Constrained Motion on Rough Table

To be clear, the particle is moving on a rough horizontal table (the XY-plane) in contact with a smooth vertical parabolic wall, so that it is constrained to move along the line $y = \tfrac12x^2$ . We shall only consider horizontal forces on the particle. When the particle has position vector $\mathbf{r} = \binom{x}{\frac12x^2}$ , it is subject to a normal reaction $\mathbf{N} = m\lambda\binom{-x}{1}$ perpendicular to the wall and a friction force $-\frac{\mu mg}{\sqrt{1 + x^2}}\binom{1}{x}$ parallel to the wall. Thus $\binom{1}{x}\ddot{x} + \binom{0}{1} \dot{x}^2 \; = \; \lambda\binom{-x}{1} - \frac{\mu g}{\sqrt{1 + x^2}}\binom{1}{x}$ Taking the inner product of this equation with $\binom{1}{x}$ , we deduce that $\begin{aligned} (1 + x^2)\ddot{x} + x\dot{x}^2 & = \; -\mu g \sqrt{1 + x^2} \\ \frac{d}{dx}\Big[\tfrac12(1 +x^2)\dot{x}^2\Big] & = \; -\mu g \sqrt{1 + x^2} \\ \tfrac12(1 + x^2)\dot{x}^2 & = \; 2 - \mu g\int_0^x \sqrt{1 + x^2}\,dx \; = \; 2 - \tfrac12\mu g\left[x\sqrt{1+x^2} + \sinh^{-1}x\right] \end{aligned}$ and so the particle will come to rest at the value $x_F$ where $x_F \sqrt{1 + x_F^2} + \sinh^{-1}x_F \; = \; \tfrac{4}{\mu g}$ In this case $\frac{4}{\mu g} = 0.8$ , and we solve this equation numerically to obtain $x_F = 0.39030523306856$ , and hence $\lfloor 100 x_F \rfloor = \boxed{39}$ .