# How does a car gain kinetic energy?

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

• If you ignore friction then the car doesn't move because the wheels spin. Jan 16, 2020 at 14:33
• Related - Work done by static friction on a car. Jan 16, 2020 at 14:37
• @DJClayworth - Q. How many physicists does it take to change a lightbulb? A. None, because neglecting the force of fiction, it can't be done. Came up with that one in my freshman year of college. Jan 17, 2020 at 16:21
• Just to be clear, is it still your position that static friction on the wheel does no work on the car? Jan 17, 2020 at 18:19
• All the Work happens inside the engine's cylinders. Each cylinder moves as the the combustion chamber expands when the combustion gases pushes it. The Work is the travel of the cylinder, times the force applied (integrated). The rest of the power train, wheels and road is just "levers and pulleys". Jan 17, 2020 at 21:19

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.

• Comments are not for extended discussion; this conversation has been moved to chat. Jan 17, 2020 at 15:51

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.

• Thanks for writing this out. I actually came to basically the identical conclusion in the chat room with Johan Liebert earlier. 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, 2020 at 22:28
• 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 Jan 17, 2020 at 12:18
• @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. Jan 17, 2020 at 14:15
• @KevinCSpeltz Since you brought up walking, take a look at this quora.com/Can-static-friction-do-work Jan 17, 2020 at 20:33
• 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. physicsforums.com/threads/… Jan 17, 2020 at 21:26

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.

• Comments are not for extended discussion; this conversation has been moved to chat. Jan 17, 2020 at 15:56

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.

• 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 Jan 17, 2020 at 12:30
• Wheels rolling without slipping would have static friction, not kinetic friction.
– JMac
Jan 17, 2020 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, 2020 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! Jan 17, 2020 at 17:20
• 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. Jan 18, 2020 at 2:19

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. 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.

• 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 Jan 17, 2020 at 15:49
• @KevinCSpeltz correct. The ground does no work because it's not moving.
– Rick
Jan 17, 2020 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. Jan 17, 2020 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. Jan 17, 2020 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, 2020 at 20:24

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.