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Discussion Overview

The discussion revolves around a physics problem involving forces, friction, and motion, specifically calculating the minimum stopping distance of a car on a rainy day given its initial speed and the coefficient of friction. The scope includes mathematical reasoning and conceptual clarification related to the application of Newton's laws of motion.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • Post 1 presents the problem setup, including initial and final velocities, gravitational acceleration, and the coefficient of friction, while seeking assistance on how to approach the calculations.
  • Post 2 asks for clarification on the normal force, friction force, and acceleration, indicating a need for foundational understanding.
  • Post 3 attempts to break down the problem, calculating the normal force and friction force, and expresses uncertainty about the correctness of the results.
  • Post 4 reiterates the equation for distance in terms of initial and final velocities and acceleration, while also providing a value for acceleration.
  • Post 5 confirms the calculations presented in Post 3, noting that the mass cancels out in the equations, which is a relevant point in understanding the dynamics involved.
  • Post 6 expresses enthusiasm and satisfaction upon receiving help, indicating a positive engagement with the topic.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the problem, but there remains some uncertainty regarding the calculations and the interpretation of forces involved. The discussion does not reach a definitive consensus on the methodology, as some participants express doubt about their understanding.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the forces acting on the car, particularly the treatment of mass and its effect on the calculations. The participants do not fully resolve the mathematical steps involved in deriving the stopping distance.

Who May Find This Useful

This discussion may be useful for students or individuals interested in understanding the application of physics principles related to motion, forces, and friction in practical scenarios.

kza62
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A car is traveling at 40.7 mi/h on a horizontal highway.
The acceleration of gravity is 9.8 m/s/s.
If the coefficient of friction between road and tires on a rainy day is 0.12, what is the minimum distance in which the car will stop?
(1 mi = 1.609 km)

So given:
vi(initial velocity): 40.7 mi/h -> 65486.3m/3600s
vf(final velocity): 0 m/s
g(gravity): -9.8
Ff(friction): -0.12
m(mass): ?
Fnet(net force): ?
x(distance): ?

Equations:
F = ma
Fnet = ma - Ff
Ff = uFn
Fn = mg
vf(^2) = vi(^2) + 2ax

Answer in back of textbook: 140.689 m

I am unsure how to set this problem up. All my attempts have had no success. Please, help would be appreciated.
 
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Welcome to PF!

Hi kza62! Welcome to PF! :smile:

(try using the X2 tag just above the Reply box :wink:)

i] what is the normal force?

ii] what is the friction force?

iii] what is the acceleration? :smile:
 
ok, i think i figured it out but I'm not sure how let me try to break it down again
so if Fgravity is -9.8 then the Fnormal is 9.8?
so is Ffriction = u * Fnormal?
which is 1.176 is a?
and then i used
vf - vi / 2a
which gave me 140.689

did i get it or was this a lucky guess?
 
*** vf^2 - vi^2 / 2a = x
and a = -1.176
 
HI kza62kza62 ! :smile:

(just got up :zzz: …)
kza62 said:
ok, i think i figured it out but I'm not sure how let me try to break it down again
so if Fgravity is -9.8 then the Fnormal is 9.8?
so is Ffriction = u * Fnormal?
which is 1.176 is a?
…
which gave me 140.689
kza62 said:
*** vf^2 - vi^2 / 2a = x
and a = -1.176

Yes, that's fine!

(btw, the forces would be mass times g, or times µFn, so what you found were the accelerations, but since the two masses cancel, it makes no difference :wink:)
 
WOOHOO! thanks for ur help! i love physics when I get it :D
 

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