Finding friction force *without* mass or coefficient

In summary, the problem involves a car moving at 50mph coming to a stop while traveling 150ft on level ground. The question asks for the size of the friction force needed to bring the car to rest. Using the equations vf^2 = vo^2 + 2a(xf-xo) and F=ma, the initial velocity of 22.35 m/s and distance traveled of 47.7 meters were converted to meters and seconds. The resulting acceleration of -5.24 m/ss and time of 4.27 seconds were used to find the coefficient of friction, which was found to be 0.53. However, since the problem only asks for the friction force and not the coefficient, more information
  • #1
Amy G.
4
0

Homework Statement


A car moving at 50mph coasts to a stop while traveling a distance of 150ft on level ground. what is the size of the friction force needed to bring this car to rest?

Homework Equations


vf^2 = vo^2 + 2a(xf-xo)
F=ma
Fk=un[/B]

The Attempt at a Solution


I first converted everything to meters and seconds.
So, initial velocity is 22.35 m/s and distance traveled is 47.7 meters.
Then, I found acceleration, which is -5.24 m/ss.
Then, I found time. 4.27 seconds

I tired to find the coefficient of friction with:
F=ma=u(mg) (n is the same as mg because the surface is level)
ma=u(mg)
a=ug
-5.24=u(-9.8)
u=0.53

I have no idea what to do now. It wants the friction force, not the coefficient.
those are pretty much the only formulas we have at this point. We haven't gotten to work or energy formulas.
 
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  • #2
Amy G. said:

Homework Statement


A car moving at 50mph coasts to a stop while traveling a distance of 150ft on level ground. what is the size of the friction force needed to bring this car to rest?

Homework Equations


vf^2 = vo^2 + 2a(xf-xo)
F=ma
Fk=un[/B]

The Attempt at a Solution


I first converted everything to meters and seconds.
So, initial velocity is 22.35 m/s and distance traveled is 47.7 meters.
Then, I found acceleration, which is -5.24 m/ss.
Then, I found time. 4.27 seconds

I tired to find the coefficient of friction with:
F=ma=u(mg) (n is the same as mg because the surface is level)
ma=u(mg)
a=ug
-5.24=u(-9.8)
u=0.53

I have no idea what to do now. It wants the friction force, not the coefficient.
those are pretty much the only formulas we have at this point. We haven't gotten to work or energy formulas.

I agree that the force needed will depend on the mass of the car. Are you sure that is the whole problem statement? Can you post a (clear) picture or screenshot of the problem statement that you were given?
 
  • #3
berkeman said:
I agree that the force needed will depend on the mass of the car. Are you sure that is the whole problem statement? Can you post a (clear) picture or screenshot of the problem statement that you were given?

I tried, but can't do it. I have photographs of it and my work saved on my computer but I can't upload them on here. I don't really have any place on the internet where I keep pictures either

I looked at it again though, and it definitely says that :frown:
 
  • #4
Amy G. said:
I tried, but can't do it. I have photographs of it and my work saved on my computer but I can't upload them on here. I don't really have any place on the internet where I keep pictures either

I looked at it again though, and it definitely says that :frown:

I'll PM you my e-mail address, in case you can e-mail them to me and I can post them for you.
 
  • #5
Maybe all they want is the force in g's.

Chet
 
  • #6
Chestermiller said:
Maybe all they want is the force in g's.

Chet

Yeah, that's kind of what I'm guessing now. There certainly is no mention of the car's mass in the problem statement, so I guess it just needs to show up in the answer as Amy did...

PF Amy Problem.jpg


PF Amy Work.jpg
 
  • #7
BTW, Amy -- your files were JPG files, so you should be able to click the UPLOAD button in the Reply window to upload a *lJPG file as an attachment.

Also, we usually do not want your work to be an image, and prefer that you type it into the PF directly. Just a note for future posts... :smile:
 
  • #8
Your answer could be expressed as "some specific fraction of the magnitude of the car's weight."
 

1. How can friction force be calculated without knowing the mass or coefficient of friction?

The friction force can be calculated using the formula F = μN, where μ is the coefficient of friction and N is the normal force. The normal force can be determined by using the equation N = mg, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, by knowing the normal force, the coefficient of friction, and the formula for friction force, the mass is not necessary for calculation.

2. What is the significance of the coefficient of friction in calculating friction force?

The coefficient of friction is a constant value that represents the amount of friction between two surfaces in contact. It is dependent on the materials of the surfaces and their roughness. The higher the coefficient of friction, the greater the friction force will be.

3. Can friction force be negative without knowing the mass or coefficient of friction?

No, friction force cannot be negative. It always acts in the opposite direction to the motion, and its magnitude cannot exceed the force applied to the object. Therefore, even without knowing the mass or coefficient of friction, the friction force will always be positive or zero.

4. How does the angle of the incline affect the calculation of friction force without mass or coefficient?

The angle of the incline does not affect the calculation of friction force without mass or coefficient. The normal force and the coefficient of friction are the only variables needed to calculate friction force, and they are not affected by the angle of the incline.

5. Can friction force be greater than the weight of the object without knowing the mass or coefficient of friction?

Yes, friction force can be greater than the weight of the object without knowing the mass or coefficient of friction. This can happen if the coefficient of friction is high enough, or if the normal force is greater than the weight of the object. In these cases, the friction force will oppose the motion and prevent the object from sliding.

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