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Question statement from David Morin's Classical Mechanics :

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.

I couldn't get the point! The ground is applying frictional force on car and it is moving in opposite direction to the car i.e the work done by the ground should be negative (is what I had thought).

Furthermore, the earth doesn't move either when it exerts gravitational force on the object to do work.

Could someone clarify this please?

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  • $\begingroup$ the ground is applying friction to the tires, which can rotate freely. try making a free body diagram to help you understand $\endgroup$ – Mohammad Athar Apr 10 '18 at 12:35
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The point where the ground is applying the force to the car is an instant center of rotation. At this point, if the tires are not slipping there is no relative movement in the horizontal direction between the tire and the road.

Since at each point there is no relative motion between the two, there is no displacement associated with that, and it does no work.

The work comes from the brakes. The brake pad slides against the contact point on the wheel, and this has the displacement and force acting with relative motion. By changing how fast the tire can spin, the brake does the work. The tires on the road just facilitate the momentum transfer.

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  • $\begingroup$ That means that every instant the tire is in contact with the ground the relative velocity of the tyre is zero. Hence, for that instant the work done is zero. Since this continues as the tyre roll, we can say that frictional force does no work on the tyre. Did I get it right? $\endgroup$ – suiz Apr 10 '18 at 13:00
  • $\begingroup$ An objection occurred to me, but I think I've resolved it. If we draw a car a few times while it is braking, and on each picture show the (backward) force acting on the tyres from the road, it seems as if we have a force whose point of application is moving in the opposite direction to the force. This accords with the force doing work. My solution: on each diagram we're drawing a $different$ force (not just a different sized force). Is this right or is it sophistry? $\endgroup$ – Philip Wood Apr 10 '18 at 13:24
  • $\begingroup$ @PhilipWood ... Why would point of application move ? Cause , every time we draw the same resisting force ...? $\endgroup$ – Nehal Samee Apr 10 '18 at 18:14
  • $\begingroup$ @Nehal Samee… Don't follow what you mean. I meant that if one drew the car at various points along the road as it came to rest we could draw a 'backward' force acting on the car tyres from the road for each position of the car that we'd chosen. So we have a force with a moving point of application, and we'd usually say that work is being done. But is it really the same force for each position of the car, or a different force each time, in which case work isn't being done (as the $Morin$ quote says). $\endgroup$ – Philip Wood Apr 10 '18 at 18:55
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The friction force from the ground on the tires is what causes the car to slow down.

No. The friction force from the brakes is what causes the car to slow down.

The fact that they can glow red-hot* should be an obvious indicator of this.

  • Check out the Nova where they show the testing of the 777's brakes. They were bright yellow.
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    $\begingroup$ In Newtonian physics a body changes its velocity only if a resultant force $from\ outside$ acts upon the body. The force (or, more correctly, forces) from the ground in the case of the car. What it is that $causes$ the car to slow down, friction in the brakes or the force from the road is a philosophical issue. $\endgroup$ – Philip Wood Apr 10 '18 at 14:49
  • $\begingroup$ Indeed, and invariably when I see questions about "who did the work?" here on SO, it's precisely one of those "philosophical issues". Although in this case, the ultimate cause is "your foot". $\endgroup$ – Maury Markowitz Apr 10 '18 at 15:50
  • $\begingroup$ "There is no cause nor effect in nature; nature simply $is$. [Ernst Mach. I can't vouch for the translation.] $\endgroup$ – Philip Wood Apr 10 '18 at 17:13

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