# Chapter 5 - Section 5.4 - Friction - Example - Page 81: 5.8

a) 36 meters b) 52 meters

#### Work Step by Step

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.

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