Determining Coefficient of Kinetic Friction w/m, t, and v

[crushedcorn]

Homework Statement

A 2.0 kg block of wood slides over a horizontal surface and comes to rest in 2.5 s. If its initial velocity was 9.0 m/s, what is the coefficient of kinetic friction?

Homework Equations

rho=mv
change in rho=F*change in t
Force of friction = mu*Force normal
Force normal=mg

The Attempt at a Solution

rho=(2.0 kg)(9.0 m/s)=18 kg*m/s
18 kg*m/s=Force of friction(2.5 s)
Force of friction=7.2
mu*Force normal=7.2
mu(2.0 kg)(9.80 m/s squared)=7.2
mu=0.37

I don't know what the correct answer to this question is and I potentially may not know until the day of my exam. Can anyone tell me if I'm doing this correctly or if I'm missing something? I really appreciated it!

 

[berkeman]

Homework Statement

A 2.0 kg block of wood slides over a horizontal surface and comes to rest in 2.5 s. If its initial velocity was 9.0 m/s, what is the coefficient of kinetic friction?

Homework Equations

rho=mv
change in rho=F*change in t
Force of friction = mu*Force normal
Force normal=mg

The Attempt at a Solution

rho=(2.0 kg)(9.0 m/s)=18 kg*m/s
18 kg*m/s=Force of friction(2.5 s)
Force of friction=7.2
mu*Force normal=7.2
mu(2.0 kg)(9.80 m/s squared)=7.2
mu=0.37

I don't know what the correct answer to this question is and I potentially may not know until the day of my exam. Can anyone tell me if I'm doing this correctly or if I'm missing something? I really appreciated it!

Looks good to me.

F = ma = μN = μmg

a = μg

μ = a/g = [(9m/s)/(2.5s)]/(9.8m/s^2) = 0.367

 

[crushedcorn]

Looks good to me.

F = ma = μN = μmg

a = μg

μ = a/g = [(9m/s)/(2.5s)]/(9.8m/s^2) = 0.367

Thanks!