Stopping Distance on dry pavement

[quick02si]

Hello everyone this is my first time in the forum and I'm actually having lot of trouble with my Physics homework. Hope someone can help me. Here's one of the problems and thanks in advanced.
If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at 29.7 ? Take the free fall acceleration to be 9.80 .

 

[Sirus]

Please include units with all your data whenever you post. Also show us how you have attempted the problem and where you got stuck.

 

[quick02si]

I'm sorry I didn't even notice that I missed the units.
Here is the problem again:
If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at 29.7m/s? Take the free fall acceleration to be 9.80m/s^2.
On this problem I don't know how to start it off. Thats where I get stuck. I don't know how to incorporate the kinetic friction to find distance.

 

[Sirus]

I think this is best done with work-energy concepts due to the lack of a value for mass of the car. What happens to the initial (kinetic) energy of the car while it skids?

 

[quick02si]

I think I see what you are saying but my problem is that I haven't completely gone through that yet. That is the next chapter. The chapter that is problem comes from is Applying Neton's Laws, but I'm stuck. Thanks for your help

 

[Sirus]

So they want you to do this with kinematics formulas?

 

[quick02si]

Yes, that's really the only thing i know. But still can't get the problem.

 

[Sirus]

You can do it if you know Newton's law $F_{net}=ma$. What is the net force acting on the car during the skid in the plane of motion (ignore normal force for now)?

 

[quick02si]

ok so the sum of the forces in the x direction would be the velocity-coefficient of kinetic friction=ma

 

[Sirus]

I don't think so. The only force acting on the car in the plane of its motion is the frictional force. Since $F_{kinetic friction}=\mu_{k}F_{N}$, and $F_{N}=mg$, you should be able to use F=ma and this information to find the acceleration of the car. Can you take it from there?

 

[quick02si]

ok so mu_{k}(mg)=ma so a=mu_{k}(g). Is this what you are trying to get me to see.

 

[Sirus]

Correct. Now consider the information you have, and apply an appropriate kinematics formula to find the distance.