m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
The coefficient of kinetic friction between the box and the ramp is 0.71.
Given data:
The angle of inclination is, [tex]\theta = 30^\circ[/tex].
The acceleration of box is, [tex]a = 1.20 \;\rm m/s^{2}[/tex].
The frictional force acting on the box is,
[tex]f= \mu \times N[/tex]
Here, [tex]\mu[/tex] is the coefficient of kinetic friction and N is the normal force.
The normal force in an inclined plane is,
[tex]N=mg cos\theta[/tex]
Since the frictional force causes the acceleration (a) of box,
[tex]F=f-mgsin\theta\\ma=\mu mgcos\theta-mgsin\theta\\a=\mu gcos\theta-gsin\theta[/tex]
Solving as,
[tex]a=\mu gcos\theta-gsin\theta\\\mu=\dfrac{a+gsin\theta}{gcos\theta} \\\mu=\dfrac{1.20+9.8sin30}{9.8cos30}\\\mu= 0.71[/tex]
Thus, the coefficient of kinetic friction between the box and the ramp is 0.71.
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