Answer
(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.
Work Step by Step
(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.
