Newtons Laws car deacceleration

[Beanie]

Homework Statement

A 1900 kg car moves along a horizontal road at speed v0 = 23.6 m/s. The road is wet, so the static friction coefficient between the tires and the road is only μs = 0.218 and the kinetic friction coefficient is even lower, μk = 0.1526.

The acceleration of gravity is 9.8 m/s2 .

Assume: No aerodynamic forces; g = 9.8 m/s2, forward is the positive direction.

What is the highest possible deceleration of the car under such conditions?

Answer in units of m/s2.

Homework Equations

Ff=mu*Fn
Sum of all Forces = ma

The Attempt at a Solution

m = 1900kg
Vi=23.6m/s = Fp
mus=0.218
muk=0.1526
ay=9.8m/s

Fn=(9.8)(1900)=18620
Ff=muk*Fn
Ff=(0.1526)(18620)
Ff=2841.412N

Sum of all forces=ma
Ff+Fp=ma
(2841.412)+Fp=(1900)a

I ran into many problems with this. First of all, in step 1 when using the Ff=mu*Fn I didn't know whether to use the coefficient of static friction or the coefficient of kinetic friction. I assumed it was the coefficient of kinetic friction because the car was already in motion. Is this right?

Also, when using the F=ma equation, I was trying to use all of the forces in the x direction to find the acceleration in the x direction. This meant that only the force of friction and the force of the object moving to the right were acting upon the object for this equation. I had already calculated the force of friction, however I couldn't calculate the force of the object moving towards the right, because I only had the initial velocity of the object moving towards the right and you can't convert velocity to force. Where do I go next from here?

 

[brainpushups]

I didn't know whether to use the coefficient of static friction or the coefficient of kinetic friction. I assumed it was the coefficient of kinetic friction because the car was already in motion. Is this right?

This is a common confusion. The term 'static' and 'kinetic' in the case of friction refers to whether there is relative motion between the surfaces in contact. Unless the car is skidding the friction between the surfaces would be static.

This meant that only the force of friction and the force of the object moving to the right were acting upon the object for this equation. I had already calculated the force of friction, however I couldn't calculate the force of the object moving towards the right, because I only had the initial velocity of the object moving towards the right and you can't convert velocity to force.

This is another common confusion. Around Newton's time people used the word force to mean various things, but the language has been cleared up. Objects do not have force by virtue of their motion - they have momentum. Forces change momentum. If the surfaces are horizontal then the net horizontal force will simply be due to friction.

 

[paisiello2]

I ran into many problems with this. First of all, in step 1 when using the Ff=mu*Fn I didn't know whether to use the coefficient of static friction or the coefficient of kinetic friction. I assumed it was the coefficient of kinetic friction because the car was already in motion. Is this right?

The car is in motion, yes, but what about the tires relative to the road surface?

 

[Beanie]

This is a common confusion. The term 'static' and 'kinetic' in the case of friction refers to whether there is relative motion between the surfaces in contact. Unless the car is skidding the friction between the surfaces would be static.
This is another common confusion. Around Newton's time people used the word force to mean various things, but the language has been cleared up. Objects do not have force by virtue of their motion - they have momentum. Forces change momentum. If the surfaces are horizontal then the net horizontal force will simply be due to friction.

Okay, this makes more sense.

So...

Ff=mus*Fn
Ff=0.218*18620
Ff=4059.16

And then because sum of all forces is the force of friction,

4059.16=(1900)a
a=2.1364m/s^2

 

[brainpushups]

Looks okay. One comment I would like to point out because I'm guessing you haven't thought of it: You have given the acceleration a positive value (which is fine), but recognize that means the initial velocity must then be given a negative value. For any puzzle involving Newton's laws I would highly recommend making a clear choice of coordinate system in your initial drawing and labeling each of the forces with the appropriate sign. I have a feeling that if you were to do this carefully you would probably give the object a positive velocity and, since friction opposes the direction of motion, a corresponding negative sign for the force of friction (and hence negative acceleration).