## University Physics with Modern Physics (14th Edition)

(a) $d = \frac{v_0^2}{2~g~\mu_k}$ (b) (i) The minimum stopping distance $d$ would be halved. (ii) The minimum stopping distance $d$ would increase by a factor of 4. (iii) The minimum stopping distance $d$ would increase by a factor of 2.
(a) $W = K_2 - K_1$ $-F_f~d = 0-\frac{1}{2}mv_0^2$ $mg~\mu_k~d = \frac{1}{2}mv_0^2$ $d = \frac{v_0^2}{2~g~\mu_k}$ (b) We can use the minimum stopping distance $d = \frac{v_0^2}{2~g~\mu_k}$ to answer the following questions. (i) If $\mu_k$ doubled, then the stopping distance $d$ would be halved. (ii) If $v_0$ doubled, then the stopping distance $d$ would increase by a factor of 4. (iii) If both $\mu_k$ and $v_0$ doubled, then the stopping distance $d$ would increase by a factor of 2.