Answer
(a) $\mu_s = 0.67$
(b) $v = 6.1 ~m/s$
(c) $v = 15~m/s$
Work Step by Step
(a) If the snow does not start sliding, let's assume that the friction force pushing up the slope is equal to the component of the weight directed down the slope:
$F_f - mg ~sin(\theta)$
$mg ~cos(\theta)\cdot \mu_s = mg ~sin(\theta)$
$\mu_s = tan(\theta) = tan(34^{\circ}) = 0.67$
(b) We can use a force equation to find the acceleration:
$ma = \sum F$
$ma = mg ~sin(\theta) - mg ~cos(\theta)\cdot \mu_k$
$a = g ~sin(\theta) - g ~cos(\theta)\cdot \mu_k$
$a = (9.80 ~m/s^2) ~sin(34^{\circ}) - (9.80 ~m/s^2) ~cos(34^{\circ})\cdot (0.10)$
$a = 4.67 ~m/s^2$
We can use the acceleration to find the velocity $v$:
$v^2 = v_0^2 + 2ax = 0 + 2ax = 2ax$
$v = \sqrt{2ax} = \sqrt{(2)(4.67 ~m/s^2)(4.0 ~m)}$
$v = 6.1 ~m/s$
(c) $v_x = v ~cos(\theta) = (6.1 ~m/s) ~cos(34^{\circ}) = 5.06 ~m/s$
$v_{y0} = v ~sin(\theta) = (6.1 ~m/s) ~sin(34^{\circ}) = 3.4 ~m/s$
We can use kinematics to find $v_y$ after it falls 10.0 meters.
$v_y^2 = v_{y0}^2 + 2ay$
$v_y = \sqrt{v_{y0}^2 + 2ay} = \sqrt{(3.4 ~m/s)^2 + (2)(9.80 ~m/s^2)(10.0 ~m)}$
$v_y = 14.4 ~m/s$
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(5.06 ~m/s)^2 + (14.4 ~m/s)^2}$
$v = 15~m/s$
