# Work done by static friction on a car

The tires of a car execute pure rolling. Therefore, the work done by friction on the tires (and hence the car) is zero. If no external work is done, how does a car's kinetic energy increase?

The increase in the car's kinetic energy comes from the internal energy of the car, stored, for example, in its gasoline or batteries.

The engine exerts torque over the wheels, which are prevented by the friction from simply rotating in place. The reaction from the ground on the car (wheels) makes it move faster.

• The internal engine cannot push the car forward. The engine pushes on the ground and the ground pushes it back. If there was no friction, how much ever you would throttle, the car wouldn't move an inch. Jul 20, 2017 at 11:26
• @Yashas That's what I meant by "which are prevented by the friction from simply rotating in place." But you are right about the ground pushing the car forward, I'll edit my answer accordingly. Jul 20, 2017 at 11:33

You are mixing up two different things.

If we have two surfaces sliding over each other, and there is a non-zero coefficient of friction between them, then energy is dissipated as friction. This lost energy is just the distance the surfaces have slid multiplied by the frictional force. In the case if the (ideal) wheels then you are quite correct that because no sliding occurs the energy lost to friction is zero.

But all this means is that no energy is lost to friction. The car and the road exert an equal and opposite force on each other, so as the car moves external work is done by the road on the car. The non-slipping of the car tyres just means all that work goes into the kinetic energy of the car and none is lost as friction.

• But for pure rolling the work done by static friction is zero. Jul 20, 2017 at 11:14
• @xasthor: no, for pure rolling the work lost to heat due to friction is zero. The work done on the car by the frictional force is not zero. Jul 20, 2017 at 11:27
• @JohnRennie, "The work done on the car by the frictional force is not zero." -- can you back up this statement? There's no displacement of the point of application of the friction force, so it's my understanding that any integral over displacement must be zero. Jul 23, 2017 at 10:42
• @stafusa: the work is not being done on the bit of the tyre in contact with the ground, it's being done on the centre of mass of the car. The force is transmitted from the ground through through the tyre to the rest of the car. Jul 23, 2017 at 12:50
• @JohnRennie, I'm convinced you are mistaken: the only work in this case is the one done by the internal forces - which convert internal (e.g., chemical) energy into kinetic energy. Remember that internal forces can't to net work only for rigid bodies, and our car has moving parts. That has already been addressed before: physics.stackexchange.com/a/2474/75633 Jul 25, 2017 at 23:35

The answer is that a static frictional force can do work.

Consider the case of you sitting in a car which is accelerating.
The seat has no back and you are not moving relative to the seat.

Now consider you as the system.
You have three forces acting on you.
In the vertical direction your weight and the normal reaction due to the seat cancel out and you are left with the horizontal static frictional force which is accelerating you and moving with you.
It is the work done by that static frictional force which causes you to increase your kinetic energy.

Of course the ultimate source of the increase in kinetic energy is the chemical energy stored in the fuel which the engine converts into a useful form.

Update as a result of a discussion with the OP.

Going back to the car with you sitting on the seat.
What does an observer on the road see?
The observer sees a constant horizontal static frictional force due to the seat acting on you.
The observer sees that force moving along with the same velocity as the car in the same way, as the observer would see somebody on the car pushing you from behind to accelerate you, or accelerate you if the seat with a frictionless bottom was place on the ground and somebody was pushing you.
In all instances there is a force acting on you and as the force undergoes a displacement work is being done by that force.

The observer on the road sees the car with a horizontal frictional force acting on the car (tyres).
That observer sees that force undergo a displacement hence that force has done work.
It matters not how that horizontal frictional force arose the fact is that the car is subjected to a constant horizontal force which we know is due to static friction.
That force accelerating the car could have been due to somebody pushing the car where the point of contact of the force is the car and so the person applying the force is not moving relative to the car.

• I agree. The forward frictional force exerted by the road on the drive tires is applied over the distance that the car travels, so it does work on the car. Jul 20, 2017 at 13:30
• I don't think that's correct. The point where the force of friction is applied doesn't move in the situation of the question (so this force cannot do work) while this point does move in your example. Besides, and more importantly, if the friction were somehow transferring energy to the car, where does this energy come from and why then does the car need fuel? Jul 20, 2017 at 23:13
• @stafusa The interesting thing is that is if you take the car as the system and ignore air resistance it has frictional forces as the only horizontal external forces acting on it. The car accelerates and its kinetic energy (and linear momentum increases). There can only be that increase if work is done on the car by an external force. So the frictional force must have done work on the car. Note that my sitting on the seat example requires some external source of energy which is the chemical energy of the fuel. Jul 21, 2017 at 8:36
• That's an subtle discussion. Let's consider the system as being Earth + car: then it becomes clear that the internal energy of the car is converted into kinetic energy of both car and Earth. If you then ignore the acceleration of Earth (which is nonzero, but tiny), it's easy to see that the internal energy of the car was converted to (its) kinetic energy: how, then, could have the Earth transmitted to the car some amount of energy that it never had? Jul 21, 2017 at 9:41
• @Farcher, going back to the basics: how could it be that "the frictional force must have done work on the car", when the point of application of this force doesn't move. From the definition of work of force it must be zero. Some references that state the same: brucesherwood.net/wp-content/uploads/2017/06/Pseudowork1983.pdf aapt.scitation.org/doi/abs/10.1119/1.10770 aapt.scitation.org/doi/abs/10.1119/1.11359 2.files.edl.io/P81MQEPOEVknx5ju55vMIsqlrP6hOS8Y9UX1b0j4f8gxstsH.pdf journals.sagepub.com/doi/abs/10.7227/IJMEE.41.1.2 iopscience.iop.org/article/10.1088/1361-6552/aa74a8 Jul 21, 2017 at 12:53

In accelerating the car from rest, static friction DOES positive work which is the change in linear kinetic energy of the car. BUT it also does negative work on the wheel(it creates counter torque) which decreases the rotational kinetic energy of the wheel. So, the NET work done by static friction is 0. It simply converts some part of the rotational kinetic energy of the engine to linear kinetic energy until the condition of no slipping is met. That makes sense if you think about it because the road does not have energy to transfer to the system.

The answer by @stafusa is correct. The increase in kinetic energy (KE) of the car is due to the decrease in internal energy of the car (from combustion, batteries, etc.) This can be understood using the first law of thermodynamics.

On a frictionless road, the increase in KE would be ever increasing rotational speed of the driven wheels. The friction of the road decreases the rotational energy of the driven wheels resulting in forward motion of the center of mass (CM) of the car.

For the no slip condition of the tires on the road, friction does no net work. There is confusion regarding this point; it needs to be recognized that friction contributes to both translational motion of the CM and rotational motion of the driven wheels. I summarize the reason friction does no work for the no slip condition using two approaches.

(1) The work done by friction is$$\int_{}^{} \vec F_{fric} \cdot \vec v \enspace dt$$. The instantaneous point of contact of the tires with the road has zero velocity for no slip; so, $$\vec v$$ is zero and friction does no work.

(2) The force of friction increases the acceleration of the center of mass (CM) and therefore does translational work that is positive, but the torque from friction relative to the CM does rotational work that is negative and equal in magnitude to the translational work. Therefore, for no slip the net work done by friction (positive translational plus negative rotational) is zero.

See the answer by @jawheele to How does a car gain kinetic energy?.

Also, for a related problem, object rolling down an incline, see the two answers by @Dale and by me to Is work done by torque due to friction in pure rolling?. For no slip, friction does no work; for slip friction does work as discussed in my answer.