In calculating work done by a constant force over a constant distance, why doesn't the subject's initial velocity matter? Assume a point-mass $m$ is travelling in a straight line, and a force $F$ will act on $m$ (in the same direction as $m$'s velocity) over a constant distance $d$; why doesn't $m$'s velocity matter to the calculation of work done on $m$ by $F$? Work is defined such that, in this example, the work done by $F$ on $m$ is equal to $Fd$, but it seems that if $m$ were moving slower, it would spend more time in the field, allowing $F$ more time to act on $m$, thereby doing more work. In fact, if $m$'s velocity were very great, it would hardly spend time in $F$'s field at all (so very little work done). Maybe I misunderstand work; can someone address this confusion of mine?
 A: Well, you simply need to accept that work is given by Force time Distance, and it doesn't matter how long it takes.
For example, the work done on a mass $m$ lifted a distance $h$ against gravity with an acceleration $g$ is given by:$$W=F\times h=mgh$$
If you are told that someone is going to drop a $1$ kilogram mass on your head from a height of $10$ metres, you may well have a lot of urgent questions, but how long the evil dropper took to get the weight up there is likely not one of them.
In the case of your example, suppose you have an object with mass $m$ travelling at velocity $v_o$, when a force $F$ is applied for a distance $D$, after which it is travelling at a velocity $v_f$, having experience an acceleration $a$.
The definition of the various constant acceleration equations give us:$$v_f^2=v_o^2+2aD$$ Multiply by $m$, divide by $2$, and we get:$$\frac12 mv_f^2=\frac12 mv_o^2+maD=\frac12 mv_o^2+FD$$The LHS is the final kinetic energy, and the RHS is the initial kinetic energy plus the work done.
A: Well, the reason it doesn't matter is that work is defined as
$$W = \int\vec{F}\cdot\mathrm{d}\vec{s}$$
so if you keep the force the same and the distance the same, this remains the same, regardless of what you do with the initial velocity.
Of course, that definition probably isn't particularly satisfying. So consider this: when an object is subject to a force, the rate at which that force changes its energy is given by the power,
$$P = \vec{F}\cdot\vec{v}$$
And since power is the rate at which the force transfers energy, the total work done will be the integral of power over time,
$$W = \int P\,\mathrm{d}t$$
or $W = P\Delta t$ for constant power.
When an object moves fast, then yes, it spends less time in the region with the force ($t = d/v$), but also that force produces more power $P = Fv$. These two effects cancel out exactly:
$$W = P\Delta t = (Fv)\biggl(\frac{d}{v}\biggr) = Fd$$
So the work done is the same either way.
A: As you note, for a constant force acting on an object which moves in one direction, the work done is equal to $Fd$. One can see from the equation that work is not dependent on time, but only on force and displacement. In order to conceptualize this, you could think about the energy involved in the situation you describe.
When work is applied by an external force to the point mass, the kinetic energy of the mass will change such that $\Delta K=W$. For a slower moving point mass, a force applied for a certain distance will act for a longer time than a force applied to a faster moving particle, as you note. It would seem, as a result, that the effect the force has on the faster mass would be less than the affect had on the slower mass, and in a way this is true. The change in velocity of the faster moving point mass will be smaller than the change in velocity of the slower moving mass, because of the difference in time that the force is applied. However, the change in kinetic energy of both cases will be identical. Because of the work kinetic energy theorem:
$$K_i + W=K_f$$we can say that if the change in kinetic energy for both cases is equivalent, then the work done in both cases must also be equivalent. No matter the amount of time the force is applied the work will be the same. In order to convince yourself of this, make up a problem for yourself where you apply a certain force to a particle over a certain distance, then calculate the resulting kinetic energy and velocity for both a particle moving slowly, and a particle moving quickly. You will find that though the change in velocity depends on the time the force is applied, the change in kinetic energy and the work are independent of this.  
A: As you describe, the definition of work is just: $W=F d$. 
What you are confusing maybe is the rate of work $P$ and the force $F$. When you move fast, $P=Fv$ is larger, however the travelling time is shorter. let's consider we are moving in a constant velocity. Then:
$$W=Pt=Fvt=Fd$$
Independent of velocity.
A: Without any math and considering only Newtonian model here, I would say that 
if you move the inertial system at the same speed and direction as your mass point is moving, than you have no initial movement of the mass point 
and the total force used for acceleration will be the same as if you calculated or measured it in the original inertial system.
A: Work done is also defined as change in kinetic energy of the body.
Since F is constant force so F/m=a is a constant acceleration of m.
So, $$v^2-u^2=2ad$$or$$mv^2/2-mu^2/2=2mad/2$$which is the work done by the force.
The body has travelled d distance with accleration a in the force field assuming u was a constant velocity when it entered the field and v is final velocity when it leaves the field. Even if it had some acceleration say, a', then also$$v^2-u^2=2(a+a')d$$So the right hand side always remains constant because a,a',d and m are constants. So work done is same for all velocities. These things comes from the definition only which is $$Fd$$ So whether body is moving or at rest, if a constant force acts and body moves a distance d in the direction of F work done is$$Fd$$
A: Lots of good answers here, but most of them are pretty mathematical and not very intuitive. Lets consider a realistic example.
You're on the moon with a six shooter and some extra bullets. You are in a uniform field, and you make two point masses travel through the same distance by dropping a bullet with one hand and firing at the lunar surface with the other.
Both bullets speed up as they approach the lunar surface, under the influence of lunar gravity. Since the force is the same and the distance is the same, the work done by the moon's gravity on the two bullets is the same, despite the fact that the fired bullet went through that meter of gravitational field orders of magnitude faster. But despite that, the kinetic energy gained by both bullets is the same and the potential energy lost by both is the same.
Does that make it intuitively any more clear?
