Is internal energy really decreasing in situation 2?

[vcsharp2003]

Homework Statement

I am trying to understand the law of conservation of energy from a very general perspective but coming across some issues. I am using the equation mentioned at end of this post, which is true for a system with no heat flows into or out of the system.

In two situations described below,if the applied force and friction force are both non-conservative then using the general equation below, the change in internal energy of block should be negative of the sum of work done by applied force and friction force. This makes sense in situation 1 where internal energy increases but not in situation 2 where I am getting that internal energy of block decreases. I am sure I am making a mistake somewhere but cannot see it.

SITUATION 1 ( without applied force)
A block of mass 1 kg moving at 10 m/s on a rough horizontal surface with a coefficient of friction of 0.25 is brought to rest by friction. We will focus from initial instant when it's speed is 10 m/s to when it comes to rest.

SITUATION 2 ( with an applied force)
A force of 200 N is applied to a block of 1 kg mass on a rough horizontal surface with a coefficient of friction of 0.25. This causes the block to move in the direction of force. We will focus on first 10 m of the block's motion.

Question
Does this reasoning for situation 2 sound correct if it's assumed that no heat flows into and out of the block?

2. Relevant equation

General equation for conservation of energy without heat flows

3. The Attempt at a Solution

SITUATION 1 ( without applied force)
The work done by friction force in this case is .5 x 1 x 10² = 50 J.
If we use the equation given at end of this post then the work done by non-conservative force of friction is converted to internal energy of the block i.e. internal energy change = - (-50) = + 50 J which causes internal energy of block to rise.

SITUATION 2 ( with an applied force)
For the first 10 m, work done by the applied force = + 200 x 10 = + 2000 J and work done by friction force is = - 1 x 10 X .25 = - 2.5 J.
So, net work done by non-conservative forces is 2000 - 2.5 = + 1997.5 J. If the equation given below is true then change in internal energy is - 1997.5 J which means the internal energy of the block has decreased by the end of this 10 m distance.

 

[TSny]

For situation 2 you didn't calculate the work done by friction correctly. (Either you didn't calculate the normal force correctly or you forgot to include the distance.)

Have you taken into account the change in KE for situation 2?

 

[Chestermiller]

The problem is with the equation $\Delta U_{int}=-W_{nonconservative}$. Where did this equation come from all of a sudden? The correct form of the first law for a closed system with Q = 0 is $$\Delta (KE)+\Delta (PE)+\Delta U=-W$$

 

[vcsharp2003]

The problem is with the equation $\Delta U_{int}=-W_{nonconservative}$. Where did this equation come from all of a sudden? The correct form of the first law for a closed system with Q = 0 is $$\Delta (KE)+\Delta (PE)+\Delta U=-W$$

Hi,
I looked up that equation from University of Rochester website at this link under the section titled 8.6. Conservation of energy
: http://teacher.pas.rochester.edu/phy121/LectureNotes/Chapter08/Chapter8.html.

 

[vcsharp2003]

For situation 2 you didn't calculate the work done by friction correctly. (Either you didn't calculate the normal force correctly or you forgot to include the distance.)

Have you taken into account the change in KE for situation 2?

Hi,
Yes, you are correct. I did not calculate situation 2 correctly. Thanks for pointing that out. But even after this correction, it still doesn't make sense.

New calculations are as below.

Friction force = μ x N = μ x mg = 0.25 x 1 x 10 = 2.5 N
Net Force = 200 - 2.5 = 197.5 N
Acceleration of block = F/m = 197.5 /1 = 197.5 m/s².
Let v be the velocity of block after 10 m.
v² = 0 + 2as = 0 + 2 x 197.5 x 10 = 3950
Final KE of block = .5 x m x v² = .5 x 1 x 3950 = 1975 J
So, ΔKE = 1975 J, ΔPE = 0, ΔU_int = - W_{nonconservative} = - (200 x 10 - 2.5 x 10) = -1975 J
So, ΔKE + ΔPE + ΔU_int = 0 should be true. Let's check this out by substituting actual values in situation 2.
1975 + 0 + (-1975) = 0
0 = 0 which is true, so the general law of conservation applies correctly.

But if this equation is true as we have seen, then it means the internal energy of block has decreased by 1975 J. The question is what has happened to this LOST internal energy of the block?

 

[TSny]

You have two non-conservative forces in your situation 2: The applied force of 200 N and the friction force. As in section 8.5 of your link, you can start with the work-energy theorem: W_net = ΔKE. Thus,
W_app + W_fric + W_grav = ΔKE. You don't have any work done by gravity, but I included it for generality.

Note that the change in internal energy is due to the work done by friction: W_fric = -ΔU_int. The work done by gravity can be written as W_grav = -ΔU_g. So, you get

W_app = ΔKE + ΔU_g +ΔU_int

If you take the block, table, and Earth as the "system", then this says that the work done on the system by the applied force equals the change in the total energy of the system.

 

[vcsharp2003]

You have two non-conservative forces in your situation 2: The applied force of 200 N and the friction force. As in section 8.5 of your link, you can start with the work-energy theorem: W_net = ΔKE. Thus,
W_app + W_fric + W_grav = ΔKE. You don't have any work done by gravity, but I included it for generality.

Note that the change in internal energy is due to the work done by friction: W_fric = -ΔU_int. The work done by gravity can be written as W_grav = -ΔU_g. So, you get

W_app = ΔKE + ΔU_g +ΔU_int

If you take the block, table, and Earth as the "system", then this says that the work done on the system by the applied force equals the change in the total energy of the system.

So, it seems that the equation mentioned under section 8.6 is missing an important piece of information even though it's always correct mathematically. It should have been like what you mentioned where internal energy is only affected by friction and not by a non-friction applied force. While the equation under section 8.6 makes mathematical sense, your equation seems to make more conceptual sense.

For the block as a system, if I use your equation then the KE of block increases and so does it's internal energy. I am not getting how to put the conservation law in words for situation 2, so it all makes sense. Or may be it's incorrect to take just block as the system.

 

[Chestermiller]

Hi,
I looked up that equation from University of Rochester website at this link under the section titled 8.6. Conservation of energy
: http://teacher.pas.rochester.edu/phy121/LectureNotes/Chapter08/Chapter8.html.

Let's focus on Situation 1. With all due respect to your reference, let's take a step back to freshman physics. What is your equation for the force balance on the block in Situation 1?

 

[vcsharp2003]

Let's focus on Situation 1. With all due respect to your reference, let's take a step back to freshman physics. What is your equation for the force balance on the block in Situation 1?

Equation of motion in horizontal direction:
The force on block in horizontal direction is friction force = 0.25 x 1 x 10 = 2.5 N = 1 x horizontal acceleration. i.e. 2.5 = 1 x a

Equation of motion in vertical direction:
Also, force in vertical direction on block is (normal reaction from table - weight of block) = 1 X 10 - R = 1 x 0 i.e. R = 10, since there is no acceleration in vertical direction for the block.

 

[Chestermiller]

Equation of motion in horizontal direction:
The force on block in horizontal direction is friction force = 0.25 x 1 x 10 = 2.5 N = 1 x horizontal acceleration. i.e. 2.5 = 1 x a

Equation of motion in vertical direction:
Also, force in vertical direction on block is (normal reaction from table - weight of block) = 1 X 10 - R = 1 x 0 i.e. R = 10, since there is no acceleration in vertical direction for the block.

I was thinking more of $$ma=m\frac{dv}{dt}=-f$$where f is the magnitude of the friction force and "a" is the acceleration of the block in the +x direction. Do you agree with this? If so, what do you get if you multiply both sides of this equation by v = dx/dt, where "v" is the velocity of the block in the +x direction?

 

[vcsharp2003]

I was thinking more of $$ma=m\frac{dv}{dt}=-f$$where f is the magnitude of the friction force and "a" is the acceleration of the block in the +x direction. Do you agree with this? If so, what do you get if you multiply both sides of this equation by v = dx/dt, where "v" is the velocity of the block in the +x direction?

Yes, I agree with the equation. If both sides of the equation are multiplied by v then
$$mv\frac{dv}{dt}=-fv$$

 

[Chestermiller]

Yes, I agree with the equation. If both sides of the equation are multiplied by v then
$$mv\frac{dv}{dt}=-fv$$

So, $$mv\frac{dv}{dt}=-f\frac{dx}{dt}$$If I integrate this from 0 to time t, I get:
$$m\frac{v^2}{2}-m\frac{v_0^2}{2}=-f(x-x_0)$$where $v_0$ is the velocity at $x_0$. What is your physical interpretation of the left- and right sides of this equation?

 

[vcsharp2003]

So, $$mv\frac{dv}{dt}=-f\frac{dx}{dt}$$If I integrate this from 0 to time t, I get:
$$m\frac{v^2}{2}-m\frac{v_0^2}{2}=-f(x-x_0)$$where $v_0$ is the velocity at $x_0$. What is your physical interpretation of the left- and right sides of this equation?

My interpretation is
Change in KE of block = Work done by the block against friction

 

[vcsharp2003]

You have two non-conservative forces in your situation 2: The applied force of 200 N and the friction force. As in section 8.5 of your link, you can start with the work-energy theorem: W_net = ΔKE. Thus,
W_app + W_fric + W_grav = ΔKE. You don't have any work done by gravity, but I included it for generality.

Note that the change in internal energy is due to the work done by friction: W_fric = -ΔU_int. The work done by gravity can be written as W_grav = -ΔU_g. So, you get

W_app = ΔKE + ΔU_g +ΔU_int

If you take the block, table, and Earth as the "system", then this says that the work done on the system by the applied force equals the change in the total energy of the system.

I think the equation mentioned in section 8.6 is only true if there is no applied force but only dissipative forces like friction or some other force that causes dissipation (i.e. conversion to heat by increasing internal energy). So, it seems that the context of equation has not been mentioned clearly by the author. What you showed in your last post is the correct equation when applied forces exist in addition to friction.

 

[vcsharp2003]

The problem is with the equation $\Delta U_{int}=-W_{nonconservative}$. Where did this equation come from all of a sudden? The correct form of the first law for a closed system with Q = 0 is $$\Delta (KE)+\Delta (PE)+\Delta U=-W$$

It appears that the context of the equation under section 8.6 has not been clearly mentioned by the author of the notes under that link. Your equation is correct in when both applied force and friction force exists, but the equation I quoted is only true if friction force is there without any applied force.

Could you please let me know what does W in your equation represent? Is it the net work done by non-conservative forces or by all forces?

 

[Chestermiller]

My interpretation is
Change in KE of block = Work done by the block against friction

Right. So, from mechanics considerations, we've shown that:$$\Delta (KE)=-W$$ If we combine this with the equation:$$\Delta (KE)+\Delta (PE)+\Delta U=-W$$we obtain:$$\Delta U=0$$

 

[Chestermiller]

It appears that the context of the equation under section 8.6 has not been clearly mentioned by the author of the notes under that link. Your equation is correct in when both applied force and friction force exists, but the equation I quoted is only true if friction force is there without any applied force.

Could you please let me know what does W in your equation represent? Is it the net work done by non-conservative forces or by all forces?

My W includes all forces.

 

[Chestermiller]

I haven't yet looked at what TSny has done with you on Situation 2, but, if you like, I can help you work through Situation 2 using the same methodology that we have used in Situation 1. Any interest?

 

[vcsharp2003]

Right. So, from mechanics considerations, we've shown that:$$\Delta (KE)=-W$$ If we combine this with the equation:$$\Delta (KE)+\Delta (PE)+\Delta U=-W$$we obtain:$$\Delta U=0$$

So, you are saying that no change in internal energy occurs in situation 1? I thought friction will cause dissipation which will produce heat in the block i.e. increase of internal energy happens.

 

[vcsharp2003]

I haven't yet looked at what TSny has done with you on Situation 2, but, if you like, I can help you work through Situation 2 using the same methodology that we have used in Situation 1. Any interest?

Sure.

 

[Chestermiller]

So, you are saying that no change in internal energy occurs in situation 1? I thought friction will cause dissipation which will produce heat in the block i.e. increase of internal energy happens.

The dissipation occurs outside the block (at the interface). Since the block is insulated, none of this heat can enter the block to increase its internal energy. So, all the heat has to go to the table and the rest of the surroundings to increase their internal energy. If the block is not insulated, then the heat can distribute itself between the block and the surroundings (but most will go to the surroundings).

 

[TSny]

I feel a need to correct my previous post.

In my previous post, I started with an application of the “work-energy” theorem as applied to the block. But at the end of the post, I said the final equation applies to the block-table-earth system. So, I must have done some “fudging” (i.e., what I did was wrong!).

It turns out that the “work-energy” equation that I wrote for the block needs careful interpretation. Start with the general law that the net external force acting on any system equals the total mass of the system times the acceleration of the center of mass of the system. Taking the system to be the block alone, this gives F_app – f = M a_CM . Taking the applied force F_app and the friction force f to be constant, and letting d_CM be the distance that the CM of the block moves, we have

F_app d_CM – f d_CM = M a_CM d_CM = (1/2)MV_f,CM² = Δ KE_CM.

The next to last inequality follows from simple kinematics. The first term on the left can be interpreted in this case as the work done by the applied force because the point of application of the force and the CM move through the same distance. (But for a deformable object, this would not necessarily be the case.) KE_CM represents the KE due to the translation of the CM.

The second term on the left is often called the “work done by friction”. But it is not the true work done microscopically by the forces that the table exerts on the block at the points of contact between the block and the table. This true work cannot, in general, be calculated directly. So, people who are careful with all of this will call the term (– fd_CM) the “pseudo-work” done by friction to distinguish it from the true work done by friction. This pseudo-work done on the block is negative. Its magnitude does not equal ΔU_int of the block. However, as will be seen below, the positive quantity fd_CM does happen to equal, in this case, the total ΔU_int of the block and the table!

So the equation F_app d_CM – f d_CM = ΔKE_CM is not a true work energy equation, it is just a rewriting of the 2nd law applied to the motion of the center of mass of the block. It is sometimes called the “center of mass” equation, or the COM equation.

Introducing the notation W_app = F_appd_CM and W_{f, pseudo} = -fd_CM, we can write the COM equation as

W_app + W_{f, pseudo} = ΔKE_CM for the block alone.

But, again, this should not be thought of as a true work-energy relation.

The true energy relation is a separate relation representing the law of conservation of energy. It states that the change in the total energy of a system equals the amount of energy transferred to the system.

If we now take the system to be the block and table, energy conservation would state that the work done by the applied 200 N force equals the change in total energy of the block-table system:

W_app = ΔKE + ΔU_int.

For the block-table system, ΔKE is just the change in KE due to the translation of the CM of the block.

Comparing the energy equation with the COM equation, you can see that the magnitude of W_{f, pseudo} in the COM equation for the block must equal ΔU_int for the total block-table system.

Here are a couple of references:
http://www4.ncsu.edu/~basherwo/docs/Pseudowork1983.pdf

http://www4.ncsu.edu/~basherwo/docs/Friction1984.pdf

 

[Chestermiller]

This is followup to TSny's post #22 regarding Situation 2. In his post, TSny addresses the total change in internal energy of the block and surroundings. I would like to address the determination of the change in internal energy of just the block, under the constraint indicated in the OP, to wit, the heat generated by friction at the boundary cannot enter the block.

TSny derived what he called the "center of mass" equation for the block. I call this equation the "mechanical energy balance equation" for the block. The mechanical energy balance equation is obtained by multiplying the equation of motion by the velocity and then integrating with respect to time. So the mechanical energy balance equation becomes: $$\Delta (KE)=m\frac{v^2}{2}-m\frac{v_0^2}{2}=(F-f)[x(t)-x(0)]=-W$$where W is the total work done by the block (i.e., our thermodynamic system) on its surroundings. If we combine this with the equation:$$\Delta (KE)+\Delta (PE)+\Delta U=-W$$we obtain:$$\Delta U=0$$
So here again we find that, if the heat generated at the interface is prevented from entering the block, the change in internal energy of the block is zero.

 

[vcsharp2003]

I feel a need to correct my previous post.

In my previous post, I started with an application of the “work-energy” theorem as applied to the block. But at the end of the post, I said the final equation applies to the block-table-earth system. So, I must have done some “fudging” (i.e., what I did was wrong!).

It turns out that the “work-energy” equation that I wrote for the block needs careful interpretation. Start with the general law that the net external force acting on any system equals the total mass of the system times the acceleration of the center of mass of the system. Taking the system to be the block alone, this gives F_app – f = M a_CM . Taking the applied force F_app and the friction force f to be constant, and letting d_CM be the distance that the CM of the block moves, we have

F_app d_CM – f d_CM = M a_CM d_CM = (1/2)MV_f,CM² = Δ KE_CM.

The next to last inequality follows from simple kinematics. The first term on the left can be interpreted in this case as the work done by the applied force because the point of application of the force and the CM move through the same distance. (But for a deformable object, this would not necessarily be the case.) KE_CM represents the KE due to the translation of the CM.

The second term on the left is often called the “work done by friction”. But it is not the true work done microscopically by the forces that the table exerts on the block at the points of contact between the block and the table. This true work cannot, in general, be calculated directly. So, people who are careful with all of this will call the term (– fd_CM) the “pseudo-work” done by friction to distinguish it from the true work done by friction. This pseudo-work done on the block is negative. Its magnitude does not equal ΔU_int of the block. However, as will be seen below, the positive quantity fd_CM does happen to equal, in this case, the total ΔU_int of the block and the table!

So the equation F_app d_CM – f d_CM = ΔKE_CM is not a true work energy equation, it is just a rewriting of the 2nd law applied to the motion of the center of mass of the block. It is sometimes called the “center of mass” equation, or the COM equation.

Introducing the notation W_app = F_appd_CM and W_{f, pseudo} = -fd_CM, we can write the COM equation as

W_app + W_{f, pseudo} = ΔKE_CM for the block alone.

But, again, this should not be thought of as a true work-energy relation.

The true energy relation is a separate relation representing the law of conservation of energy. It states that the change in the total energy of a system equals the amount of energy transferred to the system.

If we now take the system to be the block and table, energy conservation would state that the work done by the applied 200 N force equals the change in total energy of the block-table system:

W_app = ΔKE + ΔU_int.

For the block-table system, ΔKE is just the change in KE due to the translation of the CM of the block.

Comparing the energy equation with the COM equation, you can see that the magnitude of W_{f, pseudo} in the COM equation for the block must equal ΔU_int for the total block-table system.

Here are a couple of references:
http://www4.ncsu.edu/~basherwo/docs/Pseudowork1983.pdf

http://www4.ncsu.edu/~basherwo/docs/Friction1984.pdf

I think the catch/trick in applying First Law of Thermodynamics to situations is to include within the thermodynamic system all objects experiencing the dissipative force, rather than just one of them else one would end up calculating an inaccurate figure for work done by friction. In this case both table and block are involved with dissipative friction force, so by taking both as a system we eliminate the need to consider work done by friction. Does this sound correct to you? If we take just the block as our system and not a particle, then we will end up with friction work > actual friction work.

Also, it seems the friction work ( < f x d) is actually less than the friction work for a single particle ( f x d) due to surface imperfections along the area of contact. I drew the diagram below to illustrate this point. Due to surface imperfections along area of contact the friction force does work only when friction force acts which is when valley parts of the block's surface touch the table; so actual distance of friction force when determining work is less than distance traveled by the center of mass.

So when applying law of conservation of energy to a block ( not particle) as a system we could say the following:
Truly speaking, in situation 1 all the kinetic energy is not used by friction work, but a part of original KE is transformed to internal energy and the remaining part for doing work against friction.

Imperfections on friction surface

 

[Chestermiller]

I think the catch/trick in applying First Law of Thermodynamics to situations is to include within the thermodynamic system all objects experiencing the dissipative force, rather than just one of them else one would end up calculating an inaccurate figure for work done by friction. In this case both table and block are involved with dissipative friction force, so by taking both as a system we eliminate the need to consider work done by friction. Does this sound correct to you? If we take just the block as our system and not a particle, then we will end up with friction work > actual friction work.

Also, it seems the friction work ( < f x d) is actually less than the friction work for a single particle ( f x d) due to surface imperfections along the area of contact. I drew the diagram below to illustrate this point. Due to surface imperfections along area of contact the friction force does work only when friction force acts which is when valley parts of the block's surface touch the table; so actual distance of friction force when determining work is less than distance traveled by the center of mass.

So when applying law of conservation of energy to a block ( not particle) as a system we could say the following:
Truly speaking, in situation 1 all the kinetic energy is not used by friction work, but a part of original KE is transformed to internal energy and the remaining part for doing work against friction.

Imperfections on friction surface

None of this sounds correct to me.

 

[vcsharp2003]

None of this sounds correct to me.

If the block was not an insulated body then what would be the internal energy change for the block in situation 1?

 

[Chestermiller]

If the block was not an insulated body then what would be the internal energy change for the block in situation 1?

At final thermodynamic equilibrium (when the block and surroundings equilibrated), during the motion, or, at the end of the motion (but before the block and surroundings equilibrated)?

 

[vcsharp2003]

At final thermodynamic equilibrium (when the block and surroundings equilibrated), during the motion, or, at the end of the motion (but before the block and surroundings equilibrated)?

At end of motion when the block just comes to rest.

 

[Chestermiller]

At end of motion when the block just comes to rest.

If I do a thermodynamic 1st law energy balance on the interface (as a system, which has no mass), I get:
$$f\Delta x-Q_B-Q_S=0$$where $Q_B$ is the heat that has flowed into the block and $Q_S$ that has flowed into the surroundings. The split between the two heat flows cannot be determined unless we do a more detailed analysis, which includes transient heat conduction to the block and surroundings, based on the thermal properties of these entities.

 

[vcsharp2003]

If I do a thermodynamic 1st law energy balance on the interface (as a system, which has no mass), I get:
$$f\Delta x-Q_B-Q_S=0$$where $Q_B$ is the heat that has flowed into the block and $Q_S$ that has flowed into the surroundings. The split between the two heat flows cannot be determined unless we do a more detailed analysis, which includes transient heat conduction to the block and surroundings, based on the thermal properties of these entities.

So, internal energy would not change according to the thermodynamics equation you gave. Is that true?

 

[Chestermiller]

So, internal energy would not change according to the thermodynamics equation you gave. Is that true?

Well, since the interface has no mass, it would have no internal energy (or internal energy change)..

 

[vcsharp2003]

You have two non-conservative forces in your situation 2: The applied force of 200 N and the friction force. As in section 8.5 of your link, you can start with the work-energy theorem: W_net = ΔKE. Thus,
W_app + W_fric + W_grav = ΔKE. You don't have any work done by gravity, but I included it for generality.

Note that the change in internal energy is due to the work done by friction: W_fric = -ΔU_int. The work done by gravity can be written as W_grav = -ΔU_g. So, you get

W_app = ΔKE + ΔU_g +ΔU_int

If you take the block, table, and Earth as the "system", then this says that the work done on the system by the applied force equals the change in the total energy of the system.

I just found that the link I pointed to (http://teacher.pas.rochester.edu/phy121/LectureNotes/Chapter08/Chapter8.html#Heading9) stated under section 8.6 that the system being considered should be an isolated system for W_fric = -ΔU_int else W_fric ≠-ΔU_int.

In situations 1 and 2, that would mean we take block + table + Earth as the system when applying first law of thermodynamics and in that case we could say with certainty that W_fric = -ΔU_int for block + table + Earth system. If we take only block then in realistic conditions there will be some heat transfers across the block boundary which would make the system a closed system and not an isolated system, and the same would be also true for block + table system.

 

[Chestermiller]

I just found that the link I pointed to (http://teacher.pas.rochester.edu/phy121/LectureNotes/Chapter08/Chapter8.html#Heading9) stated under section 8.6 that the system being considered should be an isolated system for W_fric = -ΔU_int else W_fric ≠-ΔU_int.

In situations 1 and 2, that would mean we take block + table + Earth as the system when applying first law of thermodynamics and in that case we could say with certainty that W_fric = -ΔU_int for block + table + Earth system. If we take only block then in realistic conditions there will be some heat transfers across the block boundary which would make the system a closed system and not an isolated system, and the same would be also true for block + table system.

This is all exactly the consequence of the frictional heat distribution equation I gave in post #29.