## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) We can convert the cruising speed to units of m/s $v = (12~km/h)(\frac{1~h}{3600~s})(\frac{1000~m}{1~km})$ $v = 3.33~m/s$ We can find the force $F$ applied by the team during the acceleration phase. $\sum F = ma$ $F - F_f = ma$ $F = ma+F_f$ $F = ma+mg~\mu_k$ $F = (220~kg)(0.75~m/s^2)+(220~kg)(9.80~m/s^2)(0.080)$ $F = 337.5~N$ The maximum power output during the acceleration phase occurs when the sled reaches its cruising speed. We can find the power output. $P = F~v$ $P = (337.5~N)(3.33~m/s)$ $P = 1124~watts$ The maximum power output during the acceleration phase is 1124 watts. (b) During the cruising phase, the force applied by the team is equal in magnitude to the force of friction. We can find the magnitude of the applied force $F$. $F = F_f$ $F = mg~\mu_k$ $F = (220~kg)(9.80~m/s^2)(0.080)$ $F = 172.5~N$ We can find the power output during the cruising phase. $P = F~v$ $P = (172.5~N)(3.33~m/s)$ $P = 574~watts$ The power output during the cruising phase is 574 watts.