7-19 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \theta&=30\degree\\ F&=38\ut{lbf}\\ v&=3.0\ut{mi/h}\\ t&=8.0\ut{min}\\ 1\ut{mi}&=5280\ut{ft}\\ \end{cases} $$ $$ \begin{aligned} S&=v t\\ &=3.0\ut{mi/h}\cdot 8.0\ut{min}\cdot\frac{1\ut{h}}{60\ut{min}}\cdot\frac{5280\ut{ft}}{1\ut{mi}}\\ &=2112\ut{ft} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} W&=\vec F\cdot \vec S\\ &= F S \cos \theta\\ &=40128\sqrt3 \ut{ft\cdot lbf}\\ &\approx 69503.7348061239\ut{ft\cdot lbf}\\ &\approx 7.0\times 10^4\ut{ft\cdot lbf}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} P&=\frac{W}{t}\\ &=\frac{40128\sqrt3 \ut{ft\cdot lbf}}{8\ut{min}}\cdot \frac{1\ut{min}}{60\ut{s}}\cdot\frac{1\ut{hp}}{550\ut{ft\cdot lbf/s}}\\ &=\frac{19\sqrt3}{125}\ut{hp}\\ &\approx 0.26327172275046934\ut{hp}\\ &\approx 0.26\ut{hp}\\ \end{aligned} $$