Where does the power delivered to car's wheels go?

Question

Okay, so power is work/time. Most cases, when power is provided to something, energy is gained as kinetic energy or lost to friction.

But in a car, the engine puts power ( torque x rpm/5252) to wheels, but very little ends up in the wheels, assuming friction keeps them from spinning.

So where does the power go? Do the wheels thru the friction forces cause the energy to go to the car? Essentially the wheels do work on the car which transforms the rotational energy to kinetic energy of the car right?

Answer 1

When the engine applies a torque on the wheels via the gear mechanism, the wheels want to spin around, but the mass of the car and the friction between the road and the tires constrains it. So the only way is to move the vehicle forward. The friction between the tires and the road provides the necessary traction to do so. In the process kinetic energy of the vehicle increases because its velocity increases.

So the work done by the engine is used up in increasing the kinetic energy of the vehicle. Moreover at steady speeds, it is mostly used to overcome the loses due to sliding friction. A part of the energy is always used up in overcoming air resistance, undesired heating of the vehicle. Energy is also wasted in the tires in the form of heat(due to hysteresis loss).

Answer 2

The engine applies torque to the wheels. The wheels turns and apply friction force to the road. By Newton's third law the road applies force to the wheels which make the car moving. Engine power goes to kinetic energy of the car, dissipated heat to friction, air resistance force, battery charging and air conditioning

Answer 3

There's two compatible definitions for work and power to be used here. The one you are thinking about is $W=F\cdot \Delta x$, work is a force applied over a change in position(distance), and this $P=Fv$, power is a force applied at a velocity. There is, however, another definition, $W=\tau\cdot \Delta \theta$, work is a torque applied over a change in angle and thus $P=\tau \omega$, power is a torque applied at a given angular velocity. Sorry for the Greek letters, that's just how we roll! (yes, I went there)

It turns out these two definitions are exactly equivalent, if you break objects up into point masses. If you have a rotating object and you sum all of the forces on the point masses in all the directions they are traveling (each point is going in a different direction, since they're moving on a circle), you get the equations for work and power based on torque. If you look at torques, and look at points that are moving in a straight line (which means their radius is changing at the same time as their angles are changing), you can derive $W=F\cdot \Delta x$ and $P=Fv$ from the angular equations.

The only difference between them is that one form is very good at capturing what is happening when objects are moving linearly, while the other is very good at capturing what is happening when objects are moving circularly. If you are the kind of person who thinks in coordinate systems, forces are good in cartesian (x/y or x/y/z) coordinate systems, while torques work well in polar cordinates. Thus we can choose which definition of power to use based on what symmetries we see in the problem. We choose our approach based on what equations make the most annoying things cancel out and dissapear!

In this case, you are interested in power at the wheel. The wheel is rotating, which means the equations of motion are simpler to understand if we think in torques. You can think in forces, and get the right answer, it just takes more mental arithmetic to get there.

If we think in torques, we see what "power to the wheels" would have to mean. Your wheels are effectively applying a torque to the ground (and one step before that, your axels are applying a torque to the wheels). Your wheels are turning at some angular velocity (some number in rad/s if you're thinking like a physicist, revolutions per minute (RPM) if you're thinking like a car person). The power to the wheels is the product of these two things.

If we want to think about "power to the body of the car," as in the thing which opposes air friction, we would naturally want to think about this in a linear space. We would then wish to use forces and velocities to determine power. And, indeed, we can see that the forces applied to the contact patch of the wheels gets applied in the direction of a velocity, doing work. It's just a little trickier to see how it gets from the engine to that contact patch because all of the parts in that chain are rotating, making it a bit more difficult to think in forces.

And if we're feeling a bit masochistic, we can also solve the equations of motion for the entire car around an axis of rotation. With four wheels, the equations get a bit murky. However, if our car was a unicycle, this may be a very natural way of thinking about things. On a unicycle, it is very reasonable to think of the force of gravity pulling you down as a torque around the axel of the unicycle that needs to be balanced by a corresponding torque by your feet on the pedals going in the opposite direction.

Answer 4

So where does the power go?

Power is just like force and varies from one instant to another.(You cannot say that an object losses $3N$ force! ) What the car/object gains or loose with time energy (whose gain/loss is represented by work done). Power is the rate of work being done(i.e., it represents at what rate the energy is transferred to an object).

Do the wheels through the friction cause the energy to go to the car?

No the car doesn't gain any energy. The frictional force just acts as the converter of the internal energy stored in the car and converts it into kinetic energy( a bit of thermal too!).

Essentially the wheels do work on the car which transforms the rotational energy to kinetic energy of the car right?

Note that a net force is acting on the car which causes the car to gain translational kinetic energy. Now since the force isn't acting on the COM therefore a torque is generated in due process causing the car to gain rotational kinetic energy.

Answer 5

Like others have said, the power goes to the wheels, which push on the road surface, and the whole land/mass/road surface essentially pushes back. What isn't being said is that esp. in a rear wheel drive car, some of this must be going to springs/suspension(less concern in a hotchkiss drive leaf spring configuration, perhaps more with coil springs), and a larger percentage back thru the drivetrain, such as through a tube surrounding the driveshaft. I was contemplating on a trailing arm suspension, why it was configured the way it was if the forces were transferring up the trailing arms, and concluded a larger percentage than I'd had any concept of must be transferring to the drivetrain components and back to the transmission original transfer join location.