6-15 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=2.00\ut{kg}\\ F&=0.500mg, \theta_F&=20\degree\\ \end{cases} $$ $$ \begin{aligned} F_x&=F\cos\theta\\ &=g\cos\theta\\ &\approx 9.215236639630128\ut{N}\\ \end{aligned} $$ $$ \begin{aligned} \Sigma F_y&=F_y+F_N-mg=0\\ F_N&=mg-F\sin\theta\\ &=mg-0.5mg\sin\theta\\ &=(2-\sin\theta)g \end{aligned} $$ $$ \begin{aligned} f&=\mu F_N\\ &=\mu g(2-\sin\theta) \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} f_{sa}&=\mu_{sa} g(2-\sin\theta)\\ &\approx 9.755536896873199\gt\Sigma F_x \end{aligned} $$ $$\therefore \text{Stop}$$ $$\therefore a=0$$ $$\ab{b}$$ $$ \begin{aligned} f_{sb}&=\mu_{sb} g(2-\sin\theta)\\ &\approx 6.178506701353026\lt\Sigma F_x \end{aligned} $$ $$\therefore \text{Move}$$ $$\Sigma F_x=m a,$$ $$ \begin{aligned} F_x-f_k&=g\cos\theta-\mu_k g(2-\sin\theta)\\ &=m a\\ \end{aligned} $$ $$ \begin{aligned} a&=\frac{g\cos\theta-\mu_k g(2-\sin\theta)}{2}\\ &\approx 2.168734095596764\ut{m/s^2}\\ &\approx 2.17\ut{m/s^2}\\ \end{aligned} $$