g lower part is rough with a coefficient of friction $\mu_k$. A body starts from rest at the top and returns to rest at the bottom.

Length of the smooth part: $\frac{L}{n}$

Length of the rough part: $\frac{n-1}{n} \cdot L$

Angle of inclination: $\theta$

Coefficient of friction: $\mu_k$

The acceleration down the inclined plane due to gravity is $ a = g \sin \theta $.

The distance traveled on this smooth part is:

$ s_1 = \frac{L}{n} $

Using the kinematic equation for motion:

$ v_1^2 = u^2 + 2as_1 $

$ v_1^2 = 0 + 2(g \sin \theta) \frac{L}{n} = \frac{2gL \sin \theta}{n} $

The acceleration on this part is:

$ a' = g \sin \theta - \mu_k g \cos \theta $

The distance traveled here is:

$ s_2 = \frac{n-1}{n} \cdot L $

Final velocity at the end of this segment ($ v_2 $) is 0, and the initial velocity ($ u_2 $) is $ v_1 $.

Using the equation for this motion:

$ v_2^2 = u_2^2 + 2a's_2 $

$ 0 = v_1^2 + 2a's_2 $

Substitute the expression for $ a' $:

$ 0 = \frac{2gL \sin \theta}{n} + 2(g \sin \theta - \mu_k g \cos \theta) \cdot \frac{n-1}{n} \cdot L $

Simplify to solve for $\theta$:

$ 0 = 2gL \sin \theta + 2gL(n-1)(\sin \theta - \mu_k \cos \theta) $

$ 0 = \sin \theta + (n-1)(\sin \theta - \mu_k \cos \theta) $

$ 0 = n \sin \theta - (n-1)\mu_k \cos \theta $

$ \Rightarrow n \sin \theta = (n-1)\mu_k \cos \theta $

$ \Rightarrow \tan \theta = \frac{(n-1)\mu_k}{n} $

Thus, the angle of inclination $\theta$ is:

$ \theta = \tan^{-1}\left[\frac{(n-1) \mu_k}{n}\right] $

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