Answer
(a) The rock reaches a height of 9.27 meters above the foot of the hill.
(b) The rock will slide back down the hill.
(c) The speed at the bottom of the hill will be 11.8 m/s
Work Step by Step
(a) $K_2+U_2 = K_1+U_1+W_f$
$0+mgh = \frac{1}{2}mv^2+0 - mg~cos(\theta)~\mu_k~d$
$gh = \frac{1}{2}v^2 - g~cos(\theta)~\mu_k~\frac{h}{sin(\theta)}$
$h~(g+g~cot(\theta)~\mu_k) = \frac{1}{2}v^2$
$h = \frac{v^2}{2~(g+g~cot(\theta)~\mu_k)}$
$h = \frac{(15~m/s)^2}{2~(9.80~m/s^2+9.80~m/s^2~cot(40.0^{\circ})(0.20))}$
$h = 9.27~m$
The rock reaches a height of 9.27 meters above the foot of the hill.
(b) We can find the maximum possible force of static friction.
$F_f = mg~cos(\theta)~\mu_s$
$F_f = (28~kg)(9.80~m/s^2)~cos(40.0^{\circ})~(0.75)$
$F_f = 158~N$
We can find the component of the rock's weight directed down the hill.
$mg~sin(\theta) = (28~kg)(9.80~m/s^2)~sin(40.0^{\circ}) = 176~N$
Since the component of the rock's weight directed down the hill is greater than the maximum possible force of static friction, the rock will slide back down the hill.
(c) $K_2 +U_2 = K_1+U_1+W_f$
$\frac{1}{2}mv^2+0 = 0 + mgh - F_f~d$
$\frac{1}{2}mv^2 = mgh - mg~cos(\theta)~\mu_k~\frac{h}{sin(\theta)}$
$v^2 = 2gh~(1 - cot(\theta)~\mu_k)$
$v = \sqrt{2gh~(1 - cot(\theta)~\mu_k)}$
$v = \sqrt{(2)(9.80~m/s^2)(9.27~m)~(1 - cot(40.0^{\circ})(0.20)}$
$v = 11.8~m/s$
The speed at the bottom of the hill will be 11.8 m/s.
