A box is shoved up a plank inclined at an angle $\alpha$ above the horizontal. The plank is covered with ice so that the coefficient of friction is $\mu = Ax$ where $x$ is the distance from the bottom of the plank (in this question the static and kinetic friction are equal). The box's initial speed is $v_0$. I have to show that if the box slides up the plank and comes to rest (without sliding back down), then
$$v_0^2 \ge \frac{3g(\text{sin}\alpha)^2}{A\text{cos}\alpha}.$$
The box's weight is $mg$, we split it into the component down the plank $mg\text{sin}(\alpha)$ and the component normal to the plank $mg\text{cos}(\alpha)$, so the friction force is $Axmg\text{cos}(\alpha)$.
Assuming $v_0$ is large enough, the box will move up a distance $d$ then stop, at this point friction will be equal (and opposite) to the component of weight down the plank, so $mg\text{sin}(\alpha) = Admg\text{cos}(\alpha)$ i.e. $d = \frac{\text{tan}(\alpha)}{A}$.
By the work-energy theorem:
$$0 - \frac{1}{2}mv_0^2 = \int_0^{d} (-mg\text{sin}(\alpha) - Axmg\text{cos}(\alpha)) dx$$
$$\frac{1}{2}v_0^2 = \int_0^{d} (g\text{sin}(\alpha) + Axg\text{cos}(\alpha)) dx = dg\text{sin}(\alpha) + A\frac{d^2}{2}g\text{cos}(\alpha)$$ $$=(1/A)(g\text{sin}^2(\alpha))/(\text{cos}(\alpha)) + g(\text{cos}(\alpha))(1/2)(\text{tan}^2(\alpha))(1/A) = \frac{3g}{2A}\frac{\text{sin}^2(\alpha)}{\text{cos}{(\alpha)}}.$$
I am confused because the work-energy theorem gives $=$ not $\ge$, but $=$ doesn't really make sense because why should $v_0$ be equal to $\frac{3g(\text{sin}\alpha)^2}{A\text{cos}\alpha}?$ How do I deduce that $v_0^2 \ge \frac{3g(\text{sin}\alpha)^2}{A\text{cos}\alpha}?$
