Undergrad How to Calculate Work for a Time Varying Force?

[starstruck_]

I'm probably misunderstanding something or have confused my self, but while I was studying for calculus (work integration) I realized that we're always dealing with forces that vary with distance like gravity or a spring force, but what if you have an arbitrary force that varies with time? Like the more time that passes, the force increases? Is this a thing? Am I confusing two different things here?

 

[Dale]

I realized that we're always dealing with forces that vary with distance like gravity or a spring force, but what if you have an arbitrary force that varies with time?

Mathematically we can certainly write down such forces. However, such forces do not appear to be part of the world* as they do not conserve energy.

*one possible exception is cosmology, but it is a pretty advanced topic

 

[NFuller]

Like the more time that passes, the force increases? Is this a thing?

Sure, you already mentioned an object moving through a gravitational field. As time passes the object moves, so the force can in principle be written as a function of time.

 

[starstruck_]

Sure, you already mentioned an object moving through a gravitational field. As time passes the object moves, so the force can in principle be written as a function of time.

how would this look as an integral?

 

[NFuller]

how would this look as an integral?

If you wanted to find the work for example it would be
$$W(t)=\int_{0}^{t}\mathbf{F}(t')\cdot d\mathbf{r}(t')=\int_{0}^{t}\mathbf{F}(t')\cdot \mathbf{v}(t')dt'$$

 

[mfb]

Mathematically we can certainly write down such forces. However, such forces do not appear to be part of the world* as they do not conserve energy.

Of course they do. Consider an electron in an electromagnetic wave, for example. It will feel a force that depends on the time.

 

[Dale]

Of course they do. Consider an electron in an electromagnetic wave, for example. It will feel a force that depends on the time.

Even for such a force the Lagrangian does not depend on time. The equations of motion do, but not the Lagrangian.

 

[mfb]

It depends on how you consider the radiation. Often it is much easier to consider it as external force - consider a subsystem only where energy doesn't have to be conserved.

 

[Dale]

It depends on how you consider the radiation. Often it is much easier to consider it as external force - consider a subsystem only where energy doesn't have to be conserved.

Yes, good point. And as you say, in that subsystem energy is not conserved, so you could have an explicitly time-dependent force.

 

[starstruck_]

Ohh, so the difference between the two is that one does not conserve energy and the other does, and for forces varying with distance, your integral is just

∫F⋅dx where the force is a force that varies with distance like the spring force so you're integrating the force with respect to the distance.

And if it was varying with time you would have to integrate both, the force and the distance with respect to time?

((Sorry, this is something I'm forgetting, is it force times distance or force times displacement? If it's displacement that would mean if an object was moved and then brought back to the starting location, no net work was done right? but there would still be total work done? My physics textbook and calculus textbook have two different definitions of it (i'm going with my physics being more accurate)))

 

[PeroK]

how would this look as an integral?

There's a thread here on calculating the time taken for an object to fall, in the case where the initial height is large enough that the gravitational force varies significantly with time as it falls.

Ohh, so the difference between the two is that one does not conserve energy and the other does, and for forces varying with distance, your integral is just

∫F⋅dx where the force is a force that varies with distance like the spring force so you're integrating the force with respect to the distance.

And if it was varying with time you would have to integrate both, the force and the distance with respect to time?

((Sorry, this is something I'm forgetting, is it force times distance or force times displacement? If it's displacement that would mean if an object was moved and then brought back to the starting location, no net work was done right? but there would still be total work done? My physics textbook and calculus textbook have two different definitions of it (i'm going with my physics being more accurate)))

Force is a vector quantity, hence you must consider the displacement (which is a vector). The equation above applies in up to three dimensions:

$W = \int_{C} \vec{F}.\vec{dr} \ \$ where $C$ is the path along which the force acts.

 

[Stephen Tashi]

And if it was varying with time you would have to integrate both, the force and the distance with respect to time?

Before we say "Yes", what exactly do you mean by that?

If F(t) is the force at time t and X(t) is the distance at time t, the work done between time t = t0 and t = t1 is not $\int_{t0}^{t1} F(t)X(t)dt$.

The integral $\int_{t0}^{t1} F(t)X(t)dt$ would have units like (Newton)(meter)(second) instead of (Newton)(meter) = joule.

 

[Jnan]

Let us consider force as a function of time as F(t).
We know,
dP/dt = F(t)
m dv/dt = F(t)
m∫dv = ∫F⁽t⁾dt
m⁽v₂-v₁) = θ
(θ=∫F⁽t⁾dt with limits t₂ and t₁)

Now, for a certain situation the initial velocity⁽v₁) and mass can be easily recorded. Also the time interval for which work needs to be find out (t₁ and t₂) can be chosen accordingly.

Hence from the first relation we can easily calculate the value of v₂.

Now, to calculate the work done by this time varying force we need something more. Think! Think!

Aaahh...It's the Work Energy Theorem. If we consider only a single force F(t) to be acting then-

Change in K.E.= Work done by F(t)

And K.E. can be found easily by plugging the values of v₁ and v₂.

Thanks!
All the best!