Total distance a car moves with two coefficients of friction?

[juicyfruit]

A 1000 kg car is moving at 10 m/s, the car applies the brakes and begins to skid and leave a mark. the first 4 m of the surface has a coefficient of 0.35 and after that the coefficient of friction is 0.2. What is the total distance the car will slide when coming to rest?


Equations: F=ma and F(f)=μN
Δχ=V₀t + 1/2at²
Vf=V₀ + at
W=Fxd
W=1/2mV_F^2 - 1/2mV₀^2


Since the car is going to rest, I'm going to assume the final velocity is 0. Since there are no other Y axis forces, I think you can assume N=mg so it equals 9800N. I tried using the F=ma of the x components and got -(9800*.35)+(9800+.2)=ma so the acceleration was -5.39 m/s^2 for the whole system. Plugged that into Vf=Vo+at to get 1.86 seconds and plugged both into the second equation to get 9.27m for ΔX.

That wasn't any of the possible answer choices, any help is appreciated.

 

[Rugile]

-(9800*.35)+(9800+.2)=ma

In this equation, what are the forces on the left side? Does this equation match the problem statement? (I assume you meant to write -(9800*.35)+(9800*.2)=ma instead of the +).

By the way, I recommend to derive the final expression of your solution before putting the numbers in. That way, it is easier to find your mistake.

 

[juicyfruit]

So I took another shot at it, different approach, please correct me if I'm wrong. So since work is the change in potential energy, I think you can assume W=1/2mv^2. So W=sum of F*d in the system and I'm assuming the only forces are the frictional force since there's no external force pushing the car.

So Ff=9800(.35)(4)+9800(.2)d and that is equal to work
E=.5(1000)(10^2)
W=1/2mv0^2 since it comes to a stop? you can assume Vf=0 so you have just v0
1,960d+13,720=50,000
d=18.5m

If I made any wrong assumptions or any mistakes, please point them out. I appreciate any help or confirmations, thank you

 

[PhanthomJay]

So I took another shot at it, different approach, please correct me if I'm wrong. So since work is the change in potential energy, I think you can assume W=1/2mv^2. So W=sum of F*d in the system and I'm assuming the only forces are the frictional force since there's no external force pushing the car.

So Ff=9800(.35)(4)+9800(.2)d and that is equal to work
E=.5(1000)(10^2)
W=1/2mv0^2 since it comes to a stop? you can assume Vf=0 so you have just v0
1,960d+13,720=50,000
d=18.5m

If I made any wrong assumptions or any mistakes, please point them out. I appreciate any help or confirmations, thank you

Your first method was wrong because forces and accelerations are different during each part of the slide. Your second approach is good but you calculated d for the 2nd part. Total stopping distance is ?

 

[juicyfruit]

Your first method was wrong because forces and accelerations are different during each part of the slide. Your second approach is good but you calculated d for the 2nd part. Total stopping distance is ?

Would it just be 18.5+4? You add the distances together? Then you'd get 22.5m but the answer choices I have are:

3.5 m
24.8 m
8 m
18.4 m

Which is why I thought it had to be 18.4 since that was close to my first answer. I know it asks for total, so either it's a mistake on my part or the instructors.

 

[PhanthomJay]

Would it just be 18.5+4? You add the distances together? Then you'd get 22.5m but the answer choices I have are:

3.5 m
24.8 m
8 m
18.4 m

Which is why I thought it had to be 18.4 since that was close to my first answer. I know it asks for total, so either it's a mistake on my part or the instructors.

looks like an answer key error. Clearly from your calc the distance d is the distance traveled overt the surface with the 0.2 friction coefficient. You've for to add that value to the first 4 m traveled to get the total distance of the slide.