Force of Friction: Car Skids on Ice/Snow

[jer_hall99]

Homework Statement

13. As a car skids with its wheels locked trying to stop on a road covered with ice and snow, the force of friction between the icy road and the tires will usually be: (Please note that “normal” and “perpendicular” have the same meaning)

A. greater than the normal force of the road times the coefficient of static friction
B. equal to the normal force of the road times the coefficient of static friction
C. less than the normal force of the road times the coefficient of kinetic friction
D. greater than the normal force of the road times the coefficient of kinetic friction
E. equal to the normal force of the road times the coefficient of kinetic friction.

Homework Equations

Fk=μkn

The Attempt at a Solution

I know that it is kinetic friction since the car is moving. I think that the answer is "C" because the force of friction would be less since the car is skidding. Not sure though, could somebody either confirm my answer or clear up my confusion please?

 

[TaxOnFear]

Is this a trick question?

You have the equation right in front of you:

F_R = μ_KN

Is this not your answer? Or am I reading the question wrong?

 

[jer_hall99]

It's really not a quick question... I am really just unsure of myself... I take that from your comment you are trying to tell me that the answer then is "E"? Please confirm if the answer is E then. Thank you.

 

[TaxOnFear]

I don't like this question. Do they mean the coefficient of kinetic friction while skidding or not? It could be either or depending on the context.

 

[jer_hall99]

Not sure. I am guessing it would be "E" since the more I think about it, I put "C" on the test and got it wrong.

 

[TaxOnFear]

Harsh. They could've given you more information.

 

[TaxOnFear]

Ahh yes it is E if you think about it, the frictional force is the same as if it wasn't skidding, however the decceleration due to braking is gone because the tyres are skidding, therefore it takes longer to stop.

 

[jer_hall99]

Thank you.