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

Main Question or Discussion Point

I'm asking for clarification, but it's my understanding, that of the thread below, and my college physics book Paul A tipler, that when walking or driving a car, the force of friction from the ground does no work. This makes sense in a car becuase the engine/fuel makes the power, driving the wheels and car forward given adequate friction. Essentially friction helps convert rotational energy to kinetic energy.

Can someone tell me if that is correct? I read alot of other opinions and started to question it. Below seemed to makes sense. Thanks for help

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Dale
Mentor
when walking or driving a car, the force of friction from the ground does no work.
That is correct under the usual “no slip” condition.

That is correct under the usual “no slip” condition.
Thanks Dale! You've been great help. I was getting told on by others the ground did work. To do work something must have energy, which the ground doesn't. I've always understood that essentially the engine pushes the car thru the wheels due to friction. Simple as that. Thanks again!

Dale
Mentor
Simple as that.
Yes, simple as that!

Mister T
Gold Member
I was getting told on by others the ground did work.
Different introductory physics textbooks define work differently. That is likely why you were told different things by different people.

russ_watters
Mentor
I was getting told on by others the ground did work. To do work something must have energy, which the ground doesn't.
The key, to me, is that work is force times distance, and as far as the ground is concerned, the tire contact patch with the ground isn't moving horizontally, it is only moving up and down. It's easier to visualize when talking about a person walking. The ground doesn't know if you are walking forward or just leaning against a wall. It just feels a static force against a static contact patch.

rcgldr
Homework Helper
The question is any work done by the ground in the case of an accelerating car. Define "contact patch" as the part of a tire in contact with the ground, and one that moves with the car (the tread of the tire and the ground "flow" through the contact patch due to rolling motion. The ground exerts a forwards force onto the contact patch of the tires, but since the frame of reference is the ground itself, the ground itself doesn't perform any work. However, the combined effect of the forward force from the ground and the contact patch that moves with the car provides the ability to perform work, increasing the kinetic energy of the car during acceleration. This is offset by the decrease in potential energy of the fuel consumed by the engine, so that the net work performed on the car is zero, since any gain of kinetic energy was at the expense of potential energy.

Assume a closed system of moon and car, there are no external forces. The momentum of the closed system is conserved, and the total kinetic energy of the system increases over time. Using a reference frame that is fixed with respect to the center of mass of the closed system, the car accelerates "forwards", while the moon mostly experiences a very small "backwards" angular acceleration due to friction force. The static friction force times the distance the car moves (with respect to the reference frame) corresponds to the increase in kinetic energy of the car, while the static friction force times the distance the moons surface moves (with respect to the reference frame) corresponds to the increase of kinetic energy of the moon (which will be very small).

I don't see how this differs much from a rocket in space. The expanding force of the spent fuel accelerates the fuel backwards and the rocket forwards. That expanding force is performing work on both the rocket and the fuel. To me, this is similar to the friction force between contact patch and ground performing work on both the car and the moon.

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See post #14 from Doc Al in the following.

The force of friction x the displacement of COM will help you calculate the Kinetic energy gained but it is not from the ground/friction. It is from the engine. Remember, it is only gaining that energy because the engine is powering the wheels, creating a frictional force. SO friction is accelerating the wheel and car, delivering the energy of the engine as kinetic energy. if the car where just sitting still, no engine, the ground has no energy to do any work.

Kinda like the engine is pushing the car and helping it convert the energy via the external friction force. Friction is doing the grunt work.

maybe others can explain more, but that is my understanding

Dale
Mentor
I don't see how this differs much from a rocket in space. The expanding force of the spent fuel accelerates the fuel backwards and the rocket forwards. That expanding force is performing work on both the rocket and the fuel. To me, this is similar to the friction force between contact patch and ground performing work on both the car and the moon.
Indeed, the two scenarios are related and can smoothly transform one into the other. As the mass of the fuel increases to infinity then the acceleration of the “exhaust” goes to zero in the center of momentum frame and the work done on/by the ground goes to zero. That is the limit under discussion previous to your post.

rcgldr
Homework Helper
work done on/by the ground goes to zero.
I never mentioned work performed by the ground. I mentioned the work performed by the Newton third law pair of forces, the force exerted by the tires onto the ground, which coexists with the force exerted by the ground onto the tires. Those third law pair of forces are increasing the kinetic energy of the moon + car closed system in my example. The cars motor (since my example uses an electric motor), is what is supplying the energy to maintain the force as both moon and car are accelerated by the force. Being much more massive, the moon's acceleration is tiny compared to the car's acceleration, but it's non-zero.

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jbriggs444
Homework Helper
2019 Award
Years ago, someone (I think @Doc Al) explained the distinction between "real work" and "center of mass work". With that distinction in mind, much confusion evaporates.

For "real work", what one is considering is just the two interacting surfaces. One is not concerned with the entirety of the Earth or the entirety of the car. One is concerned only with the top surface of the pavement and its interaction with the contact patch on the bottom of the tire. The relevant motion is the motion of the two surfaces.

By contrast, for "center of mass work", one picks out objects of interest. For instance, the car and the Earth. One is concerned with the entire objects. Details of the interface or the internal motions of parts cease to be relevant. The relevant motion is the motion of the centers of masses of the objects.

For "real work" done on the tire by the ground, we focus on the motion of the contact patch. If we assume a frame of reference where the pavement and contact patch are at rest, the contact patch is at rest by definition and the work done is zero. [One could consider other reference frames. But let us not, at least for now]

For "center of mass work" done on the car by the ground, we focus on the motion of the center of mass of the car. Again, using a frame of reference where the pavement and contact patch are at rest, the center of mass of the car is moving and the work done is non-zero.

rcgldr
Homework Helper
For "center of mass work" done on the car by the ground, we focus on the motion of the center of mass of the car. Again, using a frame of reference where the pavement and contact patch are at rest, the center of mass of the car is moving and the work done is non-zero.
"Contact patch" normally refers to the part of the tire in contact with the pavement, and moves with the same direction and velocity of the car. Both tire tread and pavement "flow" through the contact patch. This is different than "point of contact".

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Dale
Mentor
I never mentioned work performed by the ground. I mentioned the work performed by the Newton third law pair of forces
One of those forces in the third law pair gives the work done on the ground and the other gives the work done by the ground.

Being much more massive, the moon's acceleration is tiny compared to the car's acceleration, but it's non-zero.
The rate of momentum transfer does not go to zero, but the rate of energy transfer does. It always is zero when the velocity of the material at the contact patch is zero, and as the mass increases the contact patch remains at rest.

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Years ago, someone (I think @Doc Al) explained the distinction between "real work" and "center of mass work". With that distinction in mind, much confusion evaporates.

For "real work", what one is considering is just the two interacting surfaces. One is not concerned with the entirety of the Earth or the entirety of the car. One is concerned only with the top surface of the pavement and its interaction with the contact patch on the bottom of the tire. The relevant motion is the motion of the two surfaces.

By contrast, for "center of mass work", one picks out objects of interest. For instance, the car and the Earth. One is concerned with the entire objects. Details of the interface or the internal motions of parts cease to be relevant. The relevant motion is the motion of the centers of masses of the objects.

For "real work" done on the tire by the ground, we focus on the motion of the contact patch. If we assume a frame of reference where the pavement and contact patch are at rest, the contact patch is at rest by definition and the work done is zero. [One could consider other reference frames. But let us not, at least for now]

For "center of mass work" done on the car by the ground, we focus on the motion of the center of mass of the car. Again, using a frame of reference where the pavement and contact patch are at rest, the center of mass of the car is moving and the work done is non-zero.
That link i attached above is where doc Al breaks down. My college Tipler Physics book picks on this example, stating friction is what propels the vehicle forward, but does no work, as all the energy gained thru friction is really via the internal energy of the engine.

Doc Al
Mentor
Years ago, someone (I think @Doc Al) explained the distinction between "real work" and "center of mass work". With that distinction in mind, much confusion evaporates.
Yep, I wrote quite a bit about that some years ago. I sometimes used the term "pseudowork" in contrast to "real work". I can dig up some of my rants on the topic, but you guys have nailed it.

Mister T
Gold Member
Yep, I wrote quite a bit about that some years ago. I sometimes used the term "pseudowork" in contrast to "real work". I can dig up some of my rants on the topic, but you guys have nailed it.
More information on the topic can be found in this AJP article written by Arnold Arons:

https://aapt.scitation.org/doi/abs/10.1119/1.19182?journalCode=ajp

Dale
Mentor
it's acceleration is always non-zero, with respect to an inertial frame of reference.
Yes, but even when the acceleration is non-zero the power is zero when the velocity of the material at the contact patch is zero. Acceleration and power are different things.

rcgldr
Homework Helper
Yes, but even when the acceleration is non-zero the power is zero when the velocity of the material at the contact patch is zero. Acceleration and power are different things.
Other than then initial start from rest, the "contact patch" velocity is non-zero. The "contact patch" moves at the same velocity as the car.

If the power output is constant, then the rate of increase of total kinetic energy of moon + car in my example is also constant. The initial state is an issue, as Δt and Δv approach 0 for time t = 0 to time = t + Δt, and velocity v = 0 to v + Δv, the associated power approaches 0. At some point in time, the acceleration is power limited instead of traction limited, and at that point, the situation can become a constant power case.

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Doc Al
Mentor
Other than then initial start from rest, the "contact patch" velocity is non-zero. The "contact patch" moves at the same velocity as the car. Over time, any point on the tire, including any point that is momentarily in contact with the pavement, has an average velocity the same as the car.
The "contact patch" is that part of the tire in momentary contact with the ground. As the car moves, a different piece of the tire becomes the next "contact patch". While the average velocity of any part of the tire must equal that of the car, the instantaneous velocity of the "contact patch" is always zero. That's what matters when considering the work done by the ground on the car.

Dale
Mentor
Other than then initial start from rest, the "contact patch" velocity is non-zero. The "contact patch" moves at the same velocity as the car.
That is why I said “material at the contact patch” and not “contact patch”. When the material at the contact patch has zero velocity the power is zero regardless of the acceleration. This is always the case in the situation under consideration in the OP.

While I agree with @Doc Al’s use of the term I recognized from your previous comments that you understood it differently. So I used unambiguous terminology.

Over time, any point on the tire, including any point that is momentarily in contact with the pavement, has an average velocity the same as the car
Not relevant.

If the power output is constant, then the rate of increase of total kinetic energy of moon + car in my example is also constant.
Agreed, but the rate of KE increase of the system is not the same as the rate that work is done at the contact patch.

I am now getting quite lost in correcting the detailed mistakes of your posts. What is your point now?

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rcgldr
Homework Helper
That if the tire is not slipping, then the ground is directly connected to the engine in the same manner as the tires, and neither ground nor tires are a source of energy, but instead transfer the force converted from the torque of the engine through the drivetrain, with that Newton 3rd law pair of forces I posted above (tires exert force onto ground, ground exerts force onto tires). There have been statements that the ground can't be the source of energy, but likewise, the tires can't be the source of energy either.

There was a prior thread about this that included an analysis of the car engine's power increasing the energy of both the planet (i used a moon in my example) and the vehicle.

Dale
Mentor
There have been statements that the ground can't be the source of energy, but likewise, the tires can't be the source of energy either.
Correct (I don't think anyone claimed otherwise here). The energy comes from the engine's fuel or battery. When the material at the contact patch is at rest then there is no work done either by the tire or the road and therefore no energy transferred from the car to the earth. All of the energy in the fuel goes into kinetic energy of the car and none to the earth.

There was a prior thread about this that included an analysis of the car engine's power increasing the energy of both the planet (i used a moon in my example) and the vehicle.
Yes, there is a range of velocities for the material at the contact patch such that this is true. It is not v=0, which is the case under consideration in this thread.

rcgldr
Homework Helper
v=0, which is the case under consideration in this thread.
The title of this thread is "... accelerating a car", and the OP didn't mention if the initial state was v=0.

Dale
Mentor
The title of this thread is "... accelerating a car", and the OP didn't mention if the initial state was v=0.
From the OP "my college physics book Paul A tipler, that when walking or driving a car, the force of friction from the ground does no work" indicates that the ground is at v=0 throughout the problem.

rcgldr
Homework Helper
From the OP "my college physics book Paul A tipler, that when walking or driving a car, the force of friction from the ground does no work" indicates that the ground is at v=0 throughout the problem.
In the case of a person walking, there's no acceleration over the long term, but since momentum of the planet + person is conserved, and if in the initial state neither planet or person were moving (both at v=0), then once the person reached a steady average speed while walking then v ≠ 0 for either ground or person.

The confusion on my part is the title mentions acceleration, while I assume the referenced text refers to non-accelerating cases.

I found one of the prior threads. One of the posts makes a comparison using a spring between two blocks, and mentions the case where one of the blocks is much more massive than the other. Another post in that prior thread mentions contact point as a virtual point that moves along the ground.