Is there work being done if no displacement occurs? So the definition of work is $W = \vec{F}\cdot\vec{s}$. Say I have a point mass which is being pushed on both sides by equal forces and therefore does not move. Does this mean that no work is being done by any force? It's apparent that there is no net force, but could I calculate the work done by each side to be the work that would have been done absent the other?
For example, assuming our point mass has a mass of 1 kg and would have been moved 1 m in a direction by our 1 N force if an equal and opposite force did not counteract it. Would our force have exerted $1N \cdot 1m = 1J$ of work, or did it not perform any work since our object didn't actually move?
 A: Your definition of work needs a little...work, it only works for forces which are constant along the path the particle moves. The more general definition is:
For any path $\gamma : [a,b] \rightarrow \mathbb{R}^3$ and any force field $\vec{F} : \mathbb{R}^3 \rightarrow \mathbb{R}^3$, the work the force does on a particle moving along $\gamma$ is $$W[\gamma,F] := \int_\gamma \vec{F} \cdot \mathrm{d}\vec{s}$$
Thus, for any particle with trajectory $x(t)$, on which the total force is $\vec{F}$, the work done from time $t_0$ to time $t_1$ is the above integral along $\gamma(t) = x(t)$ with $\gamma(a) = x(t_0)$ and $\gamma(b) = x(t_1)$. If the particle does not move at any time, then the domain of integration is a null set (the image of $\gamma$ will be a single point), and thus the work done is zero. Even if you chose $\vec{F}$ as a partial force, the work of that force is still zero since the domain of integration does not change.
In short: Yes, no work is done because the particles does not move, thus $\vec{s} = 0$.
A: If the displacement of the object is zero, then one can calculate the work done by each individual force, the work done by each force is zero.
Why? Work is not defined in terms of what would have happened to the object in the absence of other forces; it is defined in terms of the motion that actually occurred.
More concretely, if from time $t_a$ to time $t_b$ an object moves along a curve $\vec x(t)$, and if it is acted on by a force $\vec F(t)$, then (regardless of whether $\vec F$ here denotes the net force, or a single force acting on the object, or some other combination), the work done by the force $\vec F$ is defined as follows:
\begin{align}
  W(t_b, t_a) = \int_{t_a}^{t_b} \vec F(t) \cdot \frac{d \vec x}{dt}(t) \,dt.
\end{align}
If the object doesn't move during its trip, then $d\vec x/dt = 0$, and the integral vanishes, so we obtain
\begin{align}
  W(t_b, t_a) = 0.
\end{align}
A: To simply state an answer to each question:


*

*No displacement = no work done


*

*the definition of work is when a force acts on an object to displace it


*Yes, you can calculate the work done by each force. You can calculate the applied force pushing the mass one way, and the applied force pushing the mass the other way. Note that because there is no displacement, the value of the applied forces are equal.


*

*Ex. 1N pushing right, -1N pushing left - this is a negative value because it is pushing in the opposite direction


*Each force applies work to the mass. 1N X 1m = 1J and -1N X 1m = 1J


*

*Because that each force does "work" on the box in the case that the other force is absent

*We say there is no work done/there is no displacement because the work done by each force cancels out: W total = 1J + (-1J) = 0J

*In conclusion, no work is done. We can calculate the work done by each force, but usually that is only used for proof that no work is done. Since the object did not move, no work was done. 


A: You specify a point particle, which means no internal structure, which means it's not deformable.  
If the object didn't move, then why choose 1 m?  Why not 10 m, or 1,000 m ?   No displacement, no work.
A: Let's try to work this down.
$W=F \cdot d$
Now for the sake of simplicity I will say we are dealing only in one direction, namely the x-axis, so
$W=F \cdot x$
It may come in handy to derive the equation for force 
Let  $F=ma=\frac {dp}{dt}$
Let's take this a step further though, since we know that $p=mv=m \frac {dx}{dt}$
So if follows that $F=\frac \partial {\partial t}{m\frac {dx}{dt}}$
Now, remember that $W$ is a dot product, so we can state that $W=Fxcos(\theta)$
We know that the product of any number with zero is zero, so let's analyze this:


*

*If $F$ is zero then $W$ is zero

*If $x$ is zero then $W$ is zero

*If $cos(\theta)$ is zero then $W$ is zero (which occurs at 90 and 270 in terms of degrees and at $\frac {\pi}{2}$ and $\frac {3\pi}{2}$ in terms of radians)


Notice now how the first and second statements are pretty much redundant (because as we saw above $F$ depends on $x$). 
What can we conclude then? Well, we can say that, by definition, if there is no displacement (that is, if $x$ is zero), then no work is being done.
However redundant it may sound, it is also worth noticing how if force is applied perpendicularly to an object (orthogonally really), then no work will be done in a certain direction.
