How do I find the coefficient of friction?

[SpecialOps0]

Homework Statement

Please help this is due tomorrow. I was made to do an experiment and here it is.

I had to designate a fixed starting point, drive at 10 mph until that point, put the car into neutral and see how far it drifted until it stopped. Then I was instructed to go the opposite way and do the same thing.

Average rolling distance: 55.125 meters
Weight of the car: 1649.7154 kg, or 1617.2109 Newtons
Radius of tires: .333 meters
Time Taken average: 34 seconds

I'm suppose to find the average negative acceleration of my car, the rolling friction during the experiment and the coefficient of rolling friction.

Homework Equations

N/A, show if you used different..

The Attempt at a Solution

My attempt is below I'm not sure how to get much more than the average negative acceleration.

10 mph = 4.5 m/s (for convenience)

Average neg accel: (0-4.5m/s)/34 seconds = -.1324 m/s^2

Frictional force: 16167.2109N * .1324 = 2190.5387N

Coefficient of rolling friction: (10^2)/(2*9.8*55.125) = .0926N

Again I'm confused on coeff of rolling friction

 

[Sourabh N]

Homework Statement

The Attempt at a Solution

My attempt is below I'm not sure how to get much more than the average negative acceleration.

10 mph = 4.5 m/s (for convenience)

Average neg accel: (0-4.5m/s)/34 seconds = -.1324 m/s^2

Frictional force: 16167.2109N * .1324 = 2190.5387N

Coefficient of rolling friction: (10^2)/(2*9.8*55.125) = .0926N

Again I'm confused on coeff of rolling friction

Force is mass times acceleration, so instead of 16167.2109 N, you should use 1649.7154 kg.

Coefficient of friction is simply the ratio of frictional force and normal force. Normal force is equal to gravitational force (why?)

 

[SpecialOps0]

So it's 1649.7154 * .1324 = 218.4223N for the force?

and 218.422/16167.2109 = .0135 For the frictional rolling?

My teacher never explained this and I can't find it in my textbook other than in more advanced chapters I don't understand.

 

[Sourabh N]

Yes, that looks right.

 

[SpecialOps0]

Thank you all who helped :D this is now my new favorite site haha.

 

[SpecialOps0]

Ooops I meant
218.4223N is the rolling force?

.0135 is the coefficient ?

 

[PhanthomJay]

Yes, as Sourabah N noted, that looks correct. Except you should round off your answers to 2 significant figures. 220 N is the rolling friction force, and 0.014 is the rolling friction coefficient.

 

[SpecialOps0]

Thanks again sorry I mixed up my labeling before.

Does the .014 have a label as N?

 

[PhanthomJay]

Thanks again sorry I mixed up my labeling before.

Does the .014 have a label as N?

Force is mass times acceleration, so instead of 16167.2109 N, you should use 1649.7154 kg.

Coefficient of friction is simply the ratio of frictional force and normal force. Normal force is equal to gravitational force (why?)

Again as noted above, the coefficient is force unit divided by force unit, so does it have any units?

 

[SpecialOps0]

No, ok ty :).