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

[A.T.]

But without stating the "by ... on ..." (as you suggest), it's not clear at all how the resulting sign corresponds to the direction of the energy transfer.

Yes it's not easy but context usually tells you ...

If it's "not easy", requires guessing based on intuition ("context tells you") and only works "usually ", then it's definitely not the right approach for beginners.

 

[A.T.]

Why can't $v$ be the velocity of the point of application of the force? A common defintion for power is $P=F\cdot v$, where $F$ is the force on an object, and $v$ is the velocity of that object. Why can't the object be somewhat "abstract", such as the "contact patches" of the driven tires?

See post #74.

 

[jbriggs444]

Why can't vvv be the velocity of the point of application of the force?

Because, moving the point of application independent of the motion of the body does not transfer energy.

Nor, without a corresponding movement of the body, does it, along with velocity of the body, establish a time frame for the transfer of momentum.

Both versions of the work-energy theorem fail.

As I have pointed out more than once, you can model the dynamics of a car in terms of an opaque, rigid black box with a particular mass, center of mass and moment of inertia. If you apply an external force at such and such a position relative to that black box, the internal mechanical details of the car become irrelevant to the resulting position, velocity, momentum and angular momentum of the box.

In particular, it will turn out not to matter whether the car is accelerated by a wheel embedded in the road that pushes the car forward or a wheel embedded in the car that pushes the ground backward.

It will also turn out not to matter where the actual contact point is! Any point along the same line of action will do just as well. [To be perfectly clear, when I say it does not matter, I am speaking of not mattering for the position, velocity, momentum or angular momentum of the opaque blob that we call the car]

 

[rcgldr]

See post #74.

I see your point. In post #74, the object is the board, so power = force exerted on board · velocity of board. The vehicle is a real object though, so power = force exerted on vehicle · velocity of vehicle. It doesn't matter that the road is not moving, since this doesn't affect the ability for the road to exert a force on a moving vehicle, due to the rolling motion of the driven tires.

 

[A.T.]

so isn't power = ...

Again you fail to define what that power is supposed to represent.

 

[jbriggs444]

The vehicle is a real object though

The distinction between a "real object" such a car and a non-real object such as a board is not clear. How can I look at a board and tell that it is not real. It certainly feels real if it hits me in the shins as I am pushing a car out of snow bank.

 

[rcgldr]

Again you fail to define what that power is supposed to represent.

That $\int_{t0}^{t1} P(t) dt = \Delta KE$

The distinction between a "real object" such a car and a non-real object such as a board is not clear.

I should have clarified this in my prior post. The board is a real object, the spinning wheel accelerating the board is a real object, the vehicle is a real object, the road is a real object, the contact patch is an abstract object, that creates an issue because it's an interface between two real objects.

 

[sophiecentaur]

If it's "not easy", requires guessing based on intuition ("context tells you") and only works "usually ", then it's definitely not the right approach for beginners.

I wonder how many beginners you have dealt with / tried to educate. Many eventually successful Scientists are actually very Right Brained individuals.

 

[Dale]

Why can't vvv be the velocity of the point of application of the force?

Because that doesn’t give the rate of mechanical energy transferred to the system by the force, as we have shown again and again and again and again

A common defintion for power is P=F⋅vP=F⋅vP=F\cdot v, where FFF is the force on an object, and vvv is the velocity of that object.

Note that this works when it is equivalent to the above definition, I.e. when the velocity of the object is the same as the the velocity of the material at the point of application of the force.

Why can't the object be somewhat "abstract", such as the "contact patches" of the driven tires?

Why should it be? You want to know the energy transferred to a material object, so why would you think you could use a non-material velocity? The assumption that you should be able to do that seems baseless, and it demonstrably leads to incorrect values.

 

[Dale]

To be perfectly clear, when I say it does not matter, I am speaking of not mattering for the position, velocity, momentum or angular momentum of the opaque blob that we call the car

But without that extra information you cannot determine the change in the energy of the system.

 

[sophiecentaur]

The dilemma lies in conflating the change in momentum with a change in energy. Momentum and energy are closely related, but they are not the same thing. You can have a force which transfers momentum and does not transfer energy, which is the case with the car.

Well said. Our assumptions about immovable anchors in experiments is well justified, Momentum is still conserved and no Energy is 'lost'.

 

[jbriggs444]

But without that extra information you cannot determine the change in the energy of the system.

Yes, understood. Which is why I was careful not to include "energy" in the list of knowable things, given the more limited information.

 

[rcgldr]

But without that extra information you cannot determine the change in the energy of the system.

The extra information could be determined. Force exerted onto the vehicle = mass of vehicle · acceleration of vehicle. Assuming a flat road, you have a force exerted onto a vehicle and the current velocity of the vehicle. My premise is that the power = force exerted on vehicle times velocity of the vehicle. The acceleration of the vehicle is linear (still assuming a flat road). The angular acceleration of the entire drive train is due to the torque of the engine being slightly greater than the opposing torque related to the force exerted at the contact patch times the effective radius, so the "power" related to the force exerted onto the vehicle is only responsible for the energy change related to the linear acceleration.

 

[Dale]

My premise is that the power = force exerted on vehicle times velocity of the vehicle.

This premise is adequately demonstrated to be false above many many times.

 

[Dale]

Yes, understood. Which is why I was careful not to include "energy" in the list of knowable things, given the more limited information.

Ah, good point. I didn’t pick up on that the first time.

 

[jbriggs444]

The extra information could be determined. Force exerted onto the vehicle = mass of vehicle · acceleration of vehicle.

The force is not the missing information. The velocity of the material at the contact point is the missing information.

 

[rcgldr]

The force is not the missing information. The velocity of the material at the contact point is the missing information.

The velocity of the material at the contact point is zero, but's rolling motion and the tire moves at the same speed as the vehicle. If power was force times velocity of the materials, power would be zero, and over time work would be zero, but the work is known to be non-zero since the vehicle accelerates coexistent with an increase in kinetic energy.

Linear acceleration of vehicle is force / vehicle mass. So it's still my premise that the change in kinetic energy related to the linear acceleration is accounted for by the force exerted onto the vehicle times the velocity of the vehicle.

 

[Dale]

The velocity of the material at the contact point is zero, but's rolling motion and the tire moves at the same speed as the vehicle.

You can’t know that if you are treating the car as a black box.

 

[hutchphd]

So it's still my premise that the change in kinetic energy related to the linear acceleration is accounted for by the force exerted onto the vehicle times the velocity of the vehicle.

What in the world does "accounted for" mean?
Because of the mechanical system everyone agrees that the two numbers are equal.
There is nothing more interesting than that here...

 

[rcgldr]

I posted a question at stack exchange physics forum, initially about the contact patch issue which I later corrected, but the response I received is similar to what I've seen elsewhere:

"The road is, in fact, doing work on the car to propel it forward."

https://physics.stackexchange.com/questions/527495/527502#527502

I'm still looking for the thread about a small planet and car where the mass of the planet wasn't so much greater than the car that the acceleration (mostly angular) of the planet couldn't be ignored in calculating the conversion of potential energy into kinetic energy of the closed system.

 

[Dale]

the response I received is similar to what I've seen elsewhere:

"The road is, in fact, doing work on the car to propel it forward."

The response is incorrect. If the road is doing work on the car then why doesn’t the energy increase?

It is pretty funny that you have scores of correct responses here, but because that one incorrect response is what you wanted to hear you immediately grab onto it. You appear to be immune to any logical arguments and simply set on sticking with your incorrect notion regardless.

 

[rcgldr]

The response is incorrect. If the road is doing work on the car then why doesn’t the energy increase?

The kinetic energy increases due to acceleration, the potential energy decreases due to consumption of the energy source (fuel or battery). If the energy comes from an external source, such as the Sun, then the vehicle experiences a net gain in energy, the solar energy received on the solar panel is eventually converted into kinetic energy of the vehicle, using the road to supply the external force that accelerates the vehicle.

 

[Dale]

The kinetic energy increases due to acceleration, the potential energy decreases due to consumption of the energy source (fuel or battery).

Therefore the rate of change of energy for the car is zero: no power.

the road to supply the external force that accelerates the vehicle.

That is not in question. The road does supply the force that changes the momentum. The energy does not change so the road cannot supply any power.

 

[PeterDonis]

At this point the OP is long gone and the subject has been beaten to death. @Dale's final post sums up the basic points so this seems like a good time to close the thread.