System, potential energy, and nonconservative forces

[fog37]

TL;DR Summary

Systems, potential energy, and nonconservative forces

Hello,
I am trying to get my head around the idea of nonconservative forces doing work and changing the potential energy of a system.
First of all, forces acting on a system can be:

a) internal and conservative
b) internal and nonconservative (friction, pushes, pulls, thrust, etc.)
c) external and conservative
d) external and nonconservative

Potential energy is energy due to the spatial configuration assumed by two or more internal components of the system interacting via a conservative force. An external and conservative force would imply that the system and something outside of the system, in the environment, share potential energy... However, I think that potential energy can only be the energy of the SYSTEM and not between the system and something else. The correct explanation may be that external forces can be nonconservative but we cannot define potential energy in that case.

Example: a system solely composed of a rock. Planet Earth is not included on the system. The rock and Earth exert mutual gravitational and conservative forces on each other. But we cannot talk about the gravitational potential energy of the rock and Earth since Earth is outside of the system. What do you think?

In regards to the general statement $W_{nc} = \Delta KE +\Delta U$, a nonconservative force (either external or internal) can do work that changes the potential energy and the kinetic energy of a system. Assuming $\Delta KE=0$ for an object moving at constant speed, we get $W_{nc} = \Delta U$. We also know that the net work done by conservative forces $W_{c} = - \Delta U$.
How do we reconcile the fact that potential energy $U$ cannot be defined for a nonconservative force? I know that $F_c = \frac {dU} {dx}$ and this cannot be done for a nonconservative force. However, it seems that both conservative and nonconserative forces can do work that changes the potential energy $W_{nc} = \Delta U$ and $W_{c} = - \Delta U$. even if we cannot define a potential energy for a nonconservative force. Conservative forces do work that does not depend on path. However, the equation $W_{nc} = \Delta U$ seems to imply that a nonconservative force doing work $W_{nc}$ would also equal the change in $U$ regardless of the displacement over which the force acts...

I am clearly confused. I tried to re-read old threads but I am still unsure...

Thank you!

 

[kuruman]

The reconciliation is simple. You can always move $\Delta U$ from one side of the equation to the other following the rules of algebra, then change its sign again and call it $W_c$. Here is what I mean.
1. Start with your expresion: $W_{nc}=\Delta(KE)+\Delta U$.
2. Consider motion at constant speed: $W_{nc}=\Delta U$.
3. Move $\Delta U$ to the other side: $W_{nc}-\Delta U=0$.
4. Change $\Delta U$ to $-W_c$: $W_{nc}+W_c=0$.
This last equation says that the work done by the non-conservative forces plus the work done by conservative forces is zero.

Example: A block sliding down an incline at constant speed. The negative work done by friction on the block is balanced by the positive work done by gravity on the block. The fact that one can algebraically write the equation in step 2 above does not mean that one can derive the force of friction from the gravitational potential energy of the block+Earth system. It means that the Joules lost from the system's potential energy appear as heat (non-mechanical energy) at the boundary of the system.

You might wish to read this article; it could clarify your ideas and set them into perspective.

 

[fog37]

The reconciliation is simple. You can always move $\Delta U$ from one side of the equation to the other following the rules of algebra, then change its sign again and call it $W_c$. Here is what I mean.
1. Start with your expresion: $W_{nc}=\Delta(KE)+\Delta U$.
2. Consider motion at constant speed: $W_{nc}=\Delta U$.
3. Move $\Delta U$ to the other side: $W_{nc}-\Delta U=0$.
4. Change $\Delta U$ to $-W_c$: $W_{nc}+W_c=0$.
This last equation says that the work done by the non-conservative forces plus the work done by conservative forces is zero.

Example: A block sliding down an incline at constant speed. The negative work done by friction on the block is balanced by the positive work done by gravity on the block. The fact that one can algebraically write the equation in step 2 above does not mean that one can derive the force of friction from the gravitational potential energy of the block+Earth system. It means that the Joules lost from the system's potential energy appear as heat (non-mechanical energy) at the boundary of the system.

You might wish to read this article; it could clarify your ideas and set them into perspective.

Thank you. I fully understand your answer and I read the article and will re-read it again. Once we define the system and its imaginary boundary, a system is "isolated" and its total energy $E_{sys}$ when
a) there are no external forces
or
b) there are external forces but the net external force is zero
or
c) the net external force is nonzero but its work is zero, i.e. it is perpendicular to the

Also, the work done by conservative forces does not depend on the overall displacement (the path) but only on the starting and final point:
$$\int F_c dx = W_c = - (U_f - U_i ) $$

However, for the case when $\Delta KE=0$, we seem to get the same expression for the net work done by nonconservative forces and regardless of the path over which the nonconservative force does work, that work $W_{nc}$.

$$\int F_{nc} dx = W_{nc} = U_f - U_i $$

 

[ergospherical]

However, I think that potential energy can only be the energy of the SYSTEM and not between the system and something else.

This may not be the best way to think about things. For a different approach thread try https://www.physicsforums.com/threa...nconservative-forces-the-whole-story.1009267/

 

[kuruman]

However, for the case when $\Delta KE=0$, we seem to get the same expression for the net work done by nonconservative forces and regardless of the path over which the nonconservative force does work, that work $W_{nc}$.

$$\int F_{nc} dx = W_{nc} = U_f - U_i $$

Not true. Consider the case I mentioned earlier. A block slides down a rough incline at constant speed. This happens when the force of friction $f_k=\mu_k mg\cos\theta$ is equal and opposite to the the component to the weight along the incline, $f_k=\mu_k mg\cos\theta=mg\sin\theta \implies \mu_k=\tan\theta$. Note that the net force on the block is zero. Now let's say that we allow the block to slide down the incline by distance $s$, then apply an infinitesimally small force to reverse its direction and have it go back up the incline, at constant speed, to its starting point.

On the way down, the work done on the block by gravity is positive, call it $W_g$ and the work done by friction is negative and of equal magnitude, $-W_g$. On the way up, the work done on the block by gravity is negative, $-W_g$ and the work done by friction is still negative and of equal magnitude, $-W_g$. Over the round trip, the work done by gravity and the change in potential energy are zero as expected while the work done by friction is neative and equal to $-2W_g$.

 

[fog37]

Not true. Consider the case I mentioned earlier. A block slides down a rough incline at constant speed. This happens when the force of friction $f_k=\mu_k mg\cos\theta$ is equal and opposite to the the component to the weight along the incline, $f_k=\mu_k mg\cos\theta=mg\sin\theta \implies \mu_k=\tan\theta$. Note that the net force on the block is zero. Now let's say that we allow the block to slide down the incline by distance $s$, then apply an infinitesimally small force to reverse its direction and have it go back up the incline, at constant speed, to its starting point.

On the way down, the work done on the block by gravity is positive, call it $W_g$ and the work done by friction is negative and of equal magnitude, $-W_g$. On the way up, the work done on the block by gravity is negative, $-W_g$ and the work done by friction is still negative and of equal magnitude, $-W_g$. Over the round trip, the work done by gravity and the change in potential energy are zero as expected while the work done by friction is neative and equal to $-2W_g$.

Thank you kuruman for the patience. I follow your example of the inclined plane and it makes sense.

I guess I got confused by:

1) The net work of all the nonconservative forces $W_{nc}=\Delta ME= \Delta (KE+U) = \Delta KE + \Delta U$ where $ME$ is the mechanical energy $ME=KE +U$.

When $KE=0$, then $W_{nc}=\Delta U$ which looks exactly (aside from the - sign) like $W_c = - \Delta U$

In your example the change in $\Delta U$ is clearly different from the work done by friction (dissipative nonconservative force) as it does work down and up the incline. However, $\Delta KE=0$ since the block starts from rest and return to the top of the inclined plane as is again at rest...

Where is the flaw in my thinking?

THANK YOU!

 

[kuruman]

Where is the flaw in my thinking?

Given that $\Delta K=0$, when $\Delta U =0$ it must be true that $W_{nc}=0$. This is true in this particular example. Friction does negative work over on the block during the round trip. However, friction is not the only force acting on the block. Assuming that the block slides freely down the incline, it cannot reverse direction unless another force acts on it. That could be your hand that stops the block and pushes it back up the incline at constant speed. Your hand exerts a non-conservative force on the block that does work such that over the round trip $W_f+W_{hand}=0$.

The point is that you can use $\Delta K+\Delta U=W_{nc}$ for accounting purposes only, i.e. to figure out what's missing if you know everything else. For example, let's say that the block is stopped by the hand over a distance $\Delta x$ before its direction is reversed. Then you can write $\Delta U =-mg\sin\theta~\Delta x$, $\Delta K=-\frac{1}{2}mv^2$ and $W_{nc}=-F_{hand}~\Delta x-\mu_kmg\cos\theta~\Delta x$. You can use this to find the average force exerted over the interval $\Delta x$ that is needed to stop the block, $$\bar F_{hand}=mg\sin\theta+\frac{1}{2}\frac{mv^2}{\Delta x}-\mu_kmg\cos\theta.$$ This equation gives the distance-averaged force exerted by the hand that is needed to stop the block. It does so by accounting for the works that are done by all the other forces acting on the block (conservative and non-conservative) that can be cast in terms of simple algebraic expressions. The force exerted by the hand is not that simple to calculate. The equation does not imply that the force of the hand is (or can be) derived from a potential when $\Delta K=0.$