This is obviously not homework.
A force of 33.336 million newtons is pushing a constant mass of 2.9 million kg.
At what point does the velocity top out?
I asked this before and got a wide range of responses ranging from post-padding drivel to well-intentioned and possibly accurate (mostly the former), so I’m rephrasing it.
There isn’t really enough information here.
Presumably, the answer is either:
(3) being a more pedantic way of expressing (2).
If a constant force pushes a mass, it will have a constant acceleration, at least until relativistic effects take over. So the velocity will increase linearly with time (again, until the mass reaches a significant fraction of the speed of light.)
Or are you specifically asking about the relativistic effects? In that case, the velocity will get closer and closer to c, but never quite attain it.
Yup, this guy has got it. I mean I guess I could add on to what you’ve written with some classical mechanics. So F= MA (Yes, I know that’s not technically true but it’s close enough.) Anyway converting that to A = F/M we get an acceleration of
11 M/S
So then we just need the velocity equation for constant acceleration which is
Vfinal = AT + Vinitial
The OP is wondering what’s Vfinal. I can’t tell since we don’t know what the initial velocity is and we don’t know how long the acceleration was applied to the object.
Maybe I’m denser that I realized, and probably phrased the problem badly again, so I’ll try again:
Starting velocity is 0 (so to speak).
The given mass gets a one-time :WHUMP: of the given force.
How fast is it traveling?
Related question:
I guess I’m missing out somewhere badly on the physics here. If the force is being constantly applied (as I originally said), would it continue to accelerate toward c?
Define “a one-time :WHUMP:” 
The force has to act on the object for a finite time. The longer it acts, the more energy is transferred to the object, right?
A force imparts accelerates an object and the object will only accelerate whilst the force is applied, so we need to know how long the force is applied for.
Once relativistic effects come in to play the same force will result in less acceleration the faster the object gets, so an object given a constant acceleration indefintely will asympotically approach c (i.e. it’s speed will get closer to c without ever reaching it).
How fast the object ends up going is going to depend on how long the force is applied. The change in momentum of an object equals the force applied to it times the period of time over which the force is applied. This quantity is called the “impulse” delivered to the object. You can then find the object’s change in velocity by dividing the object’s momentum by its mass to get its velocity; all in all, this gives
(change in velocity) = (force)*(time applied)/(mass)
For example, if your force is applied for one second, you can go through the numbers and find that your mass will be moving at about 11 m/s afterwards. If the force is applied for ten seconds, it’ll be moving at about 110 m/s. And so forth.
Finally, if you have the force “constantly applied”, then the time in the above equation is effectively infinite, which would imply an infinite velocity. This is just because the equation above hasn’t taken relativity into account; really, the object would asymptotically approach the speed of light, getting closer and closer to it but never quite reaching it. Writing down the equation for that is a whole new kettle of fish, though, so I won’t get into that here.
well F=d/dt(mv) so if you integrate F through time f*(t[sub]2[/sub]-t[sub]1[/sub]) = m*(v[sub]2[/sub]-v[sub]1[/sub]).
33.336x10[sup]6[/sup] * small time = 2.9x10[sup]6[/sup]v[sub]2[/sub]
11.5small time = v[sub]2[/sub]
That’s for a WHUMP
Sorry, you still need to express this more carefully.
A ‘one time whump’ is exactly not constantly applied.
In my answer (1) the ‘immediately’ was for a fixed pusher where the force would disappear as soon as the object started moving and the force was no longer in contact with it.
(2) assumes that the pusher moves withe object (e.g. a rocket attached).
(3) is the same as (2) but is pedantic as you would never actually reach c.
Just to add more complexity to this there’s always the question of is friction significant in your example. If the initial “whomp” isn’t great enough to over come static friction then your object isn’t going anywhere. Furthermore even if it is great enough then you have to consider dynamic friction and subtract that from your force to get the net force and use that for acceleration.(Of course I went with the assumption that it was great enough to overcome static and that the net force was what you said.)
OK. Thanks for the responses.
I think I’ll just keep the duck.
That should be proper acceleration.
Oops, slight typo there. That should be 11 M/S^2 of course 
(The acceleration would be 11 meters per second squared.) Really all you need to know is that for every second your force is applied the velocity changes by 11 meters per second.
Below are three graphs showing the velocity of the object after applying the force for some amount of time. Each graph is at a different “zoom” level in time. I’ve taken force here to be the instantaneous change in momentum, and I’ve included relativistic effects.
If a whump is one second, you’ll reach 11.5 m/s (in agreement with notsoheavyd3). Even if you push for ten weeks, relativistic effects are still very small. After about a year, the turnover in velocity starts becoming significant, and the velocity asymptotically approaches the c.
Everything gets a lot simpler if you take a to mean proper acceleration instead of (ordinary, non-relativistic) acceleration. If you do that, then you still have F = ma, and you also have proper velocity = proper acceleration times proper time. In other words, you can just solve everything as if you were still doing Newtonian mechanics, and just (if you need to) convert proper velocity back to ordinary velocity at the end.
To the OP, if you want to describe a sudden WHUMP, then you probably want to specify the impulse of the WHUMP, not the force. A WHUMP is generally a very large (but finite) force acting for a very short (but nonzero) amount of time, but in such a way that it’s difficult to measure the force or the time easily. What you can usually measure easily, though, is the product of those two, which will be a manageable amount.
F=ma will be trivially true in the inertial frame tangential to the accelerated objects worldline, but otherwise should never be used in it’s 3 vector form as the acceleration vector of an object is not always parallel to the force vector that produces the acceleration, meaning m canot be a scalar or a scalar-valued function.
I think you meant coordinate time rather than proper time as:
w ≠ ατ
Except in the limit as proper time and proper acceleration are frame invariant whereas proper velocity is not.
I think in relativity it’salways best to express things in terms of 4-vectors as somethign like proper velocity can cause confusion as unlike proper acceleration and proper time it’s not frame invariant.
In trivial Newtonian Physics: (If I can remember it over X decades)
F=ma Force=Mass x Acceleration
W=Fd Work=Force x distance
“Work” in physics term is apply a force to a mass for a given distance (or time).
Force is ability to accelerate a mass; your question is more like “how hot will 100Watts make my hot water tank?” 100 Watts for how many milliseconds or hours? Time matters.
What is final velocity after applying a force? That’s like saying what is the final velocity if I turn on a magnet, or drop a weight? Time is still important.
the “Force” will accelerate the “Mass” by approximately 11.5 m/s^2 if I recall my SI units.
It’s approximately the same as the force of gravity (9.8m/s^2)
What you really want to do is express the “whump” in Joules.
That’s one way to do it, but it’s better to give a momentum transfer (an impulse), not an energy transfer (a work). This is because if you expend some amount of energy to get something moving, some of the energy you expend will usually end up heating the object instead of accelerating it.
If the object is encountering an instantaneous “force” then what you’re looking for is really momentum (mass x velocity), not technical physics flavored Force, which is (mass x acceleration). Whatever problem you’re trying to solve you’re probably conceiving of incorrectly - either there truly is a force, in which case it acts over a period of time, or there’s just an impact in which case you want to use momentum. I suppose you could derive instantaneous force as some kind of limit, but I can’t imagine it approaching anything other than zero.