Finding coeffiecient of friction

In summary: If you want to present your solution, you should start by stating what you did, and then justifying it.
  • #1
ross moldvoer
16
0

Homework Statement


You give a 2.85kg book an initial shove at 2.98m/s and it comes to rest after sliding 3.80m across the floor.
Find the coefficient of friction between book and floor.

Homework Equations


vf^2=vo^2+ad
f=ma

The Attempt at a Solution


0=2.98+a*3.8
=2.33
 
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  • #2
I think you'll need to explain your attempt. What were you trying to accomplish?
 
  • #3
i tried solving for a in order to use f=ma
 
  • #4
First, verify that your relevant equation is written correctly and not missing any constants.

Next, I don't see where in your attempt you accounted for squaring the initial speed. You don't seem to have applied the relevant equation.
 
  • #5
ross moldvoer said:

Homework Statement


You give a 2.85kg book an initial shove at 2.98m/s and it comes to rest after sliding 3.80m across the floor.
Find the coefficient of friction between book and floor.

Homework Equations


vf^2=vo^2+ad
f=ma

The Attempt at a Solution


0=2.98+a*3.8
=2.33
 
  • #6
I did with conservation of energy. By equating kinetic energy to the frictional work done,

1/2*mv^2=(mu)*N*d

Where m is mass
v is initial velocity
mu is coefficient of friction
N is normal force i.e. mg
d is distance travelled

I got an answer, mu = 0.16
 
  • #7
jatin9953 said:
I did with conservation of energy. By equating kinetic energy to the frictional work done,

1/2*mv^2=(mu)*N*d

Where m is mass
v is initial velocity
mu is coefficient of friction
N is normal force i.e. mg
d is distance travelled

I got an answer, mu = 0.16
Going another route, I arrived at the same expression. But - I seem to have a slightly different answer: μ≈0.12.
 
  • #8
That could be possible because of certain approximations in the calculation, like value of "g", or something else, i guess.
 
  • #9
jatin9953 said:
That could be possible because of certain approximations in the calculation, like value of "g", or something else, i guess.
No, 0.16 looks too inaccurate for that. Note that the mass is irrelevant.
But more seriously, you are not supposed to present something so much more Like a complete solution than the original poster has yet managed. Gneill's post is the right sort of hint.
 

FAQ: Finding coeffiecient of friction

What is the coefficient of friction?

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

How is the coefficient of friction determined?

The coefficient of friction can be determined experimentally by measuring the force required to move one surface over the other, divided by the weight of the object. It can also be calculated using the formula μ = F/N, where μ is the coefficient of friction, F is the force required to move the object, and N is the normal force between the two surfaces.

What factors affect the coefficient of friction?

The coefficient of friction is affected by several factors, such as the nature of the two surfaces in contact, the roughness of the surfaces, the force pressing the surfaces together, and the presence of any lubricants or contaminants on the surfaces.

How does the coefficient of friction impact motion?

The coefficient of friction plays a crucial role in determining the amount of force needed to move an object over a surface. Higher coefficients of friction result in more resistance and require a greater force to overcome, whereas lower coefficients of friction result in less resistance and require less force for motion.

What are some real-world applications of finding the coefficient of friction?

Finding the coefficient of friction is important in various fields, such as engineering, sports, and transportation. It is used to design structures with appropriate levels of friction for safety, to choose the right materials for different surfaces, and to optimize performance in sports equipment and vehicles.

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