Power of the friction force is given by the equation: $$P=\left(F,v\right)=-\left(\mu_xmg\cos^2\alpha+\mu_ymg\sin^2\alpha\right)v$$ since $\mu_x>\mu_y$, then maximum of the power is reached when $\alpha=0$
From the previous point it follows that: $$\mu_xmgv\cos^2\alpha+\mu_ymgv\sin^2\alpha=\frac{1}{1.2} \mu_xmgv$$ From where we get: $$ \cos \alpha_2 = \pm \sqrt{\frac{5\mu_x - 6\mu_y}{6(\mu_x - \mu_y)}} = \pm \frac{1}{\sqrt{2}} \\ \sin \alpha_2 = \pm \sqrt{\frac{\mu_x}{6(\mu_x-\mu_y)}} = \pm \frac{1}{\sqrt{2}} $$
Newton's equation: \begin{cases} m\dfrac{dv_x}{dt} &=-\mu_xmg\dfrac{v_x}{v}\\ m\dfrac{dv_y}{dt} &=-\mu_ymg\dfrac{v_y}{v} \end{cases} Dividing one equation by another, we get: $$\dfrac{dv_x}{dv_y}=\dfrac{\mu_xv_x}{\mu_yv_y}$$ Integrating these equations, we obtain: $$\dfrac{v_x^{\mu_y}}{v_y^{\mu_x}}=Const$$ From the initial conditions we find: $$v_x=0.125~\text{m/s}$$ So the absolute value of velosity is
Projection of the friction force onto the direction perpendicular to the velocity is given by equation: $$F_n=mg(\mu_x-\mu_y)\sin\alpha\cos\alpha$$ Newton's second law projected onto the direction perpendicular to the velocity has the form: $$\dfrac{mv^2}{R}=mg(\mu_x-\mu_y)\sin\alpha\cos\alpha,$$ where $R$ is the radius of curvature of the trajectory. It's obvious that $$R=\dfrac{v^2}{g(\mu_x-\mu_y)\sin\alpha\cos\alpha}=\dfrac{2v^2}{g(\mu_x-\mu_y)\sin2\alpha}$$ The minimum radius of curvature will be when:
If $\mu_x=\mu_y$, then the accelerations along the x and y axes will be proportional to the ratio of the initial velocities and the body will move in a straight line. If $\mu_x>\mu_y$, then the speed along the x-axis will decrease faster than in the previous case and the body will deviate from a linear motion as shown in the figure. The direction of deflection does not depend on how the initial velocity is directed. If $\mu_x<\mu_y$, then the body will deviate in the opposite direction.
Let us notice that the body will start the motion with the angle $\varphi \neq \alpha$. Then projections of the forces at this moment are given by $$\begin{cases}F_{\text{fr},x} = \mu_x mg \cos \varphi = \gamma t\cos \alpha \\ F_{\text{fr},y} = \mu_y mg \sin \varphi = \gamma t\sin \alpha \end{cases}$$ Dividing these equations by $\mu_x mg$ and $\mu_y mg$ correspondingly and summing up squares, we get $$(mg)^2 = (\gamma t)^2 \left[ \left(\frac{\cos \alpha}{\mu_x} \right)^2 + \left(\frac{\sin \alpha}{\mu_y} \right)^2 \right]$$ So the answer is
Newton's law projected onto the direction of motion of a point-like mass is: $$m\dfrac{dv}{dt}=-mg(\mu_x\cos^2\varphi+\mu_y\sin^2\varphi)$$ Since $v=L\varphi$, then: $$L\ddot\varphi=-g(\mu_x\cos^2\varphi+\mu_y\sin^2\varphi)$$ Multiply this equation by $\dot\varphi\, dt$ we get $$L\dot\varphi d\dot\varphi=-g(\mu_x\cos^2\varphi+\mu_y\sin^2\varphi)d\varphi$$ Integrating this equation, we obtain:
Newton's law for a moving body in projection onto a rod has the form: $$T+F_{\text{fr}}\sin\beta=m\dot\varphi^2L$$ where $\beta$ is the angle between frictional force and direction of velocity. Using the result from C1, we find that: $$T=\frac{mv_0^2}{L}-mg(\mu_x+\mu_y)\varphi-mg(\mu_x-\mu_y)\sin2\varphi$$ Using the result from B1, we find that: $$\frac{mv_0^2}{L}-mg(\mu_x+\mu_y)\varphi-mg(\mu_x-\mu_y)\sin2\varphi\le\dfrac{\mu_x\mu_y}{\sqrt{\mu_y^2\sin^2\varphi+\mu_x^2\cos^2\varphi}}mg$$ Note that with increasing $\varphi$, the left hand side decreases, and the right hand side increases, so the body starts to move right at $\varphi=0$. So
Substituting $v_\text{0max}$ from the previous section in the final equation of the C1, we get the equation for the $\varphi$ at the stopping point $(v=0)$: $$gL\left((\mu_x+\mu_y)\varphi+(\mu_x-\mu_y)\dfrac{\sin2\varphi}{2}\right)=\mu_y gL $$ Numerically solving this equation, we get the answer for the $\varphi$, and so the travelled distance is equal to