8-60 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} mg&=12000\ut{N}\\ F_1&=500+1.8v^2\\ v&=70\ut{km/h}=\frac{175}{9}\ut{m/s}\\ a&=0.92\ut{m/s^2}\\ 1\ut{hp}&=75\ut{kgf\cdot m/s}=75g\ut{W}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\Sigma F = ma,$$ $$F_2-F_1=ma$$ $$ \begin{aligned} F_2&=F_1+ma\\ \end{aligned} $$ $$ \begin{aligned} P&=F_2\cdot v\\ &= (F_1+ma)\cdot v\\ &= (500+1.8v^2+ma)\cdot v\\ &=\frac{875 (2125 g+19872)}{81 g}\ut{W}\\ &=\frac{875 (2125 g+19872)}{81 g}\ut{W}\cdot\frac{1\ut{hp}}{75g\ut{W}}\\ &=\frac{35 (2125 g+19872)}{243 g^2}\ut{hp}\\ &\approx 60.97244210431779\ut{hp}\\ &\approx 61\ut{hp}\\ \end{aligned} $$