6-61 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=2.8\ut{kg}\\ \theta&=30\degree\\ F&=15\ut{N}\\ \mu&=0.25\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} f&=\mu N\\ &=\mu(F\sin\theta+mg)\\ &=\frac{7 g}{10}+\frac{15}{8}\\ &=8.739655\ut{N}\\ &\approx 8.7\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$\Sigma F=ma$$ $$ \begin{aligned} a&=\frac{F\cos\theta-\mu(F\sin\theta+mg)}{m}\\ &=\frac{1}{112} \(-28 g+300 \sqrt{3}-75\)\\ &\approx 1.5181164488452068\ut{m/s^2}\\ &\approx 1.5\ut{m/s^2}\\ \end{aligned} $$