# I The coefficient of kinetic friction for a Ford Focus

1. Feb 14, 2018

### zapnthund50

Hi guys,

Yesterday I sat down, and just for fun, decided to calculate the coefficient of kinetic friction for the all electric Ford Focus. I keep getting the wrong answer, and it is driving me crazy. Here are the facts.
Battery: 23 kWh.
Curb weight: 1,651 kg.
Maximum travel distance before a recharge: 122 km

Very simply, we take 23 kWh*3600s = 82,800,000 J.
Then, the energy traveled per km is 82,800,000/122km = 679,000 J/km.
Then, force of friction for the entire car is Ff=uk*weight = 1651uk.

Then, we just say W= 679,000 = Fd = 1651uk*1000 meters,
and thus uk=0.411

But we have it from very reliable sources that the coefficient of kinetic friction for rubber on asphalt is around 0.7, so we are very low. Also, if we take into consideration wind resistance, rolling resistance, and battery discharge to 85% (I'm guessing here) instead of 100%, all these act to lower the coefficient, not raise it.

You can see my dilemma.

Thanks!

2. Feb 14, 2018

### phyzguy

What you calculated assumes the wheels are locked up and the tires are sliding across the pavement. I assume this isn't what you want. Since the wheels are rolling, there is no relative motion between the surface of the tire and the road surface, so the friction between the tire and the road does not dissipate any energy. The energy dissipated when the car is in motion comes from the other things you mentioned - mainly wind resistance and rolling resistance.

3. Feb 14, 2018

### sophiecentaur

It is not clear here what calculation you have used here and what the kinetic friction as to do directly with the energy needed to keep a car rolling. The kinetic friction is normally to do with braking distance in situations where you want to dissipate the Kinetic Energy of your moving vehicle is as few metres as possible. The rolling resistance is what's needed if you want to apportion and minimise the various losses for a moving car.
You may be able to help me with this but I suspect that your calculation just happens to have produced that figure of 0.411 but that it may not have any physical meaning. You do not seem to have considered air resistance which afaiaa, is the major loss mechanism at normal driving speed.

4. Feb 14, 2018

### CWatters

I think the force of friction you calculated is actually the total drag force on the car including air resistance, rolling resistance of the tyres, friction in motor and drive train etc.

5. Feb 14, 2018

### sophiecentaur

Agreed. What that calculation shows is that the total drag is the equivalent of having the brakes applied at a significant level, That's why a car slows down so quickly from. say 80mph to 70mph when you lift your foot off the pedal. As the speed drops, air drag reduces a lot and (surprise surprise) the mpg gets better and better!

6. Feb 14, 2018

### zapnthund50

I think you've nailed it. My problem was thinking kinetic, but I should have realized that a wheel uses the coefficient of static friction. That is, by design of the wheel, we are simply "resting" on the surface in terms of practical calculations, although we are moving forward.

Thanks, appreciate the help.

7. Feb 14, 2018

### zapnthund50

Yes, as other pointed out, I made the mistake of effectively calculated with the car brakes locked. Since a wheel uses static friction, no energy is lost during normal movement. The number I got is a result of rolling resistance, wind resistance, incomplete battery discharge, energy losses in electric motor, etc. Thanks for the help guys, problem solved.