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

(a) We can find the rate of deceleration on the patch between x = 0 and x = 2.00 m. $F_f = ma$ $mg~\mu_k = ma$ $a = g~\mu_k = (9.80~m/s^2)(0.200)$ $a = 1.96~m/s^2$ We can find the speed $v$ after the box moves from x = 0 to x = 2.00 m. $v^2 = v_0^2+2ax$ $v = \sqrt{v_0^2+2ax}$ $v = \sqrt{(4.00~m/s)^2+(2)(-1.96~m/s^2)(2.00~m)}$ $v = 2.86~m/s$ We can find the rate of deceleration on the patch between x = 2.00 m and x = 4.00 m. $F_f = ma$ $mg~\mu_k = ma$ $a = g~\mu_k = (9.80~m/s^2)(0.400)$ $a = 3.92~m/s^2$ We can find the distance $d$ the box travels past the point x = 2.00 m until it comes to rest. $d = \frac{v^2-v_0^2}{2a} = \frac{0-(2.86~m/s)^2}{(2)(-3.92~m/s^2)}$ $d = 1.04~m$ The box comes to rest at the position x = 3.04 m. (b) We can find the time $t_1$ fro the box to move from x = 0 to x = 2.00 m. $t_1 = \frac{v-v_0}{a} = \frac{2.86~m/s-4.00~m/s}{-1.96~m/s^2}$ $t_1 = 0.5816~s$ We can find the time $t_2$ for the box to move from x = 2.00 m to x = 4.00 m. $t_2 = \frac{v-v_0}{a} = \frac{0-2.86~m/s}{-3.92~m/s^2}$ $t_2 = 0.7296~s$ The total time $t$ is $t_1+t_2$ $t = 0.5816~s+0.7296~s = 1.31~s$ It takes the box 1.31 seconds to come to rest.