Answer
(a) The skier travels 5.80 meters before stopping.
(b) The speed after the rough patch would be 3.53 m/s.
(c) The toboggan stops at a vertical height of 7.35 meters above the base.
Work Step by Step
(a) $KE_1 + W = KE_2$
$\frac{1}{2}mv^2 - F_f~d = 0$
$\frac{1}{2}mv^2 = mg~\mu_k~d$
$d = \frac{v^2}{2g~\mu_k}$
$d = \frac{(5.00~m/s)^2}{(2)(9.80~m/s^2)(0.220)}$
$d = 5.80~m$
The skier travels 5.80 meters before stopping.
(b) $KE_1 + W = KE_2$
$\frac{1}{2}mv_1^2 - F_f~d = \frac{1}{2}mv_2^2$
$\frac{1}{2}mv_1^2 - mg~\mu_k~d = \frac{1}{2}mv_2^2$
$v_1^2 - 2g~\mu_k~d = v_2^2$
$v_2 = \sqrt{v_1^2 - 2g~\mu_k~d}$
$v_2 = \sqrt{(5.00~m/s)^2 - (2)(9.80~m/s^2)(0.220)(2.90~m)}$
$v_2 = 3.53~m/s$
The speed after the rough patch would be 3.53 m/s.
(c) $KE_1 + W = KE_2$
$\frac{1}{2}mv^2 + mg~d~cos(115^{\circ}) = 0$
$\frac{1}{2}v^2 = -g~d~cos(115.0^{\circ})$
$d = \frac{v^2}{-2g~cos(115.0^{\circ})}$
$d = \frac{(12.0~m/s)^2}{-(2)(9.80~m/s^2)~cos(115.0^{\circ})}$
$d = 17.4~m$
The toboggan goes a distance of 17.4 meters up the hill. We can find the vertical height above the base.
$h = (17.4~m)~sin(25.0^{\circ}) = 7.35~m$
The toboggan stops at a vertical height of 7.35 meters above the base.
