Answer
(a) (i) h = 0.925 m
(ii) h = 1.06 m
(iii) h = 2.43 m
(b) h = 1.06 m
(c) If $\theta$ increases, the speed at the bottom would increase.
Work Step by Step
(a) (i) $U+W = K$
$mgh - mg~cos(\theta)~\mu_k~\frac{h}{sin(\theta)} = \frac{1}{2}mv^2$
$gh - g~cot(\theta)~\mu_k~h = \frac{1}{2}v^2$
$h = \frac{v^2}{2~(g-g~cot(\theta)~\mu_k)}$
$h = \frac{(4.00~m/s)^2}{2~[9.80~m/s^2-9.80~m/s^2~cot(52.0^{\circ})(0.15)]}$
$h = 0.925~m$
(ii) $U+W = K$
$mgh - mg~cos(\theta)~\mu_k~\frac{h}{sin(\theta)} = \frac{1}{2}mv^2$
$gh - g~cot(\theta)~\mu_k~h = \frac{1}{2}v^2$
$h = \frac{v^2}{2~(g-g~cot(\theta)~\mu_k)}$
$h = \frac{(4.00~m/s)^2}{2~[9.80~m/s^2-9.80~m/s^2~cot(52.0^{\circ})(0.29)]}$
$h = 1.06~m$
(iii) (i) $U+W = K$
$mgh - mg~cos(\theta)~\mu_k~\frac{h}{sin(\theta)} = \frac{1}{2}mv^2$
$gh - g~cot(\theta)~\mu_k~h = \frac{1}{2}v^2$
$h = \frac{v^2}{2~(g-g~cot(\theta)~\mu_k)}$
$h = \frac{(4.00~m/s)^2}{2~[9.80~m/s^2-9.80~m/s^2~cot(52.0^{\circ})(0.85)]}$
$h = 2.43~m$
(b) Since the required height $h$ does not depend on the mass, the required height $h$ is still 1.06 meters.
(c) If $\theta$ increases, the magnitude of the work done by friction decreases, so less energy is removed from the system. Therefore, the speed at the bottom would increase.
