Answer:
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 [tex]m=5.00kg[/tex] and gravity (g) is [tex]9.81\frac{m}{s^{2}}[/tex] the weight is :
[tex]W=m.g=(5.00kg).(9.81\frac{m}{s^{2}})=49.05N\\W=49.05N[/tex]
Given that the box is sliding across the floor, the normal force is [tex]49.05N[/tex]
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 [tex]KE=\frac{1}{2}mv^{2}[/tex]
KEfinal = 0 J given that its speed is 0
ΔKE = KEfinal - KEinitial
ΔKE = 0 - [tex]\frac{1}{2}(5.00kg).(3.00\frac{m}{s})^{2}[/tex]
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 = [tex]\frac{22.5J}{(49.05N).(3.00m)}=0.1529[/tex]
The coefficient of kinetic friction is 0.1529
Notice that this coefficient is dimensionless.
