Answer:
[tex]f=mg*sin\theta[/tex]
The component of weight of the cube parallel to the plane
Explanation:
From the question we are told that
Cube slides at constant speed
Generally the equation for frictional force is given by
[tex]f=\mu.N[/tex]
where
[tex]f = frictional force\\\mu =coefficient of friction\\N=Normal force[/tex]
Generally the equation fo Normal force is mathematically given as
[tex]N=mg.cos\theta[/tex]
Therefore
[tex]f=\mu*mg*cos\theta[/tex]
Generally at constant velocity frictional force f will be
[tex]f=mg*sin\theta[/tex]
Therefore
The magnitude of the frictional force that acts on the cube due to the surface is the component of weight of the cube parallel to the plane
The magnitude of the frictional force that acts on the cube due to the surface is determined with [tex]\mu_ k \times mg \times cos(\theta)[/tex].
The normal force on the cube is calculated as;
[tex]F_n = W \times cos (\theta)\\\\F_n = mg \times cos (\theta)[/tex]
The frictional force acting on the cube due to the surface is calculated as follows;
[tex]F_f = \mu_k F_n\\\\F_f = \mu_ k \times mg \times cos(\theta)[/tex]
where;
Thus, the magnitude of the frictional force that acts on the cube due to the surface is determined with [tex]\mu_ k \times mg \times cos(\theta)[/tex].
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