Accident Reconstruction; Work-Energy Problem

[uchicago2012]

Homework Statement

Accident Reconstruction.
A car, with a mass of 1600 kg, is involved in a collision. By using techniques from later chapters, we conclude it was moving 4 m/s at impact. There are skid marks leading into the collision, 18 meters long from two of the tires; no marks from the other two. If the coefficient of friction from the pavement is .92, how fast was the car going before skidding?

Homework Equations

W_non = Change in KE + Change in PE

The Attempt at a Solution

I found the work done by the force of friction so I can use this equation:

W_non = KE₂ - KE₁ + PE₂ - PE₁
where W_non = work done by friction
but I was wondering if there were any other nonconservative forces I had to worry about. In these types of problems do you only account for things like air resistance if it specifically tells you to or should I just assume air resistance is part of the W_non?

Also, since the car is on a flat surface and stays on a flat surface, Change in PE = 0, correct?

I guess I just expected this problem to be more difficult and I'm nervous I've missed something.

 

[renee1234]

dude i think I'm in your class, that is if you go to USI. this is what i got not sure if it's right though.

Fr = μN

N = the normal force

Since only two tires were involved in the skid, the normal force is not going to equal the entire weight of the car, assuming the car is perfectly symmetrical, it will be half.

Fr = μ(½mg)

Energy = ĥd = (0.92)(1/2)(1600)(9.81)(18) = 129830.4 J

The same amount of energy expressed as a difference in kinetic energy...

½m(v1)² - ½m(v2)²

v1 = starting velocity
v2 = final velocity

129830.4 = ½(1600)(v1)² - ½(1600)(4)²

v1 = 13.4 m/s
this could be wrong and I'm not sure if we can use the idea of using 1/2 the weight of the car to find the Normal force.

 

[Porter68]

Wouldnt it be 1/4 of the weight of the car per tire instead of 1/2 of the weight of the car per tire?

 

[renee1234]

It is. It's 1/4 the weight per tire but there's two skid marks so there's two tires so it's 1/2 the weight of the car.
I'm just not sure we can assume that we use half the weight instead of all the weight.

 

[uchicago2012]

That's an interesting idea and it is a bit odd how it specifically mentions there are only two tire tracks. But for a rear wheel drive car there would only be two tire tracks and I don't know that you would calculate the weight differently for a rear wheel drive than for say an all wheel drive car.

I feel like there are two sources of friction, one from each tire. So perhaps calculate the friction for each tire using one fourth of the weight of the car as the normal force? Which is pretty much the same thing you did by halving the weight.

Or would we calculate the friction from each tire using the entire weight of the car?

Hmm.

Anyone have any suggestions?

 

[Porter68]

I believe the way renee did the problem is correct. Thats going to be my answer.