6-35 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} R&=200\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{cases} \Sigma F_R &=\cfrac{mv^2}{R}\\ \Sigma F_y &= 0\\ \end{cases} $$ $$ \begin{cases} \cfrac{mv^2}{R} &=F_N\sin\theta+\mu_s F_N\cos\theta\\ 0 &= F_N\cos\theta-mg-\mu_s F_N\sin\theta\\ \end{cases} $$ $$ \begin{aligned} v_{\max}&=\sqrt{\frac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}g R}\\ &=\frac{1}{10} \sqrt{196133} \sqrt{\frac{\tan\theta+\mu_s }{1-\mu_s \tan\theta}} \end{aligned} $$ $$\ab{b}$$ $$\mu_s=0.60,$$ $$ \begin{aligned} v_{\max}&=\frac{1}{10} \sqrt{196133} \sqrt{\frac{5 \tan\theta+3}{5-3 \tan\theta}}\\ \end{aligned} $$ $$\mu_s=0.050,$$ $$ \begin{aligned} v_{\max}&=\frac{1}{10} \sqrt{196133} \sqrt{\frac{20 \tan\theta+1}{20-\tan\theta}}\\ \end{aligned} $$ $$\ab{c}$$ $$\theta=10\degree$$ $$\mu_s=0.60,$$ $$ \begin{aligned} v&=\frac{1}{10} \sqrt{\frac{196133 (3+5 \tan10\degree)}{5-3\tan10\degree}}\ut{m/s}\\ &\approx 148.55316805684816\ut{km/h}\\ &\approx 1.5\times10^2\ut{km/h}\\ \end{aligned} $$ $$\ab{d}$$ $$\theta=10\degree$$ $$\mu_s=0.050,$$ $$ \begin{aligned} v&=\frac{1}{10} \sqrt{\frac{196133 (20 \tan10\degree+1)}{20-\tan10\degree}}\ut{m/s}\\ &\approx 76.18490604677895\ut{km/h}\\ &\approx 76\ut{km/h}\\ \end{aligned} $$