Friction problem with hardly any variables given

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Homework Help Overview

The problem involves determining the stopping distance of a car moving at 100 km/h on an incline, given a known stopping distance on level ground. The scenario includes two cases: one where the car moves down a 10-degree incline and another where it moves up the same incline. The challenge arises from the lack of specific values for the coefficient of friction and mass.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations and the need to consider gravitational components on the incline. There are questions about how to proceed without mass or friction values, and some suggest manipulating the known acceleration on level ground to find the stopping distance on the incline.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have proposed methods to derive the necessary components for solving the problem, while others express uncertainty about the requirements of the question, questioning whether a quantitative or qualitative answer is expected.

Contextual Notes

Participants note the absence of key variables such as mass and the coefficient of friction, which complicates the analysis. There is also a mention of the need to clarify the nature of the answer required by the problem statement.

mooney82
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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.
 
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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.
 
w3390 said:
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.
 
Are you sure that the question wants a quantitative answer and not a qualitative answer?
 
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?
 
w3390 said:
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!
 

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