Work done by friction when car is breaking

In summary, Morin is discussing the work done by friction. He states that because the ground isn't moving, the friction force does no work. The right side of the equation is also zero, meaning the total energy of the car remains the same.
  • #1
bksree
77
2
Hi all
Please clarify this doubt on work done by friction. The text referred is from the book 'Introduction to classical physics by Morin' - Chap 5, sec 5.1
He is referring to the work-energy theorem : Wexternal = ΔK + ΔV + ΔKinternal
"Consider a car that is braking (not skidding). The friction force from the ground on the tires is what causes the car to slow down. But this force does no work on the car, because the ground isn't moving; the force acts over zero distance. So the external work on the left side (of the above equation) is zero. The right side is therefore also zero. That is the total energy of the car doesn't change. ... ΔK = - ΔKinternal ..."

My doubt is regarding the line But this force does no work on the car, because the ground isn't moving; the force acts over zero distance
The frictional force (rolling friction since the car isn't skidding) by the ground ON the car is F = μkN where N = wt of car. This acts opposite to the direction of motion and isn't work done by friction given by ( F*dist traveled during braking)

Please clarify
TIA
 
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  • #2
bksree said:
The frictional force (rolling friction since the car isn't skidding) by the ground ON the car is F = μkN where N = wt of car.
The friction is static friction (not rolling friction, which is something else), since the tires are not slipping. The instantaneous speed of the tire patch in contact with the ground is zero, thus no work is done by that friction force. Morin is correct.

This acts opposite to the direction of motion and isn't work done by friction given by ( F*dist traveled during braking)
You can certainly calculate F*distance traveled, which will give you the change in translational KE. But the distance is not the distance through which the force acts; it is the distance traveled by the center of mass. So it is not really the work done by that force. Sometimes that quantity is called "center of mass" work or pseudo-work.
 
  • #3
bksree said:
My doubt is regarding the line But this force does no work on the car, because the ground isn't moving; the force acts over zero distance
The frictional force (rolling friction since the car isn't skidding) by the ground ON the car is F = μkN where N = wt of car. This acts opposite to the direction of motion and isn't work done by friction given by ( F*dist traveled during braking)
This situation is a classic example of why the usual F.d definition of work is difficult to apply in some scenarios. If everything is rigid then the distance, d, is clear and unambiguous, but as soon as you are dealing with non-rigid motion d becomes difficult to pin down. Is d the distance traveled during braking or is it 0? Either answer seems reasonable.

A more general and usable definition of work is that "work is the transfer of energy by means other than heat". In this case, it is clear that the work done by friction is 0 since the energy of the car is not changing. All that is happening is that kinetic energy of the car is being converted internally within the car into thermal energy in the brake pads and shoes. Since there is no transfer of energy from the car to the road there is no work.
 
  • #4
DaleSpam said:
Is d the distance traveled during braking or is it 0? Either answer seems reasonable.
Does there have to be an absolute answer to this? It seems to me, that just as work done by some force can depend on the reference frame, it can also depend on the level of detail of your free body diagram:
- If you model the entire car as one rigid body moving linearly, then work is done on the car by the ground in the ground frame.
- If you model the non slipping wheel as a separate object, then no work is done on the wheel by the ground in the ground frame.

DaleSpam said:
A more general and usable definition of work is that "work is the transfer of energy by means other than heat". In this case, it is clear that the work done by friction is 0 since the energy of the car is not changing. All that is happening is that kinetic energy of the car is being converted internally within the car into thermal energy in the brake pads and shoes. Since there is no transfer of energy from the car to the road there is no work.
Consider a block sliding over ground, slowing down due to friction. Usually we call this kinetic friction which does negative work on the block. But we actually don't know (and usually don't care) what exactly is going on at the interface between block and ground. There could be some static forces at the interface, combined with deformation (and thus energy dissipation) within the block, just like the brakes dissipate energy within the car. In the case of the block we almost always abstract the complex interaction away, and express it in terms of a friction coefficient. In the case of a car we can too abstract away the interface (wheels, brake pads, etc), if we are not interested in those details. Then the force of the ground becomes kinetic friction, which does negative work in the ground frame.
 
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  • #5
It doesn't matter that the ground doesn't move. The point of application of force does move with respect to the ground, so you have a decelerating force exerted by the pavement onto the tires over the distance it takes for the car to stop, and that equals the work done.
 
  • #6
A.T. said:
In the case of a car we can too abstract away the interface (wheels, brake pads, etc), if we are not interested in those details. Then the force of the ground becomes kinetic friction, which does negative work in the ground frame.
rcgldr said:
The point of application of force does move with respect to the ground, so you have a decelerating force exerted by the pavement onto the tires over the distance it takes for the car to stop, and that equals the work done.
Except that the car's energy doesn't change. So if there is work done and no change in energy then you have violated the conservation of energy and can make a PMM. For conservation of energy, the work must be 0.
 
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  • #7
rcgldr said:
It doesn't matter that the ground doesn't move. The point of application of force does move with respect to the ground, so you have a decelerating force exerted by the pavement onto the tires over the distance it takes for the car to stop, and that equals the work done.

The ground does move... Not much, but if we're to conserve momentum of the car+earth system as the car slows and stops relative to the earth, the Earth has to acquire some small velocity in the direction of the car's travel.

A substantial fraction of the work done happens not at the tire/road contact patch, but at the boundary between brake pads and rotors. There the frictional force is quite high, and it acts across many meters every second as the rotor moves under the pads.
 
  • #8
DaleSpam said:
Except that the car's energy doesn't change. So if there is work done and no change in energy then you have violated the conservation of energy and can make a PMM. For conservation of energy, the work must be 0.

We end up with car's kinetic energy going to zero while the brakes become hot (A few good laps on a racetrack and the rotors will glow red hot - even a drive to the grocery store will get them to where they will spit and sizzle if you splash water on them) and a tiny change in the Earth's kinetic energy as we transfer momentum from the decelerating car to the earth.
 
  • #9
DaleSpam said:
Except that the car's energy doesn't change.
The car's energy changes, but the car is part of a larger system (earth and car), as mentioned by Nugatory.

Nugatory said:
The ground does move... Not much, but if we're to conserve momentum of the car+earth system as the car slows and stops relative to the earth, the Earth has to acquire some small velocity in the direction of the car's travel.

A substantial fraction of the work done happens not at the tire/road contact patch, but at the boundary between brake pads and rotors. There the frictional force is quite high, and it acts across many meters every second as the rotor moves under the pads.
The kinetic friction converts some of the mechanical energy into heat, so that the total mechanical energy of the Earth + car is reduced by the losses.

If the car used regenerative braking charging a secondary battery, and assuming an ideal case where no energy is lost in the conversion, then the total energy of Earth + car + battery potential energy remains constant.
 
  • #10
Nugatory said:
We end up with car's kinetic energy going to zero while the brakes become hot
Yes, exactly.

Nugatory said:
and a tiny change in the Earth's kinetic energy as we transfer momentum from the decelerating car to the earth.
This tiny change, while present, is irrelevant to the question in the OP. We are likewise ignoring energy lost to air resistance, rolling resistance, acoustic emissions, and thermal emissions, etc. all of which are very small compared to the work supposedly being done.
 
  • #11
rcgldr said:
If the car used regenerative braking charging a secondary battery, and assuming an ideal case where no energy is lost in the conversion, then the total energy of Earth + car + battery potential energy remains constant.
This is a very good example. Using conservation of momentum and conservation of energy you can calculate the change in energy for each. The work done by the tires on the road is equal to the energy transferred to the earth. You can compare that to your suggestion in post 5.

With respect to the road there is no difference between regenerative braking and normal braking. The only difference is that the cars internal energy distribution is shifted between KE and chemical PE in one case and between KE and thermal energy in the other.
 
  • #12
rcgldr said:
It doesn't matter that the ground doesn't move. The point of application of force does move with respect to the ground, so you have a decelerating force exerted by the pavement onto the tires over the distance it takes for the car to stop, and that equals the work done.
The surfaces exerting the forces against each other--the ground and the contact patch of the tire--have zero relative velocity. The force of friction does no work on the car. (Similar to jumping in the air. The ground pushes you, but does no work.)

As I said before, you can certainly calculate Fnet*Δxcm to determine the change in KE. But that's not work.
 
  • #13
Getting back to the original post:

bksree said:
... the total energy of the car doesn't change.
ΔK = - ΔKinternal ...

If the car had lossless regenerative braking charging a battery, then the total energy of the car doesn't change (the mechanical energy is converted into electrical potential energy).

If the car is using regular brakes, then mechanical energy is being converted into heat, and I would consider heat as a loss instead of part of the car's internal energy.

ΔK = - ΔKheat
 
  • #14
rcgldr said:
Getting back to the original post:
If the car had lossless regenerative braking charging a battery, then the total energy of the car doesn't change (the mechanical energy is converted into electrical potential energy).

If the car is using regular brakes, then mechanical energy is being converted into heat, and I would consider heat as a loss instead of part of the car's internal energy.

ΔK = - ΔKheat

The heat is still internal to the car though - it goes into the brake disks, which are definitely part of the car. The energy is then lost as the disks cool off, so the mechanism of energy transfer is the brake disks heating the air. The act of braking does not change the car's internal energy, it's the post-braking brake cooling (that's an awkward statement if I ever typed on...) that causes the car's energy to change. If your brake disks were perfectly insulated from their surroundings, they would contain the full energy that the car had prior to stopping, and you could even possibly extract some of it with a heat engine (though that's getting a bit rube-goldbergish).
 
  • #15
Ignoring the issue of whether heat is an internal energy or a loss, the main point of the original post is correct, static friction doesn't perform work. The work is being performed by something else that is responsible for the static friction force.

As Doc Al mentioned, I was confusing the ability to calculate the change in kinetic energy as friction force times distance to brake as work, but static friction isn't the source of this work, it's the brakes.
 
  • #16
DaleSpam said:
This tiny change, while present, is irrelevant to the question in the OP.

I could have been more clear about why I mentioned that particular tiny effect...
The transfer of this tiny amount of kinetic energy is what reduces to zero the car's momentum relative to the Earth - the pad-rotor friction inside the brakes cannot affect the car's linear momentum. So despite its small size, it's essential to understanding the dynamics of the situation; if it were zero we'd see very different behavior.

The car does slow, and momentum and kinetic energy are transferred from the car to the earth, so some force must be acting between the road and the tire contact patch. It's not dynamic friction (we're all agreed about rolling without slipping here).

Imagine a block resting on a frictionless surface; I put my hand flat on the top of the block and move it around, thereby doing work on the block. The ##W=Fd## calculation works just fine here, but the force cannot exceed that allowed by the coefficient of static friction (if that were zero my hand would slide around the top of the block without doing any work at all). The car interacts with the Earth in the same way, although the value of ##d## is very small.

I got to say that I completely agree with you when you said:
This situation is a classic example of why the usual F.d definition of work is difficult to apply in some scenarios...A more general and usable definition of work is that "work is the transfer of energy by means other than heat".
Such a transfer does happen or the car wouldn't stop, it is mediated through forces between the tire contact patch and the road surface, and I find Morin's description in terms of frictional forces to be less than lucid.
 
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  • #17
rcgldr said:
If the car is using regular brakes, then mechanical energy is being converted into heat, and I would consider heat as a loss instead of part of the car's internal energy.
Thermal energy is part of a system's energy. You cannot just neglect it nor exclude it. The entire field of thermodynamics requires it. What you can do is to define the brakes to be a separate system from the rest of the car, but you cannot just wish away the thermal energy.

If you consider the brakes to be a separate system from the rest of the car then there is a transfer of energy from the car to the brakes resulting in a decrease in the car's kinetic energy and an increase of the brakes thermal energy. Thus the car does work on the brakes.

There is no significant transfer of energy between the road and the car or brakes, and therefore no work done by or on the road.
 
  • #18
bksree: All this is not at all obvious when first exposed...Just keep in mind kinetic [sliding] friction, rolling friction, and static friction are all a bit different. Which one to utilize follows from the problem description.
 
  • #19
rcgldr said:
static friction doesn't perform work
Well, static friction doesn't perform work in this case, but it can perform work in other cases. For example, consider a box on the back of an accelerating truck. Energy is being transferred to the box so the static friction force is doing work in that case.
 
  • #20
Nugatory said:
The transfer of this tiny amount of kinetic energy is what reduces to zero the car's momentum relative to the Earth
There always exists a frame where the change in the Earth's energy is 0 despite the fact that the car comes to rest relative to the earth. This is the frame where the Earth is at rest halfway through the braking. So the change of energy is not in fact what changes the momentum.

Nugatory said:
So despite its small size, it's essential to understanding the dynamics of the situation; if it were zero we'd see very different behavior
In the usual reference frame I see its only value as being for the conservation of momentum, which is not of interest here. For energy questions it only becomes "essential" in other reference frames (in my opinion).

EDIT: I should mention that in those other reference frames it can be quite large, which is why it becomes essential in those frames.
 
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  • #21
DaleSpam said:
Well, static friction doesn't perform work in this case, but it can perform work in other cases. For example, consider a box on the back of an accelerating truck. Energy is being transferred to the box so the static friction force is doing work in that case.
In this case, wouldn't the source of energy be whatever is accelerating the truck? Unlike the car on the road situation, for the box on the truck, there's no relative movment between the box and truck.

Complicating all of this is the issue that static force times distance the static force is applied (relative to some frame of reference) can be used to calculate work done on an object, even though static friction itself doesn't perform work, and instead just transfers a force between objects.
 
  • #22
rcgldr said:
In this case, wouldn't the source of energy be whatever is accelerating the truck?
Certainly. However, remember that work is a transfer of energy, so the static friction is what does the work on the box since that is what transfers the energy from the energy source (e.g. the fuel) to the box. Similarly you can say that a rope does work on an object, even though the rope is not a source of energy but merely the mechanism to transfer the energy.
 
  • #23
Thanks to all for the replies. What I have understood is that since there is no relative motion between the contact points on the tyre and ground (during the period the frictional force acts), the work done is zero.

However I have some more doubts :
Doc Al said:
The friction is static friction (not rolling friction, which is something else), since the tires are not slipping. The instantaneous speed of the tire patch in contact with the ground is zero, thus no work is done by that friction force. Morin is correct.
You can certainly calculate F*distance traveled, which will give you the change in translational KE. But the distance is not the distance through which the force acts; it is the distance traveled by the center of mass. So it is not really the work done by that force. Sometimes that quantity is called "center of mass" work or pseudo-work.

Spose you have two cases with tires of different materials have different coeff of friction. Then the frictonal force on the tyres are different and the work done by the forces transferred to the CM is different resulting in a different stopping distance. But if the work due to frictional force at the tyres is zero, how is this explained.
On the contrary suppose the tyres are skidding. Then work done by the frictional force calculated using either at the tyre contact point or by transferring the force to CM gives the same answer.



DaleSpam said:
Well, static friction doesn't perform work in this case, but it can perform work in other cases. For example, consider a box on the back of an accelerating truck. Energy is being transferred to the box so the static friction force is doing work in that case.
Is this the scenario envisaged : The box slides backward on the truck.
In such a case isn't the work done = coeff of sliding friction * distance ? (how is static friction doing the work ?)


TIA
 
  • #24
bksree said:
Thanks to all for the replies. What I have understood is that since there is no relative motion between the contact points on the tyre and ground (during the period the frictional force acts), the work done is zero.
More than that, there is no motion at all of the contact points of the tire (instantaneously, in the frame of the earth).

However I have some more doubts :


Spose you have two cases with tires of different materials have different coeff of friction. Then the frictonal force on the tyres are different and the work done by the forces transferred to the CM is different resulting in a different stopping distance.
Just because the coefficient of friction is different, does not mean that the static friction force will be different. The maximum value of static friction between surfaces, given by μN, will be different, but that may be irrelevant.

But if the work due to frictional force at the tyres is zero, how is this explained.
The same way as I explained it before. The magnitude of the static friction does not change the fact that it does no work. (And you can still calculate the stopping distance just as before.)

On the contrary suppose the tyres are skidding. Then work done by the frictional force calculated using either at the tyre contact point or by transferring the force to CM gives the same answer.
The "center of mass work" (Fnet*Δxcm) is calculated the same way regardless of the type of friction involved.


Is this the scenario envisaged : The box slides backward on the truck.
In such a case isn't the work done = coeff of sliding friction * distance ? (how is static friction doing the work ?)
No reason to think the box slides backwards. The box is sitting in the bed of the truck. The truck accelerates and the static friction pushes the box forward. Since the box is moving, work is being done. (Of course the static friction merely transfers energy from the engine.)
 
  • #25
Hi, this isn't my question, however I am wondering, what exactly 'is' static friction? I can imagine kinetic friction (somewhat), but not static.
 
  • #26
Doc Al
Thanks foir the reply. I am not sure if I explained my confusion clearly in the last post. So I'll try again.
1. In the case of a tyre rolling (w/o skidding) there is some frictional force between the tyre & ground (say F1). However, since the displacement between tyre and ground at any instant of time is zero, the work done by F1 is zero.
The effect of F1 in stopping the car is calculated by replacing it by an equivalent force at the COM of he car (and also a moment/couple). If the distance moved by the COM during braking is x1, then work done by this frictional force during stopping is F1 * x1

2. S'pose the tyre is skidding and not rolling. Then the frictonal force at tyre/ground contact is F2 (> F1). However there is displacement (say, x2)at the tyre/contact point (unlike in case 1) and the work done at the tyre /ground interface is F2 * X2.
We can also transfer F2 to the COM of the car, as inc ase 1, and calculate the work done by F2 which is again F2*x2.
In this case we get the same answer either way but in case 1 it isn't the case (F1X1 and zero).

Why is this.

TIA
 
  • #27
I am wondering, what exactly 'is' static friction?

with static friction surface irregularities prevents any relative motion between two surfaces in contact.

It's what keeps your feet from slipping out from under you when walking. Ice and shoe leather has considerably less static friction between them than concrete pavement and shoe leather, for example.
 
  • #28
bksree said:
Thanks to all for the replies. What I have understood is that since there is no relative motion between the contact points on the tyre and ground (during the period the frictional force acts), the work done is zero.
Relative motion determines if it is static friction or kinetic friction. It is not what determines if work is being done.

bksree said:
Is this the scenario envisaged : The box slides backward on the truck.
In such a case isn't the work done = coeff of sliding friction * distance ? (how is static friction doing the work ?)
No, I was discussing static friction. There is no slipping or sliding. The box is accelerated by static friction without any slipping.
 
  • #29
bksree said:
1. In the case of a tyre rolling (w/o skidding) there is some frictional force between the tyre & ground (say F1). ... If the distance moved by the COM during braking is x1, then work done by this frictional force during stopping is F1 * x1. Tha's the work done, but it's the work done by the brakes. Generally, static friction transfers forces between objects, and the work done is related to the source of those forces, not the static friction itself.

2. S'pose the tyre is skidding and not rolling. Then the frictonal force at tyre/ground contact is F2 (> F1).
In this case, F2 < F1. If the brakes are locked up, then kinetic friction at the contact patch converts all of the mechanical energy into heat. If the tires could be made to be rotating and yet still sliding, then some of the mechanical energy is converted into heat by the brakes and some by the kinetic friction at the contact patch.
 
  • #30
DaleSpam said:
This situation is a classic example of why the usual F.d definition of work is difficult to apply in some scenarios. If everything is rigid then the distance, d, is clear and unambiguous, but as soon as you are dealing with non-rigid motion d becomes difficult to pin down. Is d the distance traveled during braking or is it 0? Either answer seems reasonable.

A more general and usable definition of work is that "work is the transfer of energy by means other than heat". In this case, it is clear that the work done by friction is 0 since the energy of the car is not changing. All that is happening is that kinetic energy of the car is being converted internally within the car into thermal energy in the brake pads and shoes. Since there is no transfer of energy from the car to the road there is no work.

Hello DaleSpam

Thanks for the very nice posts you have written .I had a few doubts regarding the work and energy concepts.

Suppose the car starts on a surface having sufficient friction such that it rolls without slipping ,then in that case the work done by friction on the car is zero still the car accelerates forward .The chemical energy within the car is converted in the kinetic energy gained by the car .I am a little unclear how friction is in no way related to the conversion of energy from one form to another .Please help me understand this .

1)Is Work Energy theorem a consequence of Newtons' II Law but not Law of conservation of energy ?

2) Are law of conservation of energy and Newtons II law completely independent from each other?

3) Suppose the car starts on a frictionless surface . In this case it does not move Wnet = 0 and ΔKE = 0 as well . What does internal chemical energy convert into ?

4) Can we compare the movement of a car rolling without slipping on a surface to walking of a person ? In walking also W= 0 ,but ΔKE ≠ 0 ?

5)If we try to under the physics of walking from force perspective then we can say that friction produces acceleration . And from energy perspective we can say that work done by friction is zero and internal energy in the human body converts into kinetic energy .Is that correct ? But again as I stated above I am a not clear how external force on the system(friction in this case) is in no way related to energy conversions within a system (human body in this case ).

Pardon me if something I have written doesn't make sense or is unclear .

Kindly give your views .
 
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  • #31
Vibhor said:
1)Is Work Energy theorem a consequence of Newtons' II Law but not Law of conservation of energy ?
The work energy theorem is a consequence of Newton's 2nd law and the assumption that the body under consideration is rigid. So in Newtonian mechanics the conservation of energy is derived from Newton's laws. However, in Lagrangian mechanics Newton's laws are derived from the conservation of energy. The two formulations are equivalent, but take different starting points.

Vibhor said:
2) Are law of conservation of energy and Newtons II law completely independent from each other?
Not completely independent. Essentially, if you assume one (and some other supporting stuff) you can derive the other.

Vibhor said:
3) Suppose the car starts on a frictionless surface . In this case it does not move Wnet = 0 and ΔKE = 0 as well . What does internal chemical energy convert into ?
Thermal energy. In this case there will not be any thermal energy generated at the frictionless wheel, but only at all of the internal parts of the car.

Vibhor said:
4) Can we compare the movement of a car rolling without slipping on a surface to walking of a person ? In walking also W= 0 ,but ΔKE ≠ 0 ?
Yes.

Vibhor said:
5)If we try to under the physics of walking from force perspective then we can say that friction produces acceleration . And from energy perspective we can say that work done by friction is zero and internal energy in the human body converts into kinetic energy .Is that correct ? But again as I stated above I am a not clear how external force on the system(friction in this case) is in no way related to energy conversions within a system (human body in this case ).
I think that "in no way related" is a little too strong. It certainly is related because without it you will have a very hard time changing your internal chemical energy into kinetic energy. But nonetheless, from an energy transfer standpoint the energy is not transferred but only changed from chemical to kinetic.
 
  • #32
DaleSpam said:
I think that "in no way related" is a little too strong. It certainly is related because without it you will have a very hard time changing your internal chemical energy into kinetic energy.

Can you explain how the external force despite not doing any work is responsible/related for changing our internal chemical energy into kinetic energy ?
 
  • #33
This kind of thing happens all the time. For example, if you roll a ball up a ramp the ball's KE is converted to gravitational PE. The ramp does no work but is required for the ball's change of energy.

The ramp/ground does provide the force that is responsible for the acceleration, but work is about the transfer of energy, not merely the presence of a force. Many forces exist which do not transfer energy.
 

1. What is work done by friction when a car is breaking?

The work done by friction when a car is breaking is the energy that is dissipated as heat due to the friction between the car's tires and the road surface. This frictional force slows down the car and converts its kinetic energy into heat energy.

2. How is the work done by friction calculated?

The work done by friction can be calculated by multiplying the frictional force by the distance over which it acts. This can be expressed as W = Fd, where W is the work done, F is the frictional force, and d is the distance over which the force acts.

3. What factors affect the work done by friction when a car is breaking?

The work done by friction when a car is breaking is affected by several factors, including the speed of the car, the weight of the car, the condition of the road surface, and the type of tires on the car. A heavier car or a higher speed will result in more work done by friction.

4. Is the work done by friction always negative?

Yes, the work done by friction is always negative. This is because friction always acts in the opposite direction of motion, meaning it works against the car's movement and reduces its kinetic energy. Therefore, the work done by friction is always considered a negative value.

5. How does the work done by friction affect the car's braking distance?

The work done by friction has a direct impact on the car's braking distance. The more work done by friction, the more kinetic energy is converted into heat, resulting in a shorter braking distance. This is why it is important for cars to have good brakes and tires with high frictional properties.

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