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Force vs work

by BitWiz
Tags: force, work
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BitWiz
#1
Jan12-11, 02:15 PM
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I'm still having trouble with this:

Generically, work = m * d^2 / t ^2 (mass, displacement, time)

If to a 1kg mass at rest in space I attach a small rocket that fires with a force of one newton for one second, at the end of that period the mass will have traveled 0.5 meters, and the rocket will have done the work of 0.5 joules. (I think I've got this right(?))

Now if I do the same thing with a 1kg mass that is already traveling at 100,000 meters per second, this same rocket applies a newton over a displacement of about 100,000 meters. From an observer's frame, has the rocket now done 100,000 joules of work? If so, how are joules as quantifying *anything* useful in space travel?

Ultimately, I'm trying to resolve this assertion, seen in several locations: "Accelerating [1000 kg] to 10% of the speed of light requires 450 picojoules of work." The author used this huge number to substantiate his claim that even traveling to the middle of the Oort cloud is beyond human endurance and even physics.

I understand the arithmetic from which this number was (apparently) derived:

W = mc^2 / SQRT(1 - 10%^2) - mc^2

But I wonder about the relevance of using "work" to describe what a rocket engine does. If I'm trying to get to 10% of the speed of light, why do I care about the distance travelled during the effort except in the increments required to quantify velocity? Shouldn't I care only about newtons and seconds (impulse)? Is this 450 PJ number just smoke and mirrors?

Thanks,
Bit
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HallsofIvy
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Jan12-11, 02:47 PM
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Quote Quote by BitWiz View Post
I'm still having trouble with this:

Generically, work = m * d^2 / t ^2 (mass, displacement, time)
I think you are simply saying that "work" (energy) has units of "mass time distance squared over time squared" as in one Joule being one kg m squared per seconds squared. If that is what you are sayiung, yes, that is correct.

If to a 1kg mass at rest in space I attach a small rocket that fires with a force of one newton for one second, at the end of that period the mass will have traveled 0.5 meters, and the rocket will have done the work of 0.5 joules. (I think I've got this right(?))
Let me see. A 1 N force will accelerate a 1 kg mass at 1 m per second per second so we have dv/dt= 1, v= dx/dt= t, x= (1/2)t^2 (starting from rest at x= 0). In one second, the mass will have moved 1/2 m. and will be moving at 1 m/s and so have kinetic energy of 1/2 Joule. Yes, since the rocket has gained 1/2 Joule in kinetic energy, 1/2 Joule of work was done by the rocket engine.

Now if I do the same thing with a 1kg mass that is already traveling at 100,000 meters per second, this same rocket applies a newton over a displacement of about 100,000 meters.
If the initial speed was 100000 m/s, then after 1 second, it will have moved 1/2+ 1000000 meters. The speed has increased by 1/2 m/s, as before, and the kinetic energy has increased by 1/2 Joule as before.

From an observer's frame, has the rocket now done 100,000 joules of work?
No. The rocket engine is attached to the rocket and cannot be said to have applied the force over that distance.

If so, how are joules as quantifying *anything* useful in space travel?

Ultimately, I'm trying to resolve this assertion, seen in several locations: "Accelerating [1000 kg] to 10% of the speed of light requires 450 picojoules of work." The author used this huge number to substantiate his claim that even traveling to the middle of the Oort cloud is beyond human endurance and even physics.

I understand the arithmetic from which this number was (apparently) derived:

W = mc^2 / SQRT(1 - 10%^2) - mc^2

But I wonder about the relevance of using "work" to describe what a rocket engine does. If I'm trying to get to 10% of the speed of light, why do I care about the distance travelled during the effort except in the increments required to quantify velocity? Shouldn't I care only about newtons and seconds (impulse)? Is this 450 PJ number just smoke and mirrors?

Thanks,
Bit
cepheid
#3
Jan12-11, 02:55 PM
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The work done is equal to the object's change in kinetic energy, which is 0.5 joules in both cases. Something is wrong with the force*displacement idea in the latter case.

Although energy is conserved, it is not invariant, meaning that different observers in different frames of reference (that are in motion relative to each other) will measure different kinetic energies for the same object. That having been said, the *change* in kinetic energy should have been measured to be the same for all observers. So, I'll have to think about this some more.

By the way, a picojoule is a really small amount of energy. pico = 10-12. Perhaps you were thinking of a petajoule?

cepheid
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Jan12-11, 03:00 PM
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Force vs work

Quote Quote by HallsofIvy View Post
No. The rocket engine is attached to the rocket and cannot be said to have applied the force over that distance.
So if I'm on a train pushing a cart down the aisle, the only value for the displacement that is valid in calculating the work I did on the cart is its displacement in my rest frame (i.e. its displacment relative to the train). Its displacement relative to an observer on the side of the tracks is not relevant. Is that the answer? You have to measure the displacement in the rest frame of whatever is applying the force?

EDIT: sorry, wrong quotation initially.
BitWiz
#5
Jan12-11, 04:02 PM
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Quote Quote by cepheid View Post
By the way, a picojoule is a really small amount of energy. pico = 10-12. Perhaps you were thinking of a petajoule?
Sorry, yes. PJs, not pJs. ;-) 10^15
BitWiz
#6
Jan12-11, 04:26 PM
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Quote Quote by cepheid View Post
So if I'm on a train pushing a cart down the aisle, the only value for the displacement that is valid in calculating the work I did on the cart is its displacement in my rest frame (i.e. its displacment relative to the train). Its displacement relative to an observer on the side of the tracks is not relevant. Is that the answer? You have to measure the displacement in the rest frame of whatever is applying the force?

EDIT: sorry, wrong quotation initially.
Something like this makes more sense to me, though I don't know how to mesh it with the 450 PJ value derived from a known equation. Wasn't the concept of work originally formulated to generically measure the expected throughput of water pumps in English mines?

Instead of applying effort to overcome a persistent force (gravity), we are using it to overcome inertia. Even a small, nearly instantaneous victory in this regard, given enough time, will allow a mass initially at rest to traverse the universe (theoretically), yet exhibit no additional work (there's no force).

So is the 450 PETAjoule ;-) number inflated by an improper use of frame, or actually what is seen by an observer? Or something else?

Thanks for the replies!
Bit
cepheid
#7
Jan12-11, 06:34 PM
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I have no idea in what context the quantity called work was originally defined. But the way it is defined is the way it is defined...period. It does not need to be redefined differently to suit different situations.

That addresses your first paragraph in your most recent post. As to your second paragraph, yes, getting something moving (even at a very small speed) is enough to have it continue to move at that speed forever (in the absence of external forces). But that's not very useful for space travel, is it? So I'm not sure what you're getting at here. We're not talking about traversing the universe at "any" speed, we're talking about attaining a speed that will allow human travellers to actually reach some sort of destination in their lifetimes (which isn't going to happen, incidentally). That's why accelerating UP TO 0.1c is specifically being considered in your example.

As for your third paragraph: the energy required to attain this speed is calculated very simply by taking the difference between the total (relativistic) energy of a particle travelling at 0.1c, and that particle's rest energy. This difference is equal to the kinetic energy of the particle. So, the author is simply saying that if a particle started at rest and is now moving at 0.1c, its kinetic energy has changed by 450 PJ. SOMETHING had to do that amount of work on it in order to achieve this change in energy (edit: and all observers would agree that its energy had changed by this amount).

This also raises a very important point: we're talking about relativistic speeds here. By that, I mean speeds at which the effects of special relativity start to become important. If we were to believe classical physics, a constant force implies a constant acceleration, which means that there is no upper limit on the speed that can be achieved, given enough time. Special relativity shows that this is not true. A constant force does not imply a constant acceleration.** In fact, the closer the speed gets to c, the larger the force that is required in order to continue to accelerate the particle *at all.* The amount of force required tends towards infinity as the speed tends towards c. So, c can be approached asymptotically, but never reached. Given this information, it should be less surprising that getting anywhere near c starts to require tremendous amounts of energy. I hope that this helps.

**Yes, you heard me right. I did just claim that F = ma is not true according to special relativity, although the full special relativistic relationship between force and acceleration does of course reduce to F = ma in the limit of speeds much less than c, just as one would hope. (If the "new" theory didn't agree with classical physics under the conditions that classical physics had been shown to be experimentally valid, then that would be bad news for that theory).


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