# Force required to drive car

From what I know of Newtonian Mechanics, if an object is moving at a constant velocity, the net force acting on that object is equal to zero. If there is friction, then the applied force required to maintain a constant velocity is equal to the magnitude of the force of friction, regardless of the actual value of the velocity.

Now let's suppose a car is driving on a road at a constant velocity of 10 km/h, and the force of friction acting on the car has a magnitude of 4,000 N. The applied force [from the engines] required to maintain the speed is also, therefore, 4,000 N[as the net force is equal to zero]. If the car travels on the same road at constant speed of 100 km/h, again, it would require the same amount of applied force from the engines to maintain the speed without acceleration: 4,000 N, as this "cancels out" the force of friction, and, per Newton's Second Law, the velocity does not change.

If moving at two different constant speeds–10 km/h and 100 km/h–on the same surface requires the same constant applied force generated from the engines, then why does moving at 100 km/h use up more gasoline?

• You are assuming that the force opposing the car is a constant 4000N, in reality when you speed up several different factors will affect this force, such as higher air resistance. This probably isn't the full answer though, I'll go digging around. Nov 23 '14 at 6:05
• possible duplicate of Fuel usage at the same constant rpm at different gears Nov 23 '14 at 6:55
• Check this out: physics.stackexchange.com/questions/12128/… You're essentially asking the same question. Nov 23 '14 at 6:56
• @Pranav I think Johny was not asking this from practical point of view, but from a theoretical one. The question you linked above was concerned the technical details of the car, unlike this question.
– user49111
Nov 23 '14 at 7:15
• @JohnyDiala: "If there is friction, then the applied force required to maintain a constant velocity is equal to the magnitude of the force of friction," - Are you sure that it, is just the magnitude of the force which should be equal? Nov 23 '14 at 13:20

In a perfect vacuum, on a frictionless road, you could just turn off the engine and the car would keep moving, never slowing down. However, in the real world, there are several effects that exert a force on a moving car, slowing it down, such as:

• rolling drag between the tires and the road surface,
• fluid drag from the air that the car moves through, and
• various friction losses between moving parts in the car itself, which, unless compensated by engine power, cause the wheels to slow down and exert a torque on the roadbed slowing down the car.

To keep a car moving at a constant speed, the engine needs to exert enough force to balance all these forces.

The important thing to realize here is that, at high speeds, the main force slowing down the car is actually the fluid drag, which grows roughly in proportion to the square of the speed. Thus, to double the speed, the engine needs to exert four times as much force. (At least, that approximately holds at normal highway speeds. Things get even more interesting when you approach the speed of sound and wave drag starts to play a role.)

Because of this, minimizing the drag coefficient is a critical feature of high-speed automobile design, and is why essentially all modern cars (but especially high-speed models) feature streamlined shapes designed to minimize aerodynamic drag.

Also, as Sachin Shekhar notes in his answer, the rolling drag for pneumatic wheels is also somewhat speed dependent, mainly because the wheels are flexible, and thus deform as they rotate, losing energy as heat. These losses also increase with velocity, meaning that, even in vacuum, maintaining a higher velocity still needs more power. In principle, you could minimize these losses by making both the wheels and the road surface as hard and inflexible as possible — say, by making both out of steel, as is done for trains, which also minimize aerodynamic drag by their long and narrow shape. That's one reason why trains can travel at considerably higher speeds than would be practical for a car to maintain.

(Of course, to reduce drag even further, you could put wings on the car and have it fly high above the road, eliminating rolling drag entirely and reducing fluid drag significantly due to the lower air pressure in the upper atmosphere. Or, even better, go even higher up where the air is even thinner.)

• Finally, someone who did read the OP question! Nov 23 '14 at 17:52
• The force increases as the square of the speed. The energy is force times distance and the distance varies with the speed, so the power required varies with the cube of the speed. Nov 24 '14 at 0:11

You are wrong at assuming constant friction. Rolling Friction increases when you increase speed of the car (See the formulae at the bottom).

Also, aerodynamic drag increases with the square of speed (See the formula at the bottom).

So, at higher speed, the car engine needs to counter higher rolling friction and air drag to maintain that speed.

While the previous reason is adequate, I should add this: At higher speed the car needs to do more work per unit time. At 10Km/h, if tyres are at x RPM, the tyres need to revolve at, say, 10x RPM to maintain 100Km/h. So, engines are doing 10x more work per unit time at 100Km/h.

But, rolling friction and air drag (most dominant forces among all resistances here) are at the heart of the problem. In vacuum and frictionless environment, for example, you can turn OFF engine at 10x RPM and it'd remain at 10x RPM.

Here's the formula of Rolling Friction..

$$F_r = c_r N$$

Where $c_r$ is rolling friction coefficient (dimensionless) and $N$ is the Normal Force applied by road which is equal to weight of the car.

Now, the Rolling Friction Coefficient increases with the car speed resulting in the increase of the Rolling Friction.

For example, the rolling friction coefficients for pneumatic tyres on dry roads can be calculated with

$$c_r = 0.005 + \frac {1}{p} (0.01 + 0.0095(\frac{v}{100})^2)$$

where $p$ is tyre pressure and $v$ is speed

Here's the formula of Automobile Aerodynamic Drag..

$$F_a = A/2 × c_d × D × v^2$$

where $A$ is frontal area of the car, $c_d$ is drag coefficient, $D$ is density of air, and $v$ is speed.

Note: Weight of car goes down when gasoline burns which contributes in decrease of the rolling friction and inertia (Rocket Science), but resistance forces win by very large amount.

• Thanks, that is an adequate answer for my particular question. However, how can this phenomena be explained more broadly for situations other than rolling friction? In situations of running, for example (variation between energy used whilst running at two different constant velocities)? Moreover, isn't the friction coefficient usually a constant(expressed as μ); a dimensionless scalar value between 0 and 1 which is a material property? Nov 23 '14 at 9:47
• @Johny If your shoes aren't slipping, that's static friction. For slipping shoes, dynamic friction would be involved. There are several types of frictions with slightly different concepts. Everything can be calculated, but there's no generic solution to solve all. Nov 23 '14 at 9:51
• @Johny Rolling Friction coefficient isn't a constant if you calculate at precise level which is important for your situation. You can also see that you'll burn more gasoline with low tyre pressure. Any more question? Nov 23 '14 at 10:05
• @JohnyDiala: While running, you lose a lot more energy while running because of the hitting of bones against one another, and because the muscles dissipate energy continuously. Take a look at this. Nov 23 '14 at 17:13
• Air resistance and drive train resistance are also both major factors (for example of the latter, roll while in neutral compared to rolling while in gear; this is why down-shifting is an effective means to slow the car, and will produce significantly more resistance than air resistance or rolling friction). Nov 23 '14 at 22:54

You have ignored the car friction with air! If we assume the car as an aerodynamic body (air flows on the car surface smoothly and does not separate) then the air friction on the car is proportional to the square of car speed.

Consider that you travelled in your car at $10~{km}/{h}$ for one hour, then at $100~{km}/{h}$ for the next hour.

• First hour of the Journey: You travelled a distance of 10km, so the work done is $$W=F\times s=10F$$

• Second hour of the journey: You travelled a distance of 100km, so the work done is $$W=F\times s=100F$$

The work done is more when travelling at 100kmph for the same period, and so energy consumed is more when the velocity is 100kmph. More fuel needs to be burnt to provide the energy and keep the velocity constant at $100km/h$ for the same time duration (one hour), than at $10km/h$.

## Another Approach

It can also be stated in the following way. When the car is moving at 10kmph, it has less $KE$ than when it is moving at 100kmph. But resistance forces (friction, form drag, etc.) increase with increase in velocity. So the resistance forces are less on the car at 10kmph. So they cause less retardation in it, hence less energy is lost and less energy needs to be replaced by the car engine. But when the car is moving at 100kmph, more frictional force results in more subtraction of velocity, hence more energy is lost, and more fuel needs to be burnt to supply and replace the 'lost energy'

Note: I haven't considered various other factors responsible. What I wanted to point out was that even if the frictional force is the same in both the cases, the energy consumed isn't. Perhaps this was what the OP had asked. Refer to Sachin's and Ilmari's answers for a more practical approach.

• This answer is plain simple wrong without the use of rolling friction. If tyres are slipping, there won't be that much displacement despite the same power consumption. You ignored the heart of the problem. Nov 23 '14 at 11:10
• @SachinShekhar I haven't mentioned about rolling friction because what the OP simply asked was - "why does moving at 100 km/h use up more gasoline?"
– user49111
Nov 23 '14 at 13:18
• I don't think you've answered it. Nov 23 '14 at 16:39
• @ThePragmatick: Why don't you tell what is the force required to drive the car, as per what the OP asks ? Nov 23 '14 at 17:12
• The OP is specifically talking about constant speeds; your answer needs another question. Nov 23 '14 at 17:50

You are right

By Newtons Second law

$$\ F = m (v-u)/t$$

Once you have attained the speed of $$\ 100 m/s$$ You only need to supply the force to counteract the friction. But attaining to this speed requires more force and thus more energy.

We can conclude that riding on the same road at a constant speed of any magnitude does't require more energy.

Let the applied force be $F$ and the force of friction be $f$ the remaining force be $N$

$$\ N = F- f$$

the remaining force should be $0$ the speed of the car does not increase further gives $F=f$ which confirms that the force must be equal to friction force to maintain a constant speed.

It does require gasoline only to attain this speed. After attaining this speed the amount of gasoline used by both the cars is the same.

Adding what @wgrenard says, You would not lose your gasoline for travelling identical distances. But the case is ideal, no air resistance, the vehicles must have attained the constant speeds.

• "We can conclude that riding on the same road at a constant speed of any magnitude does't require more energy." But it does. Maintaining a larger constant speed requires more energy than maintaining a lower constant speed. Nov 23 '14 at 7:03
• @wgrenard and how ? on what basis ? neglect air resistance Nov 23 '14 at 7:05
• imakesmalltalk explains it in his answer. The work required to maintain a constant velocity for a given amount of time increases with increased velocity. Nov 23 '14 at 7:10
• @PranavHosangadi I have considered your opinion in the question. I hope it is useful. Nov 23 '14 at 7:12
• The extra work comes not because you are applying more force. We agree that (neglecting factors such as air resistance), the magnitude of the force applied in both situations is the same. Rather, the extra work shows up because over any given period of time you are applying that force over a larger distance (because you travel further in the same amount of time at a higher velocity). This is why it takes more work to maintain higher velocity as compared to lower velocity over an identical period of time. Nov 23 '14 at 7:18

power = force × velocity so at 10 km/h , power is 4000 N × 10 km/h while at 100 km/h, power is 4000 N × 100 km/h this way fuel will b consumed approx 10 times faster at 100 km/h than at 10 km/h