Answer:
The value is [tex]\mu_k = \frac{3}{4} * tan(\theta )[/tex]
Explanation:
Generally the net force acting on the smooth block is mathematically represented as
[tex]F = mgsin (\theta )[/tex]
Here this force also represented as
[tex]F = ma[/tex]
So
[tex]ma = mgsin (\theta )[/tex]
=> [tex]a = g sin(\theta )[/tex]
Generally from kinematic equation
[tex]s = ut + \frac{1}{2} a t_1^2[/tex]
before the sliding the block was at rest so u = 0 m/s
[tex]s = \frac{1}{2} a t_1^2[/tex]
=> [tex]s = \frac{1}{2} (g sin(\theta )) t^2[/tex]
=> [tex]t_1 = \sqrt{\frac{2s}{gsin(\theta )} }[/tex]
Generally the net force acting on the rough block is mathematically represented as
[tex]F = mgsin(\theta ) - F_f[/tex]
Here [tex]F_f[/tex] is the frictional force acting on the block which is mathematically represented as
[tex]F_f = \mu_k * mg * cos (\theta)[/tex]
So
[tex]F = mgsin(\theta ) - \mu_k * mg * cos (\theta) [/tex]
generally this net force is mathematically represented as
[tex]F = ma[/tex]
=> [tex] ma = mgsin(\theta ) - \mu_k * mg * cos (\theta) [/tex]
=> [tex]a = gsin(\theta ) - \mu_k * cos (\theta )[/tex]
Generally from kinematic equation
[tex]s = ut + \frac{1}{2} a t_2^2[/tex]
before the sliding the block was at rest so u = 0 m/s
[tex]s = \frac{1}{2} a t_2^2[/tex]
=> [tex]s = \frac{1}{2} [gsin(\theta ) - \mu_k * cos (\theta )} t^2[/tex]
= > [tex]t_2 = \sqrt{\frac{2s}{gsin(\theta ) - \mu_k cos(\theta )} }[/tex]
From the question we are told that
[tex]t_2 = 2 t_1[/tex]
[tex]\sqrt{\frac{2s}{gsin(\theta ) - \mu_k cos(\theta )} } = 2 * \sqrt{\frac{2s}{gsin(\theta )} }[/tex]
Squaring both sides
[tex]\frac{1}{sin(\theta ) - \mu_k cos(\theta )} = \frac{2}{sin(\theta)}[/tex]
=> [tex]sin(\theta) = 4sin(\theta ) - 4\mu_k cos(\theta )[/tex]
=> [tex]4\mu_k cos(\theta ) = 3sin(\theta )[/tex]
=> [tex]\mu_k = \frac{3}{4} * \frac{sin(\theta )}{ cos(\theta )}[/tex]
=> [tex]\mu_k = \frac{3}{4} * tan(\theta )[/tex]
