Find Displacement involving Coeff. Kin. Fr. & init. Velocity

[lynk]

Homework Statement

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.20 and the push imparts an initial speed of 4.0m/s^2?

Homework Equations

F(net)=ma will be used, but to find "a," we need the kinematic problem:
v(fin)^2=v(init)^2+2a(x(fin)-x(init))

The Attempt at a Solution

Known: v(init)=4.0m/s^2, v(fin)=0.0m/s^2, x(init)=0m, coeff kin=0.20
Unknown: a=?, x(fin)=? m=?

I'm not sure what to do, because there are too many unknowns. I know I need to find the acceleration, but am stumped because that displacement is also unknown. Without the acceleration, I cannot find the mass to figure out the normal force. I'm going in circles.
I am in an Algebra/Trig. based Physics I class with no prior physics experience. Any suggestions, please?

Thank you kindly,
Lynn

 

[Fightfish]

I think you are a little confused here. You first obtain a and then use the kinematic expression to find the total distance travelled. The push is just a one-off force that starts it off at an initial velocity, thereafter it does not affect the box anymore - friction is the only force that acts on the box throughout its 'journey'.

 

[lynk]

Ok. Thank you for your quick response.
?
I apologize. I still don't understand how to find accel, because I don't know the mass.
Am I going with: .20F(norm)=ma? I'm absolutely & definitively stumped. Am I missing some other way to determine normal force or mass?

Thank you,
Lynn

 

[Fightfish]

The box is in static equilibrium vertically, so normal force must be equal to ?

 

[lynk]

Wow...he did not teach us that (yet?), nor is it in the book I've read (so far). It looks like "Statics" are in chapter 9 of our book, but we're only on chapter 4?

My guess would be 0? But that doesn't make any sense, because it would make "mu"F(norm) = 0? Is the static eq=1, then??
I apologize, but I'm not very good at this. I am much better at straight up Calculus, but this seemingly simple problem has me very confused.

Thank you again!
Lynn