Register to reply 
Force vs work 
Share this thread: 
#1
Jan1211, 02:15 PM

PF Gold
P: 64

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 


#2
Jan1211, 02:47 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,552




#3
Jan1211, 02:55 PM

Emeritus
Sci Advisor
PF Gold
P: 5,196

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? 


#4
Jan1211, 03:00 PM

Emeritus
Sci Advisor
PF Gold
P: 5,196

Force vs work
EDIT: sorry, wrong quotation initially. 


#5
Jan1211, 04:02 PM

PF Gold
P: 64




#6
Jan1211, 04:26 PM

PF Gold
P: 64

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 


#7
Jan1211, 06:34 PM

Emeritus
Sci Advisor
PF Gold
P: 5,196

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). 


Register to reply 
Related Discussions  
Work to push a sled up a hill: Friction, inclines,force,work  Introductory Physics Homework  2  
Work and Force  Introductory Physics Homework  1  
Work Done by Force  Introductory Physics Homework  1  
Force + work  Introductory Physics Homework  3  
Work and Force  Introductory Physics Homework  6 