A5.00-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?

The coefficient of kinetic friction between the floor and the box is μ = 0.1529

Explanation:

Let's start calculating the weight of the box.

Given that and gravity (g) is the weight is :

Given that the box is sliding across the floor, the normal force is

Now, the work that the frictional force produces is equal to the kinetic energy variation that the box experiments.

ΔKE = Work frictional force

ΔKE = KEfinal - KEinitial ⇒

If

KEfinal = 0 J given that its speed is 0

ΔKE = KEfinal - KEinitial

ΔKE = 0 -

ΔKE = -22.5 J

The work that the frictional force produces is

Work frictional force = -(Frictional force).(distance)

The distance is 3.00 m

The frictional force is Nf.μ where Nf is the normal force and μ is the coefficient of kinetic friction between the box and the floor. We add a (-) given that the sense of the frictional force is opposite to the trajectory of the box.

Finally,

ΔKE = WFF

-22.5 J = - Nf.μ.d

μ =

The coefficient of kinetic friction is 0.1529

Notice that this coefficient is dimensionless.

e/c, where e is its total energy