Limits of the integral for the calculation of work

Question

To calculate the total displacement for a time dependent velocity, one can start from an infinitesimal displacement and integrate as follows $$dx=vdt$$ $$ \int_{x_i}^{x_f}dx= \int_{t_i}^{t_f}vdt $$ $$ x_f-x_i=\Delta x=\int_{t_i}^{t_f}vdt $$ My question is when we do the same thing for the calculation of work what should be the limits of the integral on the LHS? $$dW=Fdx$$ $$ \int_{?}^{?}dW= \int_{x_i}^{x_f}Fdx $$ Because in the textbook the answer is generally given as $$ W=\int_{x_i}^{x_f}Fdx $$

EDIT: Just found that this question has been asked before.

The accepted answer (by @danimal)states that the limits should be $$ \int_{W_i}^{W_f}dW=\Delta W= \int_{x_i}^{x_f}Fdx $$

The work integral is easier to understand if you think of it as ΔW, that is the change in work done, relative to some zero that you're pretty much free to choose. Energies only really care about differences, not absolutes, so some reference energy is usually chosen as 0.

I know that for potential energy the above statement is correct, one can choose the reference (zero) value as one wishes. But, for example, for kinetic energy the reference (zero) value is pretty much fixed (speed=0). So how can one be sure that work has a free reference value? Does it even make sense to talk about reference value for work?

One of the comments (by @lemon) in the original question states that if force is 0 then work must be 0 therefore the lower limit of the integral must be zero ($W_i=0$). However this statement is not correct. If the force is zero then $dW$ is zero therefore $W$ is constant!

Answer 1

One way to think of an expression like $dW = F dx$ is that it is a physicist shorthand for a differential equation. In this case, the differential equation for $W$ is a first-order ODE $$ \frac{dW}{dx} = F(x) $$ Given an initial condition (which we'll take to be $W(x_i)=0$ for simplicity), this has the solution $$ W(x) = \int_{x_i}^{x} F(x') dx' $$ This is the expression you wrote down, but it avoids needing awkward "limits" on the "work integral", and in my opinion is closer to what is meant rigorously by this kind of expression.

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In a more general context, you might see an expression like $$ dW = \vec{F} \cdot d\vec{x} = F_x d x + F_y dy + F_z dz $$ This kind of expression is implicitly also an ODE, because there is some path being traced by the particle, $\vec{x}(t)$. So $$ dW = \vec{F} \cdot \frac{d\vec{x}}{dt} dt $$ This is itself shorthand for a differential equation $$ \frac{dW}{dt} = \vec{F}\cdot\frac{d\vec{x}}{dt} $$ and the solution is a line integral (again assuming an initial condition $W(t=0)=0$ $$ W(t) = \int_0^t dt' \vec{F} \cdot \frac{d\vec{x}}{dt'} $$ For a conservative force, you can show that the work done is equal to the change in potential energy, and doesn't depend on the path, which further simplifies this problem. However, in general, you need to integrate along a specific path the particle takes.

Answer 2

I think this is one of the cases when we apply a rule beyond its applicable limits.You cannot define boundary conditions to quantities which describe energy transfer because they dont physicaly make sense.

A boundary condition means that something has a start and a end and we want to study that quantity over a closed range .There is no closed range for energy transfer quantities so boundary conditions dont make sense there...

Energy transfer quantities are described better by the area under the curve we integrate.

Answer 3

This response is specifically to this section of your question:

I know that for potential energy the above statement is correct, one can choose the reference (zero) value as one wishes. But, for example, for kinetic energy the reference (zero) value is pretty much fixed (speed=0).

It could be that you haven't taken into account that the notion of kinetic energy falls within the scope of Galilean relativity

The work-energy theorem:

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{1} $$

You can choose any inertial coordinate system: relation (1) holds good.

In that sense: We can say that in the case of linear velocity the we are free to assign any point in velocity space as the zero point of velocity. So that is the counterpart of the freedom to choose the zero point of potential energy.

For sure: it's not immediately obvious that the notion of kinetic energy is within the scope of galilean relativity.

Example: let's say you are riding a bicycle, and you compare:
-the amount of work required to go from zero units of velocity (wrt the road) to 10 units of velocity (wrt the road)
-the amount of work required to go from 10 units of velocity (wrt the road) to 20 units of velocity (wrt the road).

The latter of the two requires three times the amount of work that the former of the two does.

But that's just because when you want to accelerate you have to push off against something. To accelerate you are pushing off against the road, and as you pick up speed the road is receding from you, so pushing of against that is more work.

Answer 4

When you integrate a function, the limits of integration always have the units of the differential. We usually integrate from a starting unit to a generic unit. In that case, you would integrate: $$ \int_{W_0}^W dW = W - W_0 $$ And we usually set $W_0=0$, so we are left with $$ \int_{0}^W dW = W $$ Therefore: $$ \int_{0}^W dW = \int_{x_0}^{x_f} \vec{F} dx\\ $$ $$ W = F(x_f - x_0) $$ $$ W = F\Delta x $$

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Sidenote: Work is the integral of the force along a path. When said path is not a straight line, the differential changes: $$ W= \int_{\vec{r}_0}^{\vec{r}_f} \vec{F} \cdot d\vec{r}\\ $$ This is a line integral, and $$ d\vec{r} = (dx, dy, dz,\dots) $$