# Chapter 9 - Work and Kinetic Energy - Exercises and Problems: 49

(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

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