7-31 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec F_a&=102\ut{N}\\ m&=3.00\ut{kg}\\ \phi &= 53.0\degree\\ h&= 0.230\ut{m} \end{cases} $$ $$\put \begin{cases} \theta : \text{경사각}\\ \vec N : \text{수직항력}\\ \vec S : \text{변위} \end{cases} $$ $$\sin\theta = \frac{h}{S},$$ $$\vec S = \frac{h}{\tan\theta}\i+h\j$$ $$\Sigma \vec F = 0 \Harr \Delta \vec v=0,$$ $$ \begin{cases} \Sigma F_x &= 0\\ \Sigma F_y &= 0\\ \end{cases} $$ $$ \begin{cases} 0&= F_a\sin\phi-N\sin\theta \\ 0&= N\cos\theta -F_a\cos\phi - mg\\ \end{cases} $$ $$ 0=\(\frac{F_a\sin\phi}{\sin\theta}\)\cos\theta-F_a\cos\phi-mg $$ $$F_a\cos\phi+mg=F_a\sin\phi\frac{1}{\tan\theta}$$ $$ \begin{aligned} \frac{1}{\tan\theta}&=\frac{F_a\cos\phi+mg}{F_a\sin\phi} \end{aligned} $$ $$\therefore \vec S = \frac{F_a\cos\phi+mg}{F_a\sin\phi}h\i+h\j$$ $$ \begin{cases} \vec S &= \frac{F_a\cos\phi+mg}{F_a\sin\phi}h\i+h\j\\ \vec F_a &=F_a\sin\phi\i+F_a\cos\phi\j \end{cases} $$ $$ \begin{aligned} W&=\vec F_a\cdot\vec S\\ &=(F_a\sin\phi)\cdot\frac{F_a\cos\phi+mg}{F_a\sin\phi}h+(F_a\cos\phi)h\\ &=(F_a\cos\phi+mg)h+(F_a\cos\phi)h\\ &=mgh+2F_ah\cos\phi\\ &=\frac{69}{100}(68\sin37\degree+g)\\ &\approx 35.003749386294096\ut{J}\\ &\approx 35.0\ut{J}\\ \end{aligned} $$