#### Answer

(a) The skater's speed is 3.8 m/s
(b) The minimum value of $\mu_s$ is 0.0058

#### Work Step by Step

(a) We can find the work $W_w$ that the wind does on the skater.
$W_w = F~d~cos(\theta)$
$W_w = (4.0~N)(100~m)~cos(135^{\circ})$
$W_w = -31.1~J$
We can use the work-energy theorem to find the skater's speed after gliding 100 meters.
$KE_2 = KE_1+W_w$
$\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2+W_w$
$v_2^2 = \frac{mv_1^2+2W_w}{m}$
$v_2 = \sqrt{\frac{mv_1^2+2W_w}{m}}$
$v_2 = \sqrt{\frac{(50~kg)(4.0~m/s)^2-(2)(31.1~J)}{50~kg}}$
$v_2 = 3.8~m/s$
The skater's speed is 3.8 m/s.
(b) The force of static friction must be equal in magnitude to the horizontal component of the wind's force $F_w$. To find the minimum value of $\mu_s$, we can assume that the force of static friction is at its maximum possible value of $F_N~\mu_s$
$F_N~\mu_s = F_w~cos(45^{\circ})$
$mg~\mu_s = F_w~cos(45^{\circ})$
$\mu_s = \frac{F_w~cos(45^{\circ})}{mg}$
$\mu_s = \frac{(4.0~N)~cos(45^{\circ})}{(50~kg)(9.80~m/s^2)}$
$\mu_s = 0.0058$
The minimum value of $\mu_s$ is 0.0058