Friction problem with hardly any variables given

[mooney82]

Homework Statement

The minimum stopping distance for a car from an initial 100km/h is 60 m on level ground. What is the stopping distance when it moves (a) down a 10 degree incline; (b) up a 10 degree incline? Assume the initial speed and the surface are unchanged.

Homework Equations

Fk=uk*N

The Attempt at a Solution

I have no clue how to solve this. There is no uk value given, no mass given, or a force. I can find the acceleration of the car on a level surface (-6.431) but not sure how to find force without mass, how to find the N value without the mass or the uk value.

 

[w3390]

If that is all you are given, maybe try using the kinematic equations on the angle of the plane. What I'm saying is find the component of acceleration due to gravity in the plane of the inclined plane and try manipulating the two accelerations to get a correct stopping distance.

 

[mooney82]

If that is all you are given, maybe try using the kinematic equations on the angle of the plane. What I'm saying is find the component of acceleration due to gravity in the plane of the inclined plane and try manipulating the two accelerations to get a correct stopping distance.

But without the mass you can't find out the acceleration due to gravity.
Without the force you can't find the mass.

 

[w3390]

Are you sure that the question wants a quantitative answer and not a qualitative answer?

 

[w3390]

Actually, try this. When the car is on the incline plane, there is a force that wants to pull the car down the plane and that force is mgsin(theta). However, the car is also stopping or decelerating. We found that on flat ground, a=-6.43. So on the inclined plane to find the component along the plane, we must multiply by cos(theta). The force of this will be the cars mass, m, times the acceleration, -6.43cos(theta). Therefore, your net force will be mgsin(theta)-6.43m=ma. You can then solve this for a and the m's will cancel. Then plugging in this a into the kinematic equation, you can find the stopping distance. Understand?

 

[mooney82]

Actually, try this. When the car is on the incline plane, there is a force that wants to pull the car down the plane and that force is mgsin(theta). However, the car is also stopping or decelerating. We found that on flat ground, a=-6.43. So on the inclined plane to find the component along the plane, we must multiply by cos(theta). The force of this will be the cars mass, m, times the acceleration, -6.43cos(theta). Therefore, your net force will be mgsin(theta)-6.43m=ma. You can then solve this for a and the m's will cancel. Then plugging in this a into the kinematic equation, you can find the stopping distance. Understand?

Ya, I think so. I'll try it tomorrow, thanks a lot!