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I understand that the engine delivers power to the wheels and friction from the ground causes the wheels to roll. However, given the power (work per time) at the wheels, how does that energy become the kinetic energy of the car, since friction force from road doesn't do any work?

Is it simply becuase the wheels roll causing internal forces at the axle/frame/body of car to do work speeding the car up therefore gaining kinetic energy?

I'm assuming we ignore friction, air resistance and it's a flat road

Qmechanic
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Kevin C Speltz
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6 Answers6

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how does that energy become the kinetic energy of the car, since friction force from road doesn't do any work?

This is something I've seen several times on this site lately, and I disagree with it.

Static friction does do net translational work on the car. It applies a force in the direction of displacement; work is being done on the car due to the static friction force. I cannot see any way around this with the definition of work.

The engine (through the transmission) does work on the wheels. This is what causes them to spin. The spinning wheels are now able to do work against the road, and the road provides a nearly equal and opposite work back, with some losses. Since we are talking about the work done on the car, not on the car+road system, we can see that when you isolate the forces acting on the car, the static friction absolutely does work by the traditional definitions. It is providing a force in the direction of motion.

If we ignore friction (like your question mentions), the road obviously cannot do work on the car, and all the power from the engine just goes into rotational work of the wheels. You need the wheels to be coupled to the road by friction to actually get any translational work/kinetic energy from this rotation. This is how the static friction does work on the car.

JMac
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There have been several answers given that address the main point that friction serves to convert the energy provided by the engine into kinetic energy of the car, but none seem to address the mechanism behind this transfer of energy. The only force accelerating the car along the road is static friction, seemingly indicating that the road is doing work on the car. If the engine is supposed to be supplying the energy, what gives? In particular, you've asked in comments "where would the road get [energy] from?"

Let's imagine the scenario in which there is no friction between the wheels and road. As the engine runs and the car remains stationary, the engine still delivers energy to the car-- in the form of rotational kinetic energy of the wheels. That is, without the mediating force of friction, the direct result of the engine's work is to provide rotational kinetic energy to the wheels.

Now let's switch on friction, so the car begins to accelerate. As noted before, we're forced to admit that friction is doing translational work on the car, being the only candidate force to provide it. However, that's not all friction is doing-- the static friction force is also imparting a torque on the wheels of the car in the opposite direction of their rotation. Recall that just as forces do work according to $\int \vec{F} \cdot d\vec{s}$, torques do work according to $\int \vec{\tau} \cdot d \vec{\theta}$. The observation to make is that if a wheel has a radius $R$, the no-slip condition of the wheel's rotation (i.e. the condition that the friction is static) is that $ds = R d\theta$ as the car moves a distance $ds$ and the wheel rotates through an angle $d\theta$. Since the torque and force due to friction on a given wheel are related by $\tau_f = R F_f$, we see that $$W_f^{tr} = \int F_f ds = \int F_f Rd\theta = \int \tau_f d\theta = -W_f^{rot}.$$

That is, the translational and rotational works done by friction are equal and opposite (the negative sign in the final equality is due to the torque's being opposite the rotation of the wheels), so that the total effect of friction is to do no work on the car. In this way, we reconcile the seemingly conflicting observations that the road transfers no energy to the car, yet it does the translational work accelerating it.

Flipping this statement around, we see that the work done by the wheels on the road is equal to the translational work done by friction on the car, suggesting the interpretation that the road "gets its energy" to accelerate the car from the wheels' rotational kinetic energy, which in turn was sourced from the engine.

jawheele
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    Thanks for writing this out. I actually came to basically the identical conclusion in the chat room with Johan Liebert earlier. https://chat.stackexchange.com/transcript/message/53248613#53248613 I figured adding it to my answer would just make it a confusing mess, so it's nice to see as an answer on it's own. – JMac Jan 16 '20 at 22:28
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    Your confusing pseudowork with real work. Again, the displacement of the point of application is always zero between the friction force on tire and road. It looks like work, but isn't. Read Paul A tipler, physics for scientist and engineers – Kevin C Speltz Jan 17 '20 at 12:18
  • The velocity at the contact patch is zero so the transnational work at the contact patch is zero. The torque at the contact patch is zero so the rotational work at the contact patch is also zero. – Rick Jan 17 '20 at 13:46
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    @KevinCSpeltz The conclusion of my answer is that the road does no net work and contributes no energy itself-- we agree on those points. My goal was to communicate that this is not just due to some arbitrary sub-clause about stationary points in a heuristic definition (as your engineering text seems to imply), but instead it follows logically from the mathematical definition (the given integrals, which make no reference to the motion between points of contact). One simply has to account for all of the avenues by which the frictional force might do work to reach this conclusion. – jawheele Jan 17 '20 at 14:15
  • @KevinCSpeltz I expect that many introductory texts take the "arbitrary sub-clause" approach because they wish to communicate this fact when first discussing translational work, which usually happens long before they talk about rotational work. While the spirit of the content of introductory texts is almost always correct, keep in mind that they (especially those written for engineers/chemists/etc.) almost invariably butcher formal definitions and underlying logic in the interest of making the subject more palatable to the reader who only wants tools and results. – jawheele Jan 17 '20 at 14:25
  • I found your answer slightly confusing, as I think you're using the C.O.M. of the wheel as the analysis point. I think you're saying that the road would be adding energy to the wheel if it was pushing at the center of mass (which is moving forward) but since the it's also applying a torque about the C.O.M. that energy would be removed again. What I don't understand is why you'd analyse around the C.O.M. rather than the contact point, where the lack of movement makes it clear that there's no energy transfer. – Rick Jan 17 '20 at 17:53
  • I wonder if you could apply similar adding and subtracting energies that cancel out to the scenario I described in my answer with the spring? If not, what's different about the scenarios? – Rick Jan 17 '20 at 18:00
  • Thats why i keep pounding on my physics book. it comes down work being the force thru the displacement of point of application. Since the contact patch is not moving, road does no work. If the road did work, well, then we would have smaller engines and such. Friction is a converter, so it looks like work, but it is adding no real energy. Examples are similar to walking – Kevin C Speltz Jan 17 '20 at 19:02
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    @KevinCSpeltz Since you brought up walking, take a look at this https://www.quora.com/Can-static-friction-do-work – Bob D Jan 17 '20 at 20:33
  • @Rick We work with the displacement of the wheel's COM because we wish to decompose into rotational and linear kinetic energies (so that we can analyze the role of static friction in converting one into the other, as one is directly the energy supplied by the engine and the other is that of the car's forward motion)-- the simple rigid body relationship $KE_{tot} = KE_{tr}+KE_{rot}$ only holds if $KE_{tr}$ is the kinetic energy of the COM motion. The change in this quantity is the integral $\int \vec{F} \cdot d \vec{s}$ over the path of the COM. – jawheele Jan 17 '20 at 21:14
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    Bob, this is from physics forum and agrees with my college physics book. You cant use com for work unless you are treating it like a mass point particle which a body isnt. thanks for help I'm good now. https://www.physicsforums.com/threads/what-is-the-work-done-by-friction-when-a-man-is-walking.331641/ – Kevin C Speltz Jan 17 '20 at 21:26
  • @KevinCSpeltz Right, that's why there's the additional term for the rotational work. The COM contribution is only part of the total work (which is zero). My whole point is that, from the actual definition of work (which has nothing to do with contact points), it is not obvious that the road does no work. One must show this is the case, which is what my answer does. – jawheele Jan 17 '20 at 21:32
  • I'll say it again. Energy is ability to do work. What energy does the ground have to be able to do work? We can disagree. I got my help thanks. – Kevin C Speltz Jan 17 '20 at 22:10
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Without friction your car won't move even a bit. Even though no net work is done by the friction but it acts as an energy converter and delivers the internal energy supplied by the engine to the car in the form of translational kinetic energy.

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I think you've got the fundamentals right, the question can be resolved by considering the car as a combination of components rather than a single, point mass, and also by distinguishing between static friction while the car is stationary vs while it is moving.

EDIT : As correctly pointed in the comments, we need not resort to kinetic friction in this case. The answer has been corrected on this point.

Consider this sequence of steps in time:

(0) The car is stationary. Static friction is indeed present, and it is not doing any work.

(1) The engine burns fuel to generate some energy.

(2) This energy is converted to torque and transmitted to the wheels.

(3) The torque on the wheels is converted to tangential force at the wheel-road contact.

(4) This force is countered by static friction.

(5) When the torque on the wheels is sufficiently high, the tangential force becomes higher than the force due to static friction (this is limited by the coefficient of friction).

(6) At this point the wheels start rolling and now, the static friction generated keeps them rolling. (If there was no static friction, the wheels would rotate and the car wouldn't move. If the torque on the wheels was really high, the wheels would spin AND the car would move, or rather skid. In that case, kinetic friction would be at work).

So the force causing net movement/acceleration is the resultant of tangential force due to torque and static friction force. The energy to generate both these forces is provided by the engine. The work is therefore being done by the engine, while the road only provides a surface for generating the necessary reaction.

I think my emphasis is the following idea - static friction very much exists even without the engine, or when the car is stationary. Static friction is responsible for motion only when the engine (or some other energy source) generates a combination of forces that can cause net motion.

AppliedAcademic
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    Thank you! After discussions with fellow engineers and some published books, your description is right on and makes sense how thru the power developed by the engine, the wheel constrained by friction are driven forwards(rolled) hence causing the power from the engine to become the kinetic energy of the car. Thanks for taking the time to read thru the nonesense and straighten this out – Kevin C Speltz Jan 17 '20 at 12:30
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    Wheels rolling without slipping would have static friction, not kinetic friction. – JMac Jan 17 '20 at 14:51
  • I suggest removing the bits about kinetic/dynamic/rolling friction and just focus on how the car accelerates after it's already moving. I think it would make the answer more clear. – Rick Jan 17 '20 at 15:53
  • @JMac- You're absolutely right, I have corrected this in the answer. I was wrongly referring to static friction during rolling as kinetic friction. Thanks! – AppliedAcademic Jan 17 '20 at 17:20
  • @Rick- I get what you mean, tried to do that in the answer. Do you think its more clear now? If not, please feel free to edit! Thanks. – AppliedAcademic Jan 17 '20 at 17:20
  • @AppliedAcademic After reading your answer it is still unclear to me what your position is regarding the static friction on the car. Bottom line: So just to be clear, is it your position that static friction does work on the car, or does it not? – Bob D Jan 17 '20 at 18:21
  • @BobD- I think its unhelpful to think of forces as 'doing work', because this suggests some agency (and therefore energy etc.). Forces (may) create acceleration, while bodies possessing energy (may) do work. With that out of the way, in a moving car, there is a combination of forces acting, one of which comes from static friction. This combination of forces leads to the acceleration of the car. Take out any one force (including static friction) and the acceleration won't happen.I'm sorry I'm not giving you a yes/no answer, because I believe the question is not optimally framed for one. – AppliedAcademic Jan 17 '20 at 19:56
  • @AppliedAcademic That's OK. I suppose I shouldn't have expected a yes/no answer. – Bob D Jan 17 '20 at 22:50
  • I don't think this is quite correct. If the tangential force at the tire surface is higher than the limit of static friction to oppose it, the wheel spins - not as in spins around its axle, it spins relative to the ground. Under normal conditions, the portion of the tire in contact with the ground is always stationary with respect to the ground. – David Jan 17 '20 at 23:52
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    Note that the same kinetic energy would be gained if the engine was used to operate a winch that pulled the car at the same acceleration. Of course this would be a limited affect, since sooner or later you'd be at the end of your rope. – Hot Licks Jan 18 '20 at 02:19
  • @David- I see where you're coming from, and I agree in part. The distinction I'd like to make is again static vs kinetic friction; the spinning relative to the ground happens when kinetic friction is active. Kinetic friction becomes active when the tangential force is not only larger than that of static friction, but much larger. I don't think there is a number for how large 'much larger' is, but for perspective, it would be like a racing car skidding to a 10m halt; the force due to sudden application of brake is much larger, not slightly larger than the static friction force. – AppliedAcademic Jan 18 '20 at 10:34
  • @AppliedAcademic That's not how it works. Kinetic friction starts acting as soon as the force is too great for static friction, and the two surfaces start moving relative to each other. Kinetic friction is less force than the maximum static friction. When a racing car skids to a halt, it's less than the maximum braking power. You want your wheels to roll when you stop because it allows for more friction. That's a key part of anti-lock braking systems. – JMac Jan 28 '20 at 12:54
  • @JMac - Do you think it's better to move this to chat? (I'm not sure how to). I'd like to respectfully contest your interpretation on all but two points- which are correct, but I believe your inference thereon is not. – AppliedAcademic Jan 28 '20 at 15:59
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First lets take a look at the simpler example of a stationary object providing the force to accelerate something:

A block with a spring is held compressed to a wall, and then it's released. enter image description here

The energy to accelerate the block is stored in the spring. When the block is released, the spring expands, the end of the spring attached to the sliding block starts to move, and since it's providing a force, the work transferred from the spring to the block is equal to the velocity dotted with the force. The work at the other end of the spring is zero because the velocity is zero. The spring is using the stationary wall to convert it's internal stress/spring energy into kinetic energy of the block. No energy is transferred to or from the wall.

Now lets consider the more complicated system of the car, axle, wheel, and road. The axle transfers energy from the engine to rotational energy of the wheel. The wheel uses the road to convert it's rotational energy into translational energy. This process doesn't involve any energy transfer to or from the road because the surfaces applying the force have no velocity. The wheel transfers the translational energy to the axle, and the bearing, and to the rest of the car. All of these transfers are possible because the force dotted by the velocity (or the torque dotted by the rotational velocity) are non-zero.

Rick
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  • So Rick that makes sense on how the energy gets to be kinetic energy. So in the end to answer the other issue, the ground does no work correct? As you say the process doesn't involve any energy transfer to or from the road. If so I agree and thank you – Kevin C Speltz Jan 17 '20 at 15:49
  • @KevinCSpeltz correct. The ground does no work because it's not moving. – Rick Jan 17 '20 at 15:50
  • Rick, I went to my book and another forum yesterday to get help....how can so many people argue me, felt like I was nuts. Thanks for clarifying with positive example. – Kevin C Speltz Jan 17 '20 at 16:15
  • Technically, a small amount of energy is transferred to the road, which is why roads wear down (bits of asphalt/concrete are probabilistically eroded every time a car passes over it). Also, if no energy were transferred to the wall in your example, it wouldn't matter how big or small or dense the wall is. Compressing the spring initially also compresses the wall and makes it part of the spring, so I suppose it also contributes a trivial portion to the final result. Overall, good example, though. – Lawnmower Man Jan 17 '20 at 20:17
  • @LawnmowerMan yes, this uses typical physics assumptions like solids (other than springs) have infinite modulus of elasticity, etc. the actual situation is always more complicated. – Rick Jan 17 '20 at 20:24
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So many forms of kinetic energy in autos, so not to get to detailed a few main ones are..If its internal combustion,a certain amount of fuel/compressed air ignites producing the energy to drive piston/connecting rod down to crankshaft to rotate it and remaining internals including flywheel(manual transmission) which is weighted accordingly to provide sufficient kinetec energy with the help of gear reductions in transmission to rotate driveshaft,turning another gear reduction ring/pinion gear set to initiate movement and rotating the mass of axles/brake rotors/wheel/tire plus the mass of the vehicle, i see it as many combinations of kinetic energy which derive from igniting compressed air/fuel mixture which however initially needs electricity to turn a electric starter motor to initiate that chain reaction, variables of friction and air resistance will limit the capable kinetic energy however is still is basically a ratio of the engine efficiency(turbo,natural aspirated,Hybrid)+ displacement of engine= total power(however you want to view it, HP, Watts,etc) of the engine in combination with fuel type and mass of vehicle which will equal the kinetic energy.

Johnny
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