# Finding Kinetic Coefficient of Friction

• dzimme2
In summary, the homework statement states that a 2 kg. block is pulled up a 22° incline with a force of 10 N parallel to the incline. The equation for the force acting on the block is: F_{}N = Normal Force, F_{}k = μ_{}k F_{}N. The result of solving for μ_{}k is that it is 0.1257.
dzimme2

## Homework Statement

A 2 kg. block is pulled up a 22° incline with a force of 10N parallel to the incline. What is μ$_{}k$ if it moves up with constant speed?

## Homework Equations

F$_{}N$ = Normal Force
f$_{}k$ = μ$_{}k$ F$_{}N$

## The Attempt at a Solution

Forces perpendicular to plane: y-vector component of gravity, mgcos$\theta$ (negative) , normal force, F$_{}N$ (positive)

Forces parallel to plane: applied force, F (up plane) , kinetic frictional force, f$_{}k$ (up plane), and x-vector component of gravity, mgsin$\theta$ (down plane)

F$_{}net$$_{}y$ = F$_{}N$ - mgcos$\theta$ = 0
F$_{}N$ = mgcos$\theta$

F$_{}net$$_{}x$ = F - mgsin$\theta$ + f$_{}k$ = 0
f$_{}k$ = mgsin$\theta$ - F

μ$_{}k$mgcos$\theta$ = mgsin$\theta$ - F
μ$_{}k$ = [ mgsin$\theta$ - F ] / [ mgcos$\theta$ ]
μ$_{}k$ = ...

The given answer is μ$_{}k$ = 0.55 but I can not come up with this. Please help, thanks!

Last edited:
welcome to pf!

hi dzimme2! welcome to pf!
dzimme2 said:
F$_{}net$$_{}x$ = F - mgsin$\theta$ + f$_{}k$ = 0

(erm … the whole point of latex brackets is that you put things inside them! )

no, the friction is downhill, like the gravity, isn't it?

tiny-tim said:
hi dzimme2! welcome to pf!

Why thank you, I've frequented the site lately as a guest but finally decided to join. My thanks to you all!

(erm … the whole point of latex brackets is that you put things inside them! )

I'm not exactly sure as to what you are referring, but if it's regarding the 'Fnet' term, I was attempting to subscript "net" with the subscript "x".

no, the friction is downhill, like the gravity, isn't it?

I was under the impression that the kinetic frictional force would oppose the block's ability to slide down the plane due to gravity. While the applied force is pushing the box up the plane and gravity is pulling the block down the plane, wouldn't the frictional force be directed up the slope in opposition of the gravitational component?

hi dzimme2!
dzimme2 said:
I was under the impression that the kinetic frictional force would oppose the block's ability to slide down the plane due to gravity. While the applied force is pushing the box up the plane and gravity is pulling the block down the plane, wouldn't the frictional force be directed up the slope in opposition of the gravitational component?

ah, nope

the direction of kinetic friction is always opposite the actual motion

here, the actual motion is up, so the friction is down

(the direction of friction in the static case is more complicated, but in the kinetic case it's very simple!)

tiny-tim said:
hi dzimme2!

ah, nope

the direction of kinetic friction is always opposite the actual motion

here, the actual motion is up, so the friction is down

(the direction of friction in the static case is more complicated, but in the kinetic case it's very simple!)

Ohh, I see. Could prove to be the error with other problems as well. I'll try it out and see what I get. Thanks again!

After changing the sign of f$_{k}$ to restate : F$_{net}$$_{x}$ = F - mgsin$\theta$ - f$_{k}$ = 0;

f$_{k}$ = F - mgsin$\theta$
μ$_{k}$ F$_{N}$ = F - mgsin$\theta$
μ$_{k}$ mgcos$\theta$ = F - mgsin$\theta$
μ$_{k}$ = $\frac{F - mgsin\theta}{mgcos\theta}$
μ$_{k}$ = $\frac{(10 N) - (2kg)(9.8 m/s^2)sin 22}{ (2kg)(9.8m/s^2)cos 22}$

μ$_{k}$ = 0.1257

However, the answer key has μ$_{k}$ = 0.55 .

?

dzimme2 said:
However, the answer key has μ$_{k}$ = 0.55 .

?

i can't get 0.55

tiny-tim said:
i can't get 0.55

Do you get 0.1257?

dzimme2 said:
Do you get 0.1257?

i make it 0.146

## 1. What is the kinetic coefficient of friction?

The kinetic coefficient of friction is a measure of the force needed to overcome the resistance between two surfaces in motion. It is represented by the symbol μk and is a dimensionless quantity, typically ranging from 0 to 1.

## 2. How is the kinetic coefficient of friction determined?

The kinetic coefficient of friction can be determined experimentally by measuring the force required to keep an object moving at a constant velocity on a surface. This force is divided by the weight of the object to calculate the kinetic coefficient of friction.

## 3. What factors affect the kinetic coefficient of friction?

The kinetic coefficient of friction is affected by several factors, including the types of materials in contact, the smoothness of the surfaces, and the presence of any lubricants or contaminants.

## 4. What is the difference between kinetic and static friction?

Kinetic friction occurs when two surfaces are in motion relative to each other, while static friction occurs when two surfaces are at rest relative to each other. The kinetic coefficient of friction is typically lower than the static coefficient of friction for the same surfaces.

## 5. Why is the kinetic coefficient of friction important?

The kinetic coefficient of friction is important in many practical applications, such as designing machinery and determining the stopping distance of a moving object. It also plays a crucial role in understanding the physics of motion and friction.

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