On a specified displacement, will a force do work if velocity is constant and tends to zero while considering a time interval that tends to infinity?

Question

Let's imagine this scenario. An object that can be treated as a particle is already moving at constant speed that tends to zero (but I guess it can be said that it's not exactly zero). There are two forces with equal magnitude but opposite direction acting over it so that its velocity will never change. I know that the net work is equal to zero because velocity doesn't change. However, I want to find what is the work done by each force over the body, so, what would happen if I consider a time interval that tends to infinity? Will each force actually do work? Will the object even move at all? $$W=\int Fdx=\int Fvdt$$

Answer 1

Yes each force will do work, and the work done by each force is independent of the velocity even as it tends to zero.

Perhaps the easiest way to see this is to consider the first version of the definition of work that you present, $W = \int F d x$. The particle velocity does not enter into this expression, so the result of the integration must be independent of the velocity. If we assume that both forces are constant and the particle moves a distance $L$, then we find that $W = FL$ regardless of the velocity.

But how do we reconcile this with the second expression, $W = \int F v d t$, where it looks like the work might go to zero as $v \rightarrow 0$? Here we have to be very careful about exactly what we mean by the velocity going to zero; we have to take limits in the right way. Let us assume that the particle travels the distance $L$ with a velocity $v = \alpha v_0$ where $\alpha$ is a constant and $v_0$ a "reference velocity". We can then increase and decrease the velocity by varying $\alpha$. The total time taken is then $t_0 = L/v = L/(\alpha v_0)$. The work done is $$ W = \int_0^{t_0} F v dt = F \alpha v_0 \int_0^{L/(\alpha v_0)} d t = FL.$$ Note that this result is independent of $\alpha$ (and independent of $v_0$). As we let the velocity go to zero, for example by letting $\alpha \rightarrow 0$, the work done will still be the same because the decrease in $v$ is exactly compensated for by the increase in $t_0$.

Edit: If the forces are not constant, but always equal and opposite to each other, the math gets a little more complicated, but not enough that we can't work it out. In that case we will find the same answer as in the above, given that either

(1) the forces are a function of distance only $F = F(x)$

or

(2) the forces are functions of time $F = F(t)$, but as we decrease the velocity we also slow down the variation of the forces, i.e. when taking $v \rightarrow \alpha v$ we also take $F(t) \rightarrow F(\alpha t)$.

Note that these conditions are sufficient, not necessary, and it may be possible to generalize further.

Answer 2

Writing your equations $W=\int Fdx=\int Fvdt$ in vector notation $W=\int \vec F\cdot d\vec x=\int \vec F\cdot \vec v\,dt$ the first thing one can say is that $\vec v =0$ then no work is done by the force.

Let the object that you are considering be moving at a velocity $\vec v$ and be subjected to two forces $F_{\rm A}$ and $F_{\rm B}$ such that $\vec F_{\rm A}= -\vec F_{\rm B}$.

The net work done on the object is

$\text{work done by force } F_{\rm A} + \text{work done by force } F_{\rm B}$

$= \int \vec F_{\rm A}\cdot \vec v\,dt + \int \vec F_{\rm B}\cdot \vec v\,dt = \int (\vec F_{\rm A}+(-F_{\rm A}))\cdot \vec v\,dt =0$.

Depending on the directions of the forces and the velocity $\vec v$, one of the forces does positive work on the object and the other force does negative work on the object.