYoung
29 Jan '10 07:19
Originally posted by PBE6
It's all relative:

http://en.wikipedia.org/wiki/Kinetic_energy

"The kinetic energy of an object is...defined as the work needed to accelerate a body of a given mass from rest to its current velocity... The kinetic energy of a single object is completely frame-dependent (relative)."
Hmmm.

Let's have two frames; set one at v=0 and the other at v=V

Add enough energy to make the vehicle accelerate from v=0 to v=V

Now, with respect to the first frame, the vehicle is moving v=V. With respect to the second frame, the vehicle has v=0.

Add the same amount of energy.

With respect to the first frame, the vehicle is moving, say, v=3V/2 (the second time it accelerated it added V/2).

With respect to the second frame, the vehicle is going v=V, just as it did when going from v=0 to v=V with respect to the first frame.

This implies it's going 2V. But it's not.

Arrrgh!
6. AThousandYoung
29 Jan '10 07:352 edits
Let me try this again.

There is a little robotic space probe floating out in space that is not accelerating. We can take it's inertial frame of reference as v=0.

The engine, when on, provides 1 watt of power which is converted to kinetic energy. The probe has a mass of 2 kg. After one second, the probe has 1 J of kinetic energy, and it's velocity is

1 J = (2 kg)(v^2)/2
1 J/kg = v^2
1 (m/s)^2 = v^2
1 m/s = v

OK, so after one watt is used to accelerate the probe, it is now moving at 1 m/s with respect to the orginal frame of reference.

Now, I want to present two chains of reasoning:

A) If I run the engine for three more seconds, the probe will move at a speed of 2 m/s with respect to the original frame of reference as calculated by K=mv^2/2.

B) According to classical relativity, all inertial frames of reference are equivalent in terms of how the laws of physics acts within them. Thus, after the first second, we can choose a new inertial frame of reference that moves with the ship. Now the ship has v=0 and K=0 with respect to the new frame of reference.

We let the engine run for 1 s. Now the ship has a velocity of 1 m/s with respect to the new frame of reference. If we reset the frame of reference every second, this process will continue each second.

After four seconds, the probe is moving at 1 m/s with respect to the fourth frame of reference. The fourth frame of reference is moving at 1 m/s with respect to the third frame of reference, so the probe is mobing at 2 m/s with respect to the third frame...and 3 m/s with respect to the second frame...and the probe has a velocity of 4 m/s with respect to the probe's original velocity four seconds ago.

Why am I getting different answers?
7. wolfgang59
Mr. Wolf
29 Jan '10 11:01
The increse in kinetic energy from velocity u to velocity v is
m/2(v^2 - u^2)

you cant be surprised that its not equal to
m/2(v-u)^2
8. TheMaster37
Kupikupopo!
29 Jan '10 13:15
Originally posted by wolfgang59
The increse in kinetic energy from velocity u to velocity v is
m/2(v^2 - u^2)

you cant be surprised that its not equal to
m/2(v-u)^2
Enlighten us wolfgang, where exactly does his line of reasoning go wrong?

We alraedy know that at least one of them is wrong, seeing as they produce diferent answer.
9. 29 Jan '10 15:32
I think at least one of the problems is the assumption that you can have an engine that generates a constant power outage. Since Power = Force x Velocity, it's clear that the power must be a function of the frame of reference.
10. TheMaster37
Kupikupopo!
29 Jan '10 16:57
Originally posted by mtthw
I think at least one of the problems is the assumption that you can have an engine that generates a constant power outage. Since Power = Force x Velocity, it's clear that the power must be a function of the frame of reference.
This sounds very plausible to me ðŸ™‚
11. sonhouse
Fast and Curious
29 Jan '10 17:09
Originally posted by mtthw
I think at least one of the problems is the assumption that you can have an engine that generates a constant power outage. Since Power = Force x Velocity, it's clear that the power must be a function of the frame of reference.
I thought power was a force per unit time, like horsepower, 1 hp = lifting 555 pounds one foot in one second. Well, one foot/second is a velocity I guess. I was thinking of it in terms of acceleration. If you lift something up off the ground isn't that an acceleration?
12. 29 Jan '10 17:172 edits
Similar idea. Power is the rate of doing work. But in this situation the only work being done is accelerating the rocket.

P = d/dt(mv^2/2) = mv.dv/dt = mva = Fv
13. 29 Jan '10 17:33
One thing to note: if you look at the total kinetic energy (including the mass ejected from the rocket to make it accelerate), you'll find that although the KE is dependent on the frame of reference the rate of increase in energy is not.
14. joe shmo
Strange Egg
29 Jan '10 19:31
Originally posted by sonhouse
I thought power was a force per unit time, like horsepower, 1 hp = lifting 555 pounds one foot in one second. Well, one foot/second is a velocity I guess. I was thinking of it in terms of acceleration. If you lift something up off the ground isn't that an acceleration?
You can lift , or move something at a constant speed (theoretically). which means that the velocity isn't changing, thus acceleration of the body is 0.

And Power is a Force applied for a distance (F*d =Work= Energy) per unit time time that the force is applied.
15. AThousandYoung