Essential University Physics: Volume 1 (3rd Edition)

We use the information from the x and y components to find: $-\mu F_n = ma_x$ and $F_n=mg$ We use substitution to find: $-\mu mg = ma_x$ $-\mu g = a_x$ Thus, we find: $\Delta x = \frac{v_0^2}{-2a_x} \\ \Delta x = \frac{v_0^2}{2\mu g}$ Plugging in the different values for the friction yields an answer of 36 meters for a) and 52 meters for b), which is a difference of 16 meters.