Finding coeffiecient of friction

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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.