#### Answer

(a) $\alpha = 19.3^{\circ}$
(b) $a = 0.93~m/s^2$
(c) $v = 3.0~m/s$

#### Work Step by Step

(a) The minimum angle is when the force of static friction directed up the ramp is equal in magnitude to the component of the weight directed down the ramp.
$F_f = mg~sin(\alpha)$
$mg~cos(\alpha)~\mu_s = mg~sin(\alpha)$
$tan(\alpha) = \mu_s$
$\alpha = arctan(\mu_s) = arctan(0.35)$
$\alpha = 19.3^{\circ}$
(b) $\sum F = ma$
$mg~sin(\alpha) - mg~cos(\alpha)~\mu_k = ma$
$a = g~sin(\alpha) - g~cos(\alpha)~\mu_k$
$a = (9.80~m/s^2)~sin(19.3^{\circ}) - (9.80~m/s^2)~cos(19.3^{\circ})(0.25)$
$a = 0.93~m/s^2$
(c) We can find the speed $v$ after sliding 5.0 meters.
$v^2 = v_0^2+2ax = 0 + 2ax$
$v = \sqrt{(2)(0.93~m/s^2)(5.0~m)}$
$v = 3.0~m/s$