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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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7 Answers7

15

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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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.

2

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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2

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.

Eph
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This answer is correct, but I'd like to emphasize a fundamental point that seems to be forgotten in many answers to this and similar questions:

Work and total energy depend on the reference frame.

Unfortunately there's a tendency to think about the energy in a system (or better, in a volume at a given time) as something objective, which all observers agree upon, say like the amount of matter or of electric charge. But that is not the case. The total amount of energy depends on the adopted reference frame. From this point of view it is very similar to momentum: the momentum of an object can be zero in one reference frame (the one fixed with the object), and non-zero in another.

The same goes for work: a force can be doing zero work in one reference frame, positive work in another frame, and negative in yet another. This is clear if we write an infinitesimal amount of work in terms of power: $\pmb{F}\cdot\pmb{v}\,\mathrm{d}t$. The velocity $\pmb{v}$ of the application point depends on the reference frame, so the work or power depends on the reference frame as well (the force doesn't, see below).

Because of this frame-dependence, the energy analysis for the car can look very different in different frames:

  • Take a frame fixed with the car, say from the driver's perspective. In this frame, the car has zero kinetic energy, and never gains or loses kinetic energy. On the other hand, in this frame the road (and the whole Earth) gains kinetic energy. The car is appplying a tangential force – friction – on the road (and vice versa), and the road moves. The car is therefore doing work on the road. In this reference frame, the internal energy of the car is decreasing because of this work, and the kinetic energy of the road is increasing because of this work. Since this frame is non-inertial (see @Dale's comment below), there are also inertial forces, and we must count in the work done by them. On the car this work is zero because it isn't moving. On the road it is positive, and contributes to the road's increase in kinetic energy.

  • Take a frame fixed with the road. In this frame, the car is gaining kinetic energy, whereas the road has zero kinetic energy and doesn't lose or gain any. The car is applying a tangential force on the road (and vice versa), but the wheel's rubber and the road have zero velocity at the point of contact, so this force does no work. In this reference frame, the internal energy of the car is decreasing because it's converted into kinetic energy of the car.

Unfortunately this frame-dependence of energy and work is usually not emphasized in introductory physics courses, so we build a somewhat wrong intuition that increases the difficulty in understanding this kind of physical situations.

A good discussion about this point is given by
• Bridgman: The Nature of Thermodynamics (Harvard University Press 1943; also freely available here);
see around page 30, "Work and the flow of mechanical energy".


Some additional notes:

  • If it is not intuitive that there's no relative motion at the point of contact between wheel and road, think of a tank instead, and of its caterpillar tracks; the car's wheel is a sort of very short track. The situation is also analogous in walking: there's no relative motion between the shoe and the pavement.

  • The frame-dependence analogy between energy and momentum shouldn't be surprising: we know from relativity theory that energy, flux of energy, momentum, and force are actually different aspects of one entity (the energy-momentum tensor). This connection is also present in Newtonian mechanics. In fact there are works that introduce, in a completely rigorous way, an energy-momentum tensor in Newtonian mechanics, which becomes a sort of Galilean relativity theory. See for instance
    • Lévy-Leblond: Galilei Group and Galilean Invariance (Academic Press 1971)
    or
    • Truesdell & Toupin: The Classical Field Theories (Springer 1960), sections 211 and 284.

  • Internal energy is frame-independent, because it is defined as the energy measured in a frame at rest with the system (or more precisely with a given volume element).

  • Good books in mechanics are explicit about which quantities are frame-independent and which aren't. Examples of frame-independent quantities: mass, internal energy, contact forces (and some body forces as well), heat flux, temperature. Examples of frame-dependent quantities: velocity, momentum, kinetic energy, work and power.


Extra references that discuss these matters:

pglpm
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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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