The discussion revolves around determining the coefficient of friction between a book and the floor after it slides to a stop. The problem involves concepts from kinematics and energy conservation.
There is ongoing exploration of different approaches to the problem, with some participants providing feedback on the original poster's attempts. Guidance has been offered regarding the need to verify equations and consider approximations in calculations.
Participants note potential issues with approximations in calculations and the relevance of certain variables, such as mass and gravitational acceleration, in the context of the problem.
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
Going another route, I arrived at the same expression. But - I seem to have a slightly different answer: μ≈0.12.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
No, 0.16 looks too inaccurate for that. Note that the mass is irrelevant.jatin9953 said:That could be possible because of certain approximations in the calculation, like value of "g", or something else, i guess.