# Finding the coefficient of frictional force

• drinkingstraw
In summary, the question is asking for the coefficient of sliding friction between the tires and the icy road when a 2500 Kg car traveling at a constant speed of 14 m/s comes to a halt in 25 m after the brakes are slammed. Using the equation F=ma, the acceleration is calculated to be 3.92 m/s^2. The force of friction is then determined to be 700 N. Using the formula F=uFn, the coefficient of sliding friction is calculated to be 0.4.

#### drinkingstraw

You are driving 2500 Kg car at a constant speed of 14 m/s along an icy, but stright and lvel road. While approaching a traffic light, it turns red. You slam on the brakes. You wheels lock, the tires begin skidding and the car slides to a halt in a distance of 25 m. What is the coefficient of sliding friction between the tires and the icy road?

-------------------------------------
Calculations:

m = 2500 Kg
Vi = 14 m/s
Vf = 0 m/s
deltaD = 25 m
a = (vf^2 - vi^2)/(2 x deltaD)
= -14 m/s / 50
= 2.8 m/s^2

W = (2500) (9.81)
= 24525 N

Fnet = (2500)(-0.28)
= -700 N

I`m not sure what to do after this.

Looks good so far.

You know that friction is the force producing this acceleration (and thus net force), so what is the equation for the force of friction?

Nabeshin said:
Looks good so far.

You know that friction is the force producing this acceleration (and thus net force), so what is the equation for the force of friction?

F = uFn

700 = u(24525)

Is that what I am supposed to be doing?

drinkingstraw said:
F = uFn

700 = u(24525)

Is that what I am supposed to be doing?

Yup. Have a little confidence :)

Nabeshin said:
Yup. Have a little confidence :)

I have an answer sheet and it says the answer is supposed to be 0.4 so I'm a little confused.

Sorry I wasn't checking your arithmetic, the approach was correct though.

The acceleration should be 3.92m/s^2:
$$14^2/(50)=3.92$$

Also, for some reason in your f=ma you used your a value divided by ten instead of the a you derived...

But yeah, see if you can get the .4 now.

Nabeshin said:
Sorry I wasn't checking your arithmetic, the approach was correct though.

The acceleration should be 3.92m/s^2:
$$14^2/(50)=3.92$$

Also, for some reason in your f=ma you used your a value divided by ten instead of the a you derived...

But yeah, see if you can get the .4 now.

Thank you so much :)

## 1. What is the coefficient of frictional force?

The coefficient of frictional force is a measure of the amount of resistance between two surfaces in contact. It is a dimensionless value that represents the ratio of the force required to move one surface over the other, to the force pressing the surfaces together.

## 2. How is the coefficient of frictional force calculated?

The coefficient of frictional force is calculated by dividing the force required to move one surface over the other by the force pressing the surfaces together. This can be done by using a frictional force sensor or by measuring the weight of an object on an inclined plane.

## 3. What factors affect the coefficient of frictional force?

The coefficient of frictional force can be influenced by factors such as the texture and roughness of the surfaces in contact, the type of material, and the presence of any lubricants or contaminants. It can also vary depending on the applied force and the temperature.

## 4. Why is it important to find the coefficient of frictional force?

The coefficient of frictional force is important in understanding the behavior of objects in contact and how much force is needed to move them. It is also useful in designing and improving the efficiency of machines and tools, as well as predicting and preventing potential accidents caused by friction.

## 5. How can the coefficient of frictional force be used in real-life applications?

The coefficient of frictional force is used in a variety of real-life applications, including designing car tires with optimal grip on roads, creating non-slip surfaces for shoes and floors, and determining the maximum weight a person can safely lift. It is also important in industries such as manufacturing, construction, and transportation, where minimizing friction is crucial for efficiency and safety.