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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?

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Work is frame dependent. –  Iota Jun 26 at 23:18
    
If a box is on the ground, there is a force on the bottom of the box towards the centre of the Earth due to it's weight, while there is an equal and opposite reaction from the ground in the opposite direction towards the box. In this example as well, forces are acting, but there is no movement and from the observer's point of view, the box is just sitting there. In such a case we would not consider work being done. Your example of two forces pushing on an object in the opposite directions is also the same thing, and we cannot say that work is being done. –  user13267 Jun 27 at 10:03
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4 Answers 4

up vote 6 down vote accepted

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}

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+1 for addressing explicitly that work is not what one would calculate in the absence of other forces. –  BMS Jun 26 at 23:07
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+1 Nice. Couldn't have put it more clearly :) –  Iota Jun 26 at 23:19
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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.

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I don't understand your quibble with the particle being a point. Whether or not it's deformable or really physical doesn't seem relevant to the question. You could leave the whole bit about the object being a point and it would still make sense. –  Brandon Enright Jun 27 at 5:49
    
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. –  Brandon Enright Jun 27 at 5:50
    
The last sentence does provide an answer. –  BMS Jun 27 at 6:44
    
@BrandonEnright If the object were a tennis ball, it might be compressed from both sides, but not displaced. Work would be done with no displacement. Also, you are probably not aware that the OP is a follow-on to a question the OP asked earlier, where deformability and internal structure was an issue. (Which is why my answer is brief.) –  garyp Jun 27 at 11:57
    
@BrandonEnright I take that back! This question was so similar to this post, including the same MathJax errors, that I assumed this was a follow up, but this is by a different author! –  garyp Jun 27 at 12:04
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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 $\vec{s} = 0$.

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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:

  1. If $F$ is zero then $W$ is zero
  2. If $x$ is zero then $W$ is zero
  3. 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.

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