Finding coeffiecient of friction

AI Thread Summary
To find the coefficient of friction for a 2.85kg book sliding at an initial speed of 2.98m/s and coming to rest after 3.80m, the relevant equations include kinematic equations and the work-energy principle. The initial attempt to calculate acceleration using the equation vf^2 = vo^2 + ad was criticized for not properly accounting for the squared initial speed. A solution using conservation of energy yielded a coefficient of friction (μ) of approximately 0.16, while another method suggested a slightly lower value of around 0.12 due to calculation approximations. The mass of the book is deemed irrelevant in determining the coefficient of friction. Accurate calculations and proper application of formulas are essential for obtaining the correct coefficient of friction.
ross moldvoer
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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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I think you'll need to explain your attempt. What were you trying to accomplish?
 
i tried solving for a in order to use f=ma
 
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.
 
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
 
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
 
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.
 
That could be possible because of certain approximations in the calculation, like value of "g", or something else, i guess.
 
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.
 
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