Six dogs pull a two-person sled with a total mass of $220kg$. The coefficient of kinetic friction between the sled and the snow is $0.080$. The sled accelerates at $0.75m/s^2$until it reaches a cruising speed of $12km/h$. What is the team’s

(a) maximum power output during the acceleration phase and

(b) power output during the cruising phase?

The power output of dog can be calculated as

$P_{dog}=\frac{W_{dog}}{t}$

The distance travelled during the acceleration phase is

$$\begin{gathered} {v_f}^2={v_i}^2+2a∆s \\ (12km/hr\times \frac{1000m}{1km}\times \frac{1h}{3600s})=(0m/s^2)+2(0.75m/s^2)∆s \\ ∆s=\frac{(12km/hr\times \frac{1000m}{1km}\times \frac{1h}{3600s})}{2(0.75m/s^2)} \\ ∆s=7.4m \end{gathered}$$

The time taken to touch the cruising speed starting from rest.

$$\begin{gathered} ∆s=\frac{1}{2}a{t}^2 \\ t=\sqrt{\frac{2∆s}{a}} \\ t=\sqrt{\frac{2(7.4m)}{0.75m/s^2}} \\ t=4.44s \end{gathered}$$

The force applied by the dog during acceleration phase is

$$\begin{gathered} n=mg \\ Also, \\ {F}_{dog}=ma+{\mu }_{k}mg \\ {F}_{dog}=220kg(0.75m/s^2+0.08\times 9.8m/s^2) \\ =337N \end{gathered}$$

Now the power output of dog is

$$\begin{gathered} P_{dog}=\frac{W_{dog}}{t} \\ {P}_{dog}=\frac{(337N)(7.4m)}{4.44s} \\ =560W \end{gathered}$$

At cruising speed there is no acceleration. Therefore, the force applied by the dog during the acceleration phase is

$$\begin{gathered} n=mg \\ Also, \\ {F}_{dog}={\mu }_{k}mg \\ {F}_{dog}=(220N)(0.08\times 9.8m/s^2) \\ =172\hspace{0.17em}N \end{gathered}$$

Now, the power output of the dog

$$\begin{gathered} P_{dog}=\frac{W_{dog}}{t} \\ {P}_{dog}=(172N)(12km/h\times \frac{1000m}{1hr}\times \frac{1h}{3600s}) \\ =570W \end{gathered}$$