University Physics with Modern Physics (14th Edition)

(a) $\alpha = 19.3^{\circ}$ (b) $a = 0.93~m/s^2$ (c) $v = 3.0~m/s$
(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. $v^2 = v_0^2+2ax = 0 + 2ax$ $v = \sqrt{(2)(0.93~m/s^2)(5.0~m)}$ $v = 3.0~m/s$