Box on Two Surfaces (Part 2)

Consider an instant in which the box overlaps into the rough region by a distance $y$ . Now as it moves slightly farther into the rough region, an infinitesimal amount of work is done. The total work done by friction is therefore (box length is $L$ ):

$W = \int_0^L F(y) \, dy$

Note that the friction force is a function of the overlap. We must calculate this friction force using calculus, since the coefficient of friction is variable over space.

$dm = m \frac{dx}{L} \\ dF = \mu \, g \, dm = \alpha x \, g \, dm = \alpha x \, g \, m \frac{dx}{L} = \frac{m g \alpha}{L} x \, dx \\ F(y) = \frac{m g \alpha}{L} \int_0^y x \, dx = \frac{m g \alpha}{2 L} y^2$

Plugging into the work integral:

$W = \int_0^L F(y) \, dy = \frac{m g \alpha}{2 L} \int_0^L y^2 \, dy = \frac{m g \alpha}{2 L} \frac{L^3}{3} = \frac{m g \alpha L^2}{6}$

Set the work done by friction equal to the initial kinetic energy:

$\frac{m g \alpha L^2}{6} = \frac{1}{2} m v_0^2 \\ \frac{g \alpha L^2}{3} = v_0^2 \\ \alpha = \frac{3 v_0^2}{g L^2} = \frac{3 \times 2^2}{10 \times \frac{4}{9}} = \boxed{2.7}$