A 5.00-kg box slides 3.00 m across the floor before coming to rest. what is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?

Let's call the coefficient of kinetic friction k.

We can use the equation v^2 = v (initial) ^2 + 2a (x-x (initial)) to find the acceleration of the box.

v=0 m/s

v (initial) = 3 m/s

x = 3 m

x (initial) = 0 m

a = (v^2-v (initial) ^2) / 2 (x-x (initial))

a = (-9 m^2/s^2) / 3 m

a = - 3 m/s^2

Since F = ma, and the only horizontal force on the box is friction, we can say that F (friction) = k*F (normal) = k*F (gravitational) = m*a

k = m*a/F (gravitational)

F (gravitational) = m*g

k = (m*a) / (m*g)

k = a/g

k = (-3 m/s^2) / (-9.81 m/s^2)

k = 0.31