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

  • #151
sophiecentaur
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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'.
 
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  • #152
jbriggs444
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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.
 
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  • #153
rcgldr
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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.
 
  • #154
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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.
 
  • #155
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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.
 
  • #156
jbriggs444
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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.
 
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  • #157
rcgldr
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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.
 
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  • #158
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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.
 
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  • #159
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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.......
 
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  • #160
rcgldr
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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.
 
  • #161
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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.
 
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  • #162
rcgldr
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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.
 
  • #163
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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.
 
  • #164
PeterDonis
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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.
 
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