Work done for round trip when force is function of velocity

[Ebby]

Homework Statement

If the force is a function of *velocity*, show that the work for a round trip along a closed path can be different from zero

Relevant Equations

W=Fx etc

I'd have no problem with this sort of problem if the force were a function of position. But here, I'm not sure where to go. Perhaps I'd start with an expression for the work done over an arbitrary distance if the force is given by $g(v)$:$$W = \int_a^b g(v) \, dx$$
Not sure what to do next. How do I integrate this?

 

[jbriggs444]

Homework Statement: If the force is a function of *velocity*, show that the work for a round trip along a closed path can be different from zero
[...]
Not sure what to do next. How do I integrate this?

The question essentially asks for a counter-example. You are free to choose a $g(x)$ that makes your life easy.

So pick a simple $g(x)$ and a simple trajectory.

 

[Ebby]

But it's a $g(v)$...

 

[erobz]

But it's a $g(v)$...

Do you have any ideas for a closed trajectory and an applied force (direction and magnitude)?

 

[jbriggs444]

But it's a $g(v)$...

That is a valid objection. But if $g(x)$ is simple and if it is different for the outbound trip than for the return trip then it should be easy to come up with a $g(v)$ that matches.

 

[wrobel]

use viscous friction

 

[jbriggs444]

use viscous friction

The problem with viscous friction (and a stationary medium) is that you do not get a round trip.

I have in mind some things simpler than viscous friction. One idea is physical with a moving medium and one is simply mathematical. Either should be viable. As is viscous friction with a moving medium.

 

[wrobel]

The problem with viscous friction (and a stationary medium) is that you do not get a round trip.

Just shoot straight up and wait when the bullet comes back

 

[jbriggs444]

Just shoot straight up and white when the bullet comes back

Fair enough. With linear drag, that's just a coordinate transformation away from viscous friction with a moving medium.

 

[Herman Trivilino]

Homework Statement: If the force is a function of *velocity*, show that the work for a round trip along a closed path can be different from zero
Relevant Equations: W=Fx etc

View attachment 329772

I'd have no problem with this sort of problem if the force were a function of position. But here, I'm not sure where to go. Perhaps I'd start with an expression for the work done over an arbitrary distance if the force is given by $g(v)$:$$W = \int_a^b g(v) \, dx$$
Not sure what to do next. How do I integrate this?

Why bother with the integral? Pick a constant force. Then pick a closed path. Most textbooks will have a worked example involving sliding friction.

 

[haruspex]

Why bother with the integral? Pick a constant force. Then pick a closed path. Most textbooks will have a worked example involving sliding friction.

A constant force won't do it. I assume you meant constant magnitude.

 

[Herman Trivilino]

A constant force won't do it. I assume you meant constant magnitude.

I was thinking piecewise constant, but you are correct.

 

[Ebby]

OK so let's say $g(v) = kv = k \frac {dx} {dt}$, then:$$W = \int_a^b k \frac {dx} {dt} \, dx$$
I don't know how to integrate this.

 

[haruspex]

OK so let's say $g(v) = kv = k \frac {dx} {dt}$, then:$$W = \int_a^b k \frac {dx} {dt} \, dx$$
I don't know how to integrate this.

You continue to write scalar equations, but to integrate around a closed loop you will need to think more vectorially.
But if, in scalars, $F=kv$ and the mass is $m$, what equation can you write for the acceleration? Then use $\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}$.

 

[wrobel]

By definition any dissipative force $\boldsymbol F$ satisfies the inequality $(\boldsymbol F,\boldsymbol v)\le 0.$ If in addition we know that for some period of time this inequality is strict (<) then that is enough to conclude that the work done$\ne 0$. No explicit integrations and changes of a variable needed.

 

[Steve4Physics]

Hi @Ebby. Can I chip in, as I think you are over-complicating this.

As already suggested (@jbriggs444 and @erobz), you need only consider a simple case, e.g. moving in a circle, radius R, at constant speed, $v$ (e.g. swimming).

A convenient simple velocity-dependent force is $\vec F = -k\vec v$. Note that (for $k \gt 0$) the directions of $\vec F$ and $\vec v$ are antiparallel. (Think of $\vec F$ as the 'drag'.)

Can you derive an expression for the work done by $\vec F$ going once around? The fact that it is non-zero answers the question.

 

[haruspex]

To add to @Steve4Physics' post #16, note that the path can be anything you like because you can introduce other forces. All that matters is that the work done by the given force is nonzero.
And if it's easier, don't worry about making the path a circle. Out and back in a straight line will do.

 

[Ebby]

Thanks to all. I've taken a little from each reply.

Even though I started off in a scalar way, I was intending to form an equation for work done on the outbound journey and an equation for the inbound one, and then sum them. Would have helped if I'd said so, of course.

I struggled to imagine what the details of the trajectory might be though. I kept thinking that it somehow mattered, although I see now that it doesn't, and that a circular or even simple up/down trajectory would do. I also struggled to pick a simple function for $g(\vec v)$. I should have remembered the general friction formula $\vec F = -(c_1 v + c_2 v^2) \hat i$ - although I don't have to use it in its complete form. I find I often forget loads of stuff I know when confronted by a maths or physics question that says "show" or "prove". These words are intimidating.

I got hung up on the integration because I don't know how to integrate a velocity function over a displacement $dx$, when there's really no need even to perform the integration! I can just show that we're left with two integrals that don't sum to zero.

So I'll choose a simple up/down trajectory from point $A$ to point $B$ and back again, as if I were throwing a rock up into the air.

I should have started off with, say:$$W_{trip} = W_{AB} + W_{BA}$$
Where:$$W_{AB} = \int_A^B \vec F \cdot d \vec r$$
And $\vec F$ is some $g(\vec v)$. If we choose $\vec F = -c \vec v$, then:$$W_{AB} = \int_A^B -cv \, dr$$
Similarly, the work done by friction for the return trip is:$$W_{BA} = \int_B^A cv \, dr$$
So:$$W_{trip} = \int_A^B -cv \, dr + \int_B^A cv \, dr= \int_A^B -cv \, dr + \int_A^B -cv \, dr$$
$$= -2c\int_A^B v \, dr$$
This expression $\neq 0$ assuming $c > 0$ and that the trip length $> 0$, and therefore I have shown that friction (my chosen $g(\vec v))$ is a non-conservative force.