Friction

[msesulka]

A 1500-kilogram automobile travels at a speed of 90 km/h along a straight concrete highway. Faced with an emergency situation, the driver jams on the brakes , and the car skids to a stop. What will the stopping distance be for (a) dry pavement and (b) wet pavement?

[Hootenanny]

In future could you please post your homework questions in the homework forums, thanks.

With respect to this question, what are your thoughts? What have you attempted? Is this all the data given in the question?

[cristo]

This looks like homework. If it is, please note that you are required to show working before we can help; so, what are your thoughts?

N.B. In future, please post in the homework forum.

[msesulka]

sorry about that i don't really know how to get around on this site really. So far i know that the given coefficiants of kinetic friction for rubber on dry concrete is 0.85 and for rubber on wet concrete is 0.60. I know you somehow have to add all the forces in the problem using F=(m)(g) and F=(m)(a) but i dont know how to get those components

[cristo]

F=mg gives the weight of the truck. Do you know the equation F=\muR, where \mu is the coefficient of friction, and R is the normal contact force. This gives you the force of friction.

[msesulka]

Yea I actually have that formula. So would I find the force of friction for the wet pavement and then the force of friction for the dry pavement? Then how would I incorporate those answers into finding the stopping distance?

[cristo]

Try using conservation of energy. To start the car has kinetic energy, at the end it has zero (since it's at rest). How can you relate this change in energy to the frictional force?