# Finding coeffiecient of friction

## 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.

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

0=2.98+a*3.8
=2.33

## Answers and Replies

gneill
Mentor
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

gneill
Mentor
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.

## 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.

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

Svein
Science Advisor
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.

haruspex
Science Advisor
Homework Helper
Gold Member
2020 Award
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.