Respuesta :
Answer:
A) 1.2
B) 1.1
Explanation:
A)
F = force required to start the box moving = 36.0 N
m = mass of the box = 3 kg
[tex]F_{g}[/tex] = weight of the box = mg = 3 x 9.8 = 29.4 N
[tex]F_{n}[/tex] = normal force acting on the box by the floor
normal force acting on the box by the floor is given as
[tex]F_{n}[/tex] = [tex]F_{g}[/tex] = 29.4
[tex]F_{s}[/tex] = Static frictional force = F = 36.0 N
[tex]\mu _{s}[/tex] = Coefficient of static friction
Static frictional force is given as
[tex]F_{s}[/tex] = [tex]\mu _{s}[/tex] [tex]F_{n}[/tex]
36.0 = [tex]\mu _{s}[/tex] (29.4)
[tex]\mu _{s}[/tex] = 1.2
B)
a = acceleration of the box = 0.54 m/s²
F = force applied = 36.0 N
[tex]f_{k}[/tex] = kinetic frictional force
[tex]\mu _{k}[/tex] = Coefficient of kinetic friction
force equation for the motion of the box is given as
F - [tex]f_{k}[/tex] = ma
36.0 - [tex]f_{k}[/tex] = (3) (0.54)
[tex]f_{k}[/tex] = 34.38 N
Coefficient of kinetic friction is given as
[tex]\mu _{k}=\frac{f_{k}}{F_{n}}[/tex]
[tex]\mu _{k}=\frac{34.38}{29.4}[/tex]
[tex]\mu _{k}[/tex] = 1.1
