Chapter 6 - Applications of Newton's Laws - Problems and Conceptual Exercises - Page 178: 3

$1.8m$

We can find the required distance as follows: $a=\frac{f_K}{m}$ $\implies a=\frac{-\mu_k mg}{m}$ $\implies a=-\mu_k g$ We know that $\Delta x=\frac{v^2-v_{\circ}^2}{2a}$ $\implies \Delta x=\frac{0-v_{\circ}^2}{-2\mu_K g}$ We plug in the known values to obtain: $\Delta x=\frac{(4.0)^2}{2(0.46)(9.81)}$ $\Delta x=1.8m$