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Answer :
To solve this problem we will use the equations of kinetic and mechanical energy.
We start for the first angle where the given angle is 36.5 ° and the acceleration is zero.
The force equations would be given by:
[tex]F= mgsin\theta\\N= mgcos\theta\\F_r = \mu_s N[/tex]
Where [tex]\mu[/tex] is coefficient of friction (static)
If we make a sum of forces in the direction in which the block moves we will have the SinΘ Component and that of the friction force, so
[tex]\sum F = 0\\F-F_r=0\\F=F_r \\mg sin\theta = \mu N\\mg sin\theta= \mu mgcos\theta[/tex]
solving for the coefficient,
[tex]\mu = \frac{sin\theta}{cos\theta}=tan\theta \\\mu = tan (36.5) \\\mu = 0.740[/tex]
We apply the same consideration for the angle of 30.5°, then we know that the coefficient (dinamic) is[tex]\mu_k = tan\theta \\\mu_k = tan (30.5) \\\mu_k = 0.589[/tex]
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