Is there any work done by static friction when accelerating a car?

[jbriggs444]

I'm leaning against a wall, pushing with a force of 1N, for 60 seconds, so that's 60 kg-m/s. How does this tell us anything useful? I'm not moving and didn't specify my mass.

If we focus on you, we know by Newton's third law that the wall has been pushing on you with a force of 1N for those 60 seconds. We assume that you started the 60 seconds at rest. You've told us that you ended the 60 seconds at rest.

We can conclude that there were other forces acting on you whose integral over 60 seconds added up to an average of 1N in the opposite direction that the wall was pushing.

Possibly your feet were pushing the floor in a direction away from the wall and consequently, the floor was pushing you toward the wall.

 

[Dale]

the total energy of the car remains constant, but the potential energy of the car's fuel or batteries is being converted into mechanical kinetic energy

Precisely. This is why the power at the contact patch is zero. No energy change means no power.

so the power output by the engine during the conversion of potential energy into mechanical energy is not zero

Obviously not, but that does not imply that any power is delivered to the car through the contact patch.

the Earth also gains a tiny amount of kinetic energy from the force exerted by the tires time the distance that the ground moves with respect to that same inertial frame of reference

In this problem the work is given to be zero which, as you correctly point out here, implies that the distance is zero. In other words, the Earth is considered to be so massive that it does not move under the force.

Ignoring the very tiny amount of acceleration of the earth, the Earth could be use as an approximately inertial frame of reference, in which case all of the energy converted by the cars engine or motor goes into increasing the kinetic energy of the car.

Yes, you got it!

 

[russ_watters]

If we focus on you, we know by Newton's third law that the wall has been pushing on you with a force of 1N for those 60 seconds. We assume that you started the 60 seconds at rest. You've told us that you ended the 60 seconds at rest.

We can conclude that there were other forces acting on you whose integral over 60 seconds added up to an average of 1N in the opposite direction that the wall was pushing.

Possibly your feet were pushing the floor in a direction away from the wall and consequently, the floor was pushing you toward the wall.

Yes, that's the scenario. My question is, do physicists really use momentum to describe such static forces?

 

[Dale]

I'm leaning against a wall, pushing with a force of 1N, for 60 seconds, so that's 60 kg-m/s. How does this tell us anything useful? I'm not moving and didn't specify my mass.

It is very useful. Since you had momentum flowing in from the wall and yet you didn’t accelerate it tells us that there must have been an opposite momentum flow somewhere else. This principle is used extensively in statics.

 

[russ_watters]

It is very useful. Since you had momentum flowing in from the wall and yet you didn’t accelerate it tells us that there must have been an opposite momentum flow somewhere else. This principle is used extensively in statics.

I honestly don't think I've ever heard the term "momentum flow" before. It never would have occurred to me to use the concept of momentum to describe a static force. I may need some time to digest...

 

[Dale]

I honestly don't think I've ever heard the term "momentum flow" before.

That may be my terminology. I think other sources say “transfer” or “rate of transfer”

 

[russ_watters]

That may be my terminology. I think other sources say “transfer” or “rate of transfer”

Well I googled it and it's definitely a thing. But regardless of the term used to label it, it's not something I would have expected to be useful.

 

[jbriggs444]

Well I googled it and it's definitely a thing. But regardless of the term used to label it, it's not something I would have expected to be useful.

I don't think it's anything more than a different mental bucket to use. A force called a "momentum flow" does the same thing as a force called a "force".

If one defines force using $F=\frac{dp}{dt}$ then the mental model of a flow fits well.

 

[A.T.]

Well I googled it and it's definitely a thing. But regardless of the term used to label it, it's not something I would have expected to be useful.

For example: When you think about forces as transfer (rate) of momentum, it becomes immediately clear how Newton's 3rd Law implies momentum conservation.

 

[rcgldr]

Change in momentum = net force · time. In the case of pushing against the wall, and assuming not slipping on the floor, then the sum of the forces exerted by the wall and floor onto the person is zero, so no net force.

power at the contact patch is zero.

Perhaps this is an issue with terminology. My interpretation is the same as the tire dynamics guys, the "contact patch" can be moving with respect to the ground (even though the tread isn't, since it's moves backwards through the contact patch), and power = force · speed. The contact patch static friction force from the ground times the speed of the contact patch equals the power that accelerates the car. The car's engine (or motor) is the source of that power and is responsible for the Newton 3rd law pair of static friction forces at the contact patch.

internal forces can't change energy

Internal forces can't change a systems linear velocity or energy, but can change angular velocity and energy (angular momentum is conserved).

 

[alkaspeltzar]

In my automobile, all work is being done by the four gerbils who live inside my engine block and who push the pistons through a distance when stimulated by the distributor. Those gerbils (and the similar stopping gerbils in the wheels) do all the "work", positive and negative. The rest of the machine simply redirects the force vector...just like on a bicycle.
And force from the road causes acceleration...why is this confusing?

i like this!

 

[jbriggs444]

power = force · speed

Right. But power times the speed of what, exactly? It it not speed of the contact patch. That speed is not relevant. It can be speed of the center of mass of the car. That speed is relevant. The figure you get for power using that speed will reflect the rate at which the bulk kinetic energy of the car increases due to the force.

The speed of the contact patch is only relevant to the extent that it matches the speed of the car.

 

[A.T.]

The contact patch static friction force from the ground times the speed of the contact patch equals the power that accelerates the car.

This only happens to be true in this special case, where:

speed of the contact patch = speed of the object (center of mass)

In general you cannot compute the power that accelerates the object based on the speed of the contact patch. Consider the Flintstones' car where the contact patch under the foot has a speed of zero, but the power that accelerates the car is not zero.

 

[A.T.]

The speed of the contact patch is only relevant to the extent that it matches the speed of the car.

Exactly, but so does the speed of the rear view mirror. That doesn't make the rear view mirror key to power calculations.

 

[rcgldr]

The speed of the contact patch is only relevant to the extent that it matches the speed of the car.

This only happens to be true in this special case, where:
speed of the contact patch = speed of the object (center of mass)

Which by the usage of the term contact patch as I (and the tire dynamics guys) is always true (ignoring contact patch deformations due to load factors) (update and except as noted by A.T. if the road is not flat, but goes over a hill or through a dip).

In general you cannot compute the power that accelerates the object based on the speed of the contact patch. Consider the Flintstones' car where the contact patch under the foot has a speed of zero, but the power that accelerates the car is not zero.

In the case of the Flintstone's car, the average speed of the driver's feet is the same as the average speed of the car. In the prior thread, a wheel composed of feet for the contact patches was used as an example.

 

[A.T.]

Which by the usage of the term contact patch as I (and the tire dynamics guys) is always true (ignoring contact patch deformations due to load factors).

No it isn't always true. When you go over a hill, the contact patch moves slower than the center of mass.

In the case of the Flintstone's car, the average speed of the driver's feet is the same as the average speed of the car.

Power is an instantaneous quantity that can be computed at any instant, not just on average.

 

[rcgldr]

Which by the usage of the term contact patch as I (and the tire dynamics guys) is always true (ignoring contact patch deformations due to load factors).

No it isn't always true. When you go over a hill, the contact patch moves slower than the center of mass.

True, I was only considering a flat road. I assume that the power would sill be contact patch force times contact patch speed (which as you noted, would be slower than the cars's center of mass speed), since the contact patch is the point of application of the force that is generated by the engine.

Power is an instantaneous quantity that can be computed at any instant, not just on average.

True, but I was trying to average the overall effect of a "foot" powered car.

 

[jbriggs444]

Which by the usage of the term contact patch as I (and the tire dynamics guys) is always true (ignoring contact patch deformations due to load factors) (update and except as noted by A.T. if the road is not flat, but goes over a hill or through a dip).

In the case of the Flintstone's car, the average speed of the driver's feet is the same as the average speed of the car. In the prior thread, a wheel composed of feet for the contact patches was used as an example.

Again, irrelevant. Compute the integral of the work done by Fred's feet using the [erroneous] contact patch formulation...

Fred puts his right foot down and applies force F. The contact patch is stationary. The power delivered to the right foot is $0 \times F = 0$ while the right foot is on the ground.

Fred lifts his right foot. The foot and whatever contact patch you want to ascribe to it moves forward at 2v but the forward force on the contact patch by the road is zero. The power delivered to the right foot is $2v \times 0 = 0$ while the right foot is off the ground.

Similarly, the power delivered to the left foot is zero.

However, since the material at the contact patch in this case is moving at the same speed as the contact patch, this calculation of power does give a correct and relevant result -- the rate of energy transfer from the road to the soles of Fred's feet: Zero.

 

[rcgldr]

I see the points being made here. The static friction force at the contact patch is transmitted by the rolling tire, resulting in a forward force exerted by tire and wheel onto the axle of the tire, which does move at the same speed of the car (ignoring issued like hills and valleys), and the forward force on the axle times the speed of the axle (wrt grond) correlates to the power that accelerates the car and increases it's kinetic energy.

For the Fred Flinstone car, Fred transmits the forward force from the ground to the car, by moving his feet backwards relative to the car, and the forward force times the speed at where Fred exerts the forward force correlates to the power that accelerates the car and increases it's kinetic energy.

Going back to the rolling tire case, for a flat road, the speed of the contact patch matches the speed of the axle on a flat road. If going over a hill, the car rotates in the same direction as the tire, but at a much slower rate, but I suspect that this reduces the force, so that the forward force at the axle times the axle speed will equal the forward force at the contact patch times the contact patch speed.

Part of my issue with all of this, is I see the forward force as originating from the ground, and then transmitted to the car via the tire, wheel, and other components of the car. Maybe using power = contact patch force times contact patch speed is an over simplification, but my thinking is that it produces the same power result as power = forward force on the axle (or other component) times the speed of the axle. The only force external to the car is the forward force from the contact patch.

 

[A.T.]

True, I was only considering a flat road.

On a flat road you can also use the speed of the rear view mirror. The contact patch speed is no more special than that.

I assume that the power would sill be contact patch force times contact patch speed (which as you noted, would be slower than the cars's center of mass speed),

The acceleration power must match the increase in kinetic energy, for which the speed of the center of mass is relevant, not the speed of the contact patch.

Maybe using power = contact patch force times contact patch speed is an over simplification,

It's not a simplification, but an obfuscation.

 

[rcgldr]

Still, the only force external to the car is the contact patch static friction force, and the point of application of that force advances forwards as the car moves forwards over time (and at the same speed if on a flat road). The remaining forward forces that are transmitted from the tire to components of the vehicle are internal to the vehicle, so the argument here is that internal forces are responsible for the power that accelerates a car?

 

[jbriggs444]

Still, the only force external to the car is the contact patch static friction force, and the point of application of that force advances forwards as the car moves forwards over time (and at the same speed if on a flat road). The remaining forward forces that are transmitted from the tire to components of the vehicle are internal to the vehicle, so the argument here is that internal forces are responsible for the power that accelerates a car?

Well yeah. Fred's legs. The car's engine. Significantly, neither is rigid. If you focus in even more tightly, the non-rigidity in the muscles in Fred's legs and in the combustion gasses in the engine cylinders makes it possible for those entities to act as mechanical energy sources.

Flogging the horse a bit more, the motion of the contact patch has nothing to do with anything.

 

[A.T.]

Still, the only force external to the car is the contact patch static friction force, and the point of application of that force advances forwards as the car moves forwards over time (and at the same speed if on a flat road).

That it happens to work out right only in special cases (flat road), should give you the hint that it's a red herring.

 

[A.T.]

Here a simple example to see that for work & power calculations the velocity of the physical material at the point of force application is relevant, not the advance of the geometrical contact point:

A wheel is turned by a motor, while its axis is held in place. If you place the wheel on a board which lies on ice, it will accelerate the board, by doing work on it. But the geometrical contact point is static, so it's obviously irrelevant to the work done.

 

[jbriggs444]

A wheel is turned by a motor, while its axis is held in place. If you place the wheel on a board which lies on ice, it will accelerate the board, by doing work on it. But the geometrical contact point is static, so it's obviously irrelevant to the work done.

Oh nice, I see what you did there. Instead of looking at work done on the car, you looked at work done on the road. And even though the red herring happens to works out nicely for the car (since the car is more or less stationary relative to the point of contact), it works out poorly for the road (since the road is not).

 

[rcgldr]

Here a simple example to see that for work & power calculations the velocity of the physical material at the point of force application is relevant, not the advance of the geometrical contact point

A wheel is turned by a motor, while its axis is held in place. If you place the wheel on a board which lies on ice, it will accelerate the board, by doing work on it. But the geometrical contact point is static, so it's obviously irrelevant to the work done.

In this case, the force has a velocity (other than the initial state where t=0, v=0). It's still my opinion that the work done can be calculated in the normal way for both cases. Where the surface distance of the work done equation is based on the point of application. Either the force is moving or the surface is moving, or in the case of my small moon + car example, both.

$$ \int F(s) \cdot ds $$

 

[Dale]

the "contact patch" can be moving with respect to the ground (even though the tread isn't, since it's moves backwards through the contact patch), and power = force · speed. The contact patch static friction force from the ground times the speed of the contact patch equals the power that accelerates the car.

This just doesn’t work. Yes, the speed of the contact patch has units of speed, so if you multiply it by the force at the contact patch you do get a quantity that has units of power. But what does this quantity mean?

One possibility is that it is the rate of energy transfer across the contact patch, this is the most physically important quantity since it is the one related to the conservation of energy. However, that quantity is 0, so your quantity is the wrong value.

Ok, so maybe your quantity is the power output of the engine. For the original scenario that coincidentally works, but if the tire is spinning in place (peeling out) then the power output can be quite high while your quantity is zero. So it doesn’t work that way either.

Ok, so maybe your quantity represents the increase in KE. Again, for the original scenario this works coincidentally, but consider a runner accelerating. In that case your quantity is zero while the KE is increasing. So it doesn’t work for that either.

I cannot think of any meaning that you can consistently give to your quantity. Yes, it has units of power, but it does not represent the power transferred at the contact patch, nor the power produced by the engine, nor the rate of increase of the KE. As far as I can tell it has no physical significance. It is simply a quantity with units of power.

It's still my opinion that the work done can be calculated in the normal way for both cases. Where the surface distance of the work done equation is based on the point of application.

How many times in how many different ways does this need to be proven wrong before your opinion on the matter will change? If you don’t have an exact number a ballpark estimate would be fine.

Look, I completely understand where you are coming from. The human mind is excellent at generalizing. You have a concept that in this one scenario works pretty well. So it is easy to believe that it is a general approach. It is not. It works for a car on level road with no slipping. It doesn’t work for many non-cars, it doesn’t work for some non-roads, it doesn’t work for hills, it doesn’t work during slipping. Your opinion here is simply an over generalization.

In contrast to the force times the velocity of the contact patch, the force times the velocity of the material at the contact patch has a well defined and consistent physical meaning. It is the power transferred to or from the object through the contact patch. That definition works regardless of if the surface is stationary or moving and regardless of if there is slipping or not. It works for tires or feet, roads or boards, hills or level road, slipping or not.

 

[rcgldr]

In contrast to the force times the velocity of the contact patch, the force times the velocity of the material at the contact patch has a well defined and consistent physical meaning. It is the power transferred to or from the object through the contact patch. That definition works regardless of if the surface is stationary or moving and regardless of if there is slipping or not. It works for tires or feet, roads or boards, hills or level road, slipping or not.

All chassis dynamometers measure driven wheel power using static friction force times speed at the contact patch of the driven tires, and yet there's no relative motion at the contact patch between the moving tires tread and the drum(s). Typically the force is calculated by sensing the angular acceleration of the drum(s), knowing the drum(s) radius and angular inertia. How does this significantly differ from the case where a car is accelerating on a road?

Note that there are no sensors in the straps that hold a car in place. The measurement of power is solely based on the force at the contact patch times the speed.

It wouldn't take much to change the software to calculate force times distance moved by the surface of the drum(s), in order to measure the work performed on the drum(s), and compare it to the increase in angular kinetic energy of the drum(s).

 

[A.T.]

It's still my opinion that the work done can be calculated in the normal way for both cases.

What is the "normal way"? What "both cases"? What is the work done on the board by the wheel in post #75, according to your method based on contact patch advance speed?

 

[jbriggs444]

speed at the contact patch of the driven tires

Bingo. They are using the speed of the material at the contact patch, not the speed of the contact patch.

 

[Dale]

All chassis dynamometers measure driven wheel power using force times speed at the contact patch ... How does this significantly differ from the case where a car is accelerating on a road?

It differs pretty substantially and shows another way that your approach fails.

In this case the speed of the contact patch is 0, so according to your method the power is also 0. This is incorrect.

In contrast, the speed of the material at the contact patch is nonzero as is the force, correctly giving a nonzero power transferred at the contact patch. This correctly indicates that energy is leaving the car since the battery or fuel is being depleted without increase in KE.

So now that you are providing examples where the “velocity of the contact patch” method fails do I correctly understand that you now recognize that it doesn’t work?

 

[rcgldr]

They are using the speed of the material at the contact patch, not the speed of the contact patch.

I thought the issue in prior posts was that the force being used to determine power or work done is a static friction force, where the surface of the tire and the surface of the road travel at the same speed (which could be 0) at the contact patch.

For the dynamometer, the drum surface is moving, for the original case, the car and what I refer to as contact patch are moving. In both cases, you have a static friction force and a speed, which can be multiplied to calculate the power. There have been dynamometer comparisons versus "real world", comparing what the dynamometer shows as power, versus the power calculated from video and/or radar capture of a car's acceleration and speed, where the car's acceleration · car's mass is used to calculate the static friction force. If the dynamometer is accurate (and the conditions similar), the resulting power calculations via the two methods are the same (within reasonable tolerance).

 

[A.T.]

For the dynamometer, the drum surface is moving, for the original case, the car and what I refer to as contact patch are moving.

There is no consistency in your approach. In one case you use the contact material speed, in the other the geometric contact point speed.

Your error, as explained many times, is replacing the car's CoM with contact patch in the original case. It happens to work out numerically in special cases, but has no physical meaning and is pure obfuscation.

 

[jbriggs444]

I thought the issue in prior posts was that the force being used to determine power or work done is a static friction force, where the surface of the tire and the surface of the road travel at the same speed (which could be 0) at the contact patch.

The force being discussed throughout this post has been static friction. It is static because there is no relative motion between tire and road. Both the surface of the tire and the surface of the road move at the same speed. No one has disputed that. There is no confusion on that point.

The problem is with the speed that is used to determine power. Which speed is to be used?

1. The speed of the contact patch?
2a. The speed of the tire material at the contact patch?
2b. Equivalently, the speed of the road material at the contact patch?
3. The speed of the car's center of mass?

It appears that your approach is "whatever speed gives a calculated result that matches the engine power it is supposed to match". This leads you to pick the single incorrect choice from the list: the speed of the contact patch.

All of the other choices can be used to compute results that reflect physically meaningful power measurements.

2a: Rate at which energy is being delivered by road to tire surface.
2b: Rate at which energy is being delivered by tire to road surface.
3: Rate at which bulk kinetic energy of car is increasing due to external force from road.

 

[hutchphd]

I ask @rcgldr to explain the motion of an "inchworm" using this theory.

 

[Dale]

I thought the issue in prior posts was that the force being used to determine power or work done is a static friction force, where the surface of the tire and the surface of the road travel at the same speed (which could be 0) at the contact patch.

Nobody disagrees about the force. The issue is the speed. You want to use the contact patch speed which is incorrect in many cases, including the dynamometer.

Note that with the use of the speed of the material at the contact patch it is possible to handle both static friction and sliding friction.

 

[rcgldr]

The force being discussed throughout this post has been static friction. It is static because there is no relative motion between tire and road. Both the surface of the tire and the surface of the road move at the same speed. No one has disputed that. There is no confusion on that point.

I apparently mistakenly thought that the issue was with a static friction force being able to perform work, since that is the question being asked in the title of this thread: "Is there any work done by static friction when accelerating a car?"

1. The speed of the contact patch?
...
3. The speed of the car's center of mass?

On a flat road, 1 and 3 are the same. For a car on a hill situation, I could change my small moon + car example to use a car on stilts that raises the cars center of mass above the contact patch so that the car's center of mass is located 25% further away from the center of the moon than the contact patch. So what speed (1 or 3) should I use to calculate the power based on static friction force time speed?

 

[Dale]

So what speed (1 or 3) should I use to calculate the power based on static friction force time speed?

IMO, neither. You should use the speed of the material at the contact patch. That gives the correct answer that the static friction force does no work on the car meaning no power is transferred from the ground to the car.

The net force on the whole car times the speed of the center of mass of the whole car can be used to calculate the rate of change of the KE. You cannot attribute that to work done by any force or even by the combination of forces. It has units of power, but it does not represent a transfer of energy (work) to or from the system.

 

[Dale]

Which speed is to be used? ...

2a. The speed of the tire material at the contact patch?
2b. Equivalently, the speed of the road material at the contact patch?

... reflect physically meaningful power measurements.

2a: Rate at which energy is being delivered by road to tire surface.
2b: Rate at which energy is being delivered by tire to road surface.

I am glad that you separated out 2a and 2b. This distinction is particularly useful in cases of slipping where you are using dynamic friction and the speed of the road material is different from the speed of the tire material. In that case the forces are equal and opposite (per Newton's 3rd law) but since the velocities are not the same the power is different. The difference in power is the amount of power that remains at the interface as thermal energy. It is a good exercise to calculate this in different reference frames. The work done by each force depends strongly on the reference frame, but the power dissipated is a (Galilean) invariant.

 

[rcgldr]

You should use the speed of the material at the contact patch. That gives the correct answer that the static friction force does no work on the car meaning no power is transferred from the ground to the car.

So what about chassis dynamometers? The only force they use to calculate power is static friction force (times the speed of the tire/drum(s) surfaces at the contact patch). in this case, the static friction force is performing work on the drum(s), increasing their angular kinetic energy.

 

[Dale]

So what about chassis dynamometers? The only force they use to calculate power is static friction force (times the speed of the tire/drum(s) surfaces at the contact patch). in this case, the static friction force is performing work on the drum(s), increasing their angular kinetic energy.

I already addressed this explicitly above in post 82. The “velocity of the material at the contact patch” correctly calculates the power for a dynamometer. The “contact patch velocity” does not.

 

[rcgldr]

The “velocity of the material at the contact patch” correctly calculates the power for a dynamometer. The “contact patch velocity” does not.

Just to be clear, static friction force can perform work on the drum(s) in a dynamometer, but not on car on a road?

 

[jbriggs444]

Just to be clear, static friction force can perform work on the drum(s) in a dynamometer, but not on car on a road?

Static friction can perform work on a box of car parts in the bed of your pickup truck when you accelerate after a stop.

Static friction from the highway can perform "center of mass" work on a car on the highway.

 

[Dale]

Just to be clear, static friction force can perform work on the drum(s) in a dynamometer, but not on car on a road?

Yes. On the drums the velocity of the material is non-zero so the work is non-zero. On the road the velocity of the material is zero so the work is zero.

 

[rcgldr]

Static friction from the highway can perform "center of mass" work on a car on the highway.

On the road the velocity of the material is zero so the work is zero.

This seems like a conflict, what am I missing here?

The only external force acting on a car on the road is the static friction force from the road, created as part of a Newton 3rd law pair of forces that originates with the car's engine or motor. If that is not the external force that performs work on an accelerating car, then what force is responsible for the increase in mechanical kinetic energy that was converted from potential energy of fuel or battery by the car's engine? Internal forces can't increase linear kinetic energy.

 

[jbriggs444]

This seems like a conflict, what am I missing here?

The need to be clear on what definition of "work" one is using. Which means the need to be clear about which velocity (or displacement) is being used.

 

[Dale]

This seems like a conflict, what am I missing here?

The "center of mass work" is not work since it can occur without any transfer of energy. It is a somewhat common misuse of language. I prefer $\Delta KE$.

 

[Doc Al]

This seems like a conflict, what am I missing here?

"Center of mass" work is just an application of Newton's 2nd law; the actual "real" work done (as in the first law of thermodynamics) is zero. As @jbriggs444 says, be careful what version of "work" you're using. (I would prefer that center of mass "work" not be called work at all.)

The only external force acting on a car on the road is the static friction force from the road, created as part of a Newton 3rd law pair of forces that originates with the car's engine or motor. If that is not the external force that performs work on an accelerating car, then what force is responsible for the increase in mechanical kinetic energy that was converted from potential energy of fuel or battery by the car's engine? Internal forces can't increase linear kinetic energy.

Sure, an external force is needed to convert internal energy into mechanical kinetic energy. Yet no work is done by that force.

 

[jbriggs444]

I agree with @Dale on this. Center of mass work is as much about momentum as about energy. You just apply the SUVAT equations and get something that looks like energy.

Note that when you apply it to the case of a car cresting a hill where the center of mass and the contact patch have significant differences in speed, you will find that the normal force on front and back tires have a significant discrepancy in magnitude and angle. That combines to produce a retarding horizontal force. Fail to account for that force and your measure of center-of-mass work will be off.

That is a case where using contact patch force times contact patch speed leads to a simple computation and a correct answer.

 

[Dale]

The only external force acting on a car on the road is the static friction force from the road, created as part of a Newton 3rd law pair of forces

Yes.

If that is not the external force that performs work on an accelerating car

There is no work on the car (under the usual assumptions). Note, I consistently use work meaning an actual transfer of energy, not fictitious “center of mass work”.

what force is responsible for the increase in mechanical kinetic energy

Forces are the rate of change of momentum, not energy. Power is the rate of change of energy. So this question is wrong. It incorrectly assumes that forces do something that power actually does.

The power that is responsible for the increase in KE is the engine/motor power.

Internal forces can't increase linear kinetic energy.

There is no such thing as linear KE, and internal forces most certainly can increase KE.

Force is the rate of change of momentum. The external force provides the momentum, not the kinetic energy. Both increase but they have different sources.