Work done on accelerating car is zero?
- Context: Graduate
- Thread starter BrainSalad
- Start date
-
- Tags
- Accelerating Car Work Work done Zero
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
83 replies · 13K views
Physics news on Phys.org
OmCheeto
Gold Member
- 2,530
- 3,625
Doc Al said:Right here:
Oh...
Umm...
It never crossed my mind that west coast, internally powered hamster balls:
might be externally powered.
We call those:
elephant balls.
My bad.
OmCheeto
Gold Member
- 2,530
- 3,625
Doc Al said:Just because the energy source is internal does not remove the need for an external force to accelerate.
Hmmm... My boss and I were discussing "mathematical word problems" the other day, and I showed him that internet meme about aliens, hats, and purple. It was inspired by this thread.
If you are Hercules, positioned between Earth and BrainSalads car, and pushed really hard, in equal and opposite directions, what happens?
If you removed gravity, the car would go one way, and the Earth would move the other way.
Mathematically, the resulting momentum of the car and Earth would have the same magnitude, but the kinetic energy would go almost exclusively to the car.
If you're not willing to ignore gravity, then model this as a "Car in Spaaaaaace...", mechanically linked, but free to move, via fancy gears and such, to the midpoint of a million kilometer long, 6E24 kg, roadway. Floor the accelerator pedal for about a second.
Where now is this "external force" that you require?
--------------------------------
ps. Everything I've learned about momentum and impulse, I've learned in the last two days.
BrainSalad
- 53
- 1
OmCheeto said:Where now is this "external force" that you require?
--------------------------------
ps. Everything I've learned about momentum and impulse, I've learned in the last two days.
Taking the car to be the system, the external force is the static friction force of the road on the tires.
If you've really only learned about momentum the past two days, add to your knowledge that "external" and "internal" are relative terms. Any force exerted by an object outside the system in question, arbitrarily defined, is external. Thus, an external force is needed to change momentum (accelerate). Of course, if you change the size of the system, car+Earth for example, a previously external force may now be internal and cause no net change of momentum at all (Earth gains as much negative momentum as the car gains positive).
OmCheeto
Gold Member
- 2,530
- 3,625
BrainSalad said:Taking the car to be the system...
I learned long ago, to ignore everything past the first wrong premise.
You're kidding, right?OmCheeto said:If you're not willing to ignore gravity, then model this as a "Car in Spaaaaaace...", mechanically linked, but free to move, via fancy gears and such, to the midpoint of a million kilometer long, 6E24 kg, roadway. Floor the accelerator pedal for about a second.
Where now is this "external force" that you require?
For the car to accelerate, the road must exert an external force on it. Same for your hamster wheel.
--------------------------------
ps. Everything I've learned about momentum and impulse, I've learned in the last two days.
So it seems.
BrainSalad
- 53
- 1
OmCheeto said:I learned long ago, to ignore everything past the first wrong premise.
What's wrong with the premise? We can't take the whole universe as the object in question.
You don't have any real physics education, do you? Every force is an internal force if you define the system large enough, but that isn't helpful a lot of the time.
- 13,338
- 4,236
That is true in the rest frame of the objects in contact. But when you model the entrie car as one translating block, then there is no rest frame of both objects: ground and car.rcgldr said:I've often read here that static friction does no work.
I know, in reality there is a wheel that has a contact point at rest. But when you do calculations in physics, you do them based on your model, not on reality. And if you apply the F*d definition of work here, then the external force is doing work, because it is not static in this particular coarse model of reality.
- 13,338
- 4,236
DaleSpam said:The correct answer is that the road performs 0 work on the car. Any method that disagrees with that is simply wrong.
Rather than "wrong", I would say inconsistent with the thermodynamic definition of work. I like this definition, and I used per default in post #2 to give a straight forward answer.
But still, I would like to know how/if the F*d definition can be applied when non-rigid bodies are modeled as rigid blocks. Or when such modelling is permissible. Therefore I would like to hear you thoughts on my question from post #55:
A.T. said:So in respect to Newtons 3rd Law force pairs and work we we can say:
If kinetic energy is converted into other energy forms at the interface, then the positive work done by one force can be less than the negative work done by the other force.
What about the opposite situation?
If kinetic energy is generated from other energy forms at the interface, then the positive work done by one force can be more than the negative work done by the other force.
OmCheeto
Gold Member
- 2,530
- 3,625
Yes we can. The universe is the system. The car is a part of the system.BrainSalad said:What's wrong with the premise? We can't take the whole universe as the object in question.
OmCheeto
Gold Member
- 2,530
- 3,625
Doc Al said:I suggest that you reread this thread from the beginning.
I've read it at least 6 times. I will let things digest, and reread it next weekend, as the thread's title still makes me want to poke someone in the eye.
BrainSalad
- 53
- 1
OmCheeto said:Yes we can. The universe is the system. The car is a part of the system.
Newton's laws would be a good place to start, my friend.
- 36,992
- 15,979
I'm still thinking about it. It made me reconsider my stance, and I am still not sure where I will wind up.A.T. said:. Therefore I would like to hear you thoughts on my question from post #55:
OmCheeto
Gold Member
- 2,530
- 3,625
No.Doc Al said:You're kidding, right?
Are you referring to the normal force? A.T. mentioned that in post #2 as I recall.For the car to accelerate, the road must exert an external force on it. Same for your hamster wheel.
I didn't chose my name lightly.So it seems.
- 30,548
- 7,530
This question and the way it is being discussed seems totally crazy to me. If I jump up in the air, before I actually leave the ground, my foot is in contact all the time and there is no movement. Are you bothered that you can't say that the ground is doing work on me - or that I am doing work on the ground? Why do you choose to be arguing about the 'wheel' question rather than the 'jumping' question? They are, in essence, exactly the same in that you need to look at the whole mechanism and not just the contact point for the answer. The only difference is that the contact point on the wheel keeps changing but your foot is on the same spot until you actually leave the ground by jumping.
BrainSalad
- 53
- 1
OmCheeto said:No.Are you referring to the normal force?
No. It's a static friction force in the forward direction that drives the car forward. It could not accelerate from rest without that force, as per Newton's 1st law. A normal force always acts perpendicular to the surface.
Last edited:
catellus
- 3
- 0
You're all making this more complicated than necessary, and oversimplified at the same time.
The OP started with a very simple model of a car as a point mass and a force being applied to it somehow, but then asked a question about a more complex system (one with rotating wheels and a road). That simple model doesn't capture enough about the real world to answer that question.
If you want to model what happens on a car with rotating tires, you need a more complex model. Let's take a point mass car with a single wheel. The important part of the wheel is the part from the stationary contact point to the axle, which forms a lever, rotating about the contact point. The torque of this lever, supplied by the drive train, applies a force to the axle, where we will assume the point mass of the car is located. Now you have a force applied to the car over a distance and hence doing work.
p.s. Yes, the distance involved in the motion of this lever can only be small, but when you consider the rotating wheel, you always have that same lever applying the same force. And I changed "patch" to "point' to keep the lever model simple. And the wheel has inertia if you're looking for a numerical answer. If you want to look at other aspects of the car, expand your model to include them.
Catellus
The OP started with a very simple model of a car as a point mass and a force being applied to it somehow, but then asked a question about a more complex system (one with rotating wheels and a road). That simple model doesn't capture enough about the real world to answer that question.
If you want to model what happens on a car with rotating tires, you need a more complex model. Let's take a point mass car with a single wheel. The important part of the wheel is the part from the stationary contact point to the axle, which forms a lever, rotating about the contact point. The torque of this lever, supplied by the drive train, applies a force to the axle, where we will assume the point mass of the car is located. Now you have a force applied to the car over a distance and hence doing work.
p.s. Yes, the distance involved in the motion of this lever can only be small, but when you consider the rotating wheel, you always have that same lever applying the same force. And I changed "patch" to "point' to keep the lever model simple. And the wheel has inertia if you're looking for a numerical answer. If you want to look at other aspects of the car, expand your model to include them.
Catellus
OmCheeto
Gold Member
- 2,530
- 3,625
BrainSalad said:No. It's a static friction force in the forward direction that drives the car forward. It could not accelerate from rest without that force, as per Newton's 1st law. A normal force always acts perpendicular to the surface.
Thank you for pointing out that both forces ultimately point in the same direction.
I really like this thread.
BrainSalad
- 53
- 1
catellus said:If you want to model what happens on a car with rotating tires, you need a more complex model. Let's take a point mass car with a single wheel. The important part of the wheel is the part from the stationary contact point to the axle, which forms a lever, rotating about the contact point. The torque of this lever, supplied by the drive train, applies a force to the axle, where we will assume the point mass of the car is located. Now you have a force applied to the car over a distance and hence doing work.Catellus
I like this idea, and have thought about it before, but it's wrong. The drive train does just as much positive work on the axle as the axle does negative work on the drive train, so no net work is done (equal, opposite forces exerted over same distance). This is the very reason that only external forces may do work on an object. Since the only external force acting on the car does no work, no work is done on the car. All this means is that no kinetic energy is transferred to the car via outside sources. All its energy comes from the engine.
- 8,947
- 687
As a counter example to this, take the case of a spinning ice staker that pulls their arms inwards. Although the work done pulling the arms inwards is internal, the angular kinetic energy of the spinning ice skater increases.BrainSalad said:This is the very reason that only external forces may do work on an object.
- 36,992
- 15,979
That isn't a counter example. The skater has exchanged internal chemical potential energy for kinetic energy. No transfer of energy has happened and therefore no work.
BrainSalad
- 53
- 1
rcgldr said:As a counter example to this, take the case of a spinning ice staker that pulls their arms inwards. Although the work done pulling the arms inwards is internal, the angular kinetic energy of the spinning ice skater increases.
No work is done on the ice skater. NO internal forces can do work, because they transfer no energy from outside the system.
Similar threads
- amandela
- · Replies 5 ·
- Introductory Physics Homework Help
- Replies
- 5
- alkaspeltzar
- · Replies 163 ·
- Mechanics
- Replies
- 163
High School
Work done in lifting and lowering an object
- rudransh verma
- · Replies 34 ·
- Mechanics
- Replies
- 34
- kuruman
- · Replies 7 ·
- Mechanics
- Replies
- 7
- bbbbb
- · Replies 1 ·
- Introductory Physics Homework Help
- Replies
- 1
- hgandh
- · Replies 6 ·
- Mechanics
- Replies
- 6
- NoahCygnus
- · Replies 6 ·
- Mechanics
- Replies
- 6
- Kwcty6888
- · Replies 4 ·
- Introductory Physics Homework Help
- Replies
- 4
- Bennett Haselton
- · Replies 15 ·
- Mechanics
- Replies
- 15
- bksree
- · Replies 32 ·
- Mechanics
- Replies
- 32