8-52 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} mg&=560\ut{N}\\ R&=20\ut{m}\\ v_B&=7.0\ut{m/s}\\ \theta&=20\degree\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$ \begin{aligned} \Delta h &= R-R\cos\theta\\ &=R(1-\cos\theta) \end{aligned} $$ $$\ab{a}$$ $$\Sigma \Delta E=0,$$ $$\Delta \KE+\Delta \GE=0$$ $$ \begin{aligned} \Delta \KE&=-\Delta \GE\\ \Delta \(\frac{1}{2}mv^2\)&=-\Delta (mgh)\\ \Delta \(v^2\)&=-2g\Delta h\\ {v_A}^2-{v_B}^2&=-2gR(1-\cos\theta)\\ \end{aligned} $$ $$ \begin{aligned} v_A&=\sqrt{{v_B}^2-2gR(1-\cos\theta)}\\ &\approx 5.034229393383376\ut{m/s}\\ &\approx 5.0\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v_A&=\sqrt{{v_B}^2-2gR(1-\cos\theta)}\ge0\\ 0&=\sqrt{{v_B}^2-2gR(1-\cos\theta)}\\ \end{aligned} $$ $$ \begin{aligned} v_B&=\sqrt{2 g R (1-\cos \theta)}\\ &=4\sqrt{5g}\sin10\degree\\ &\approx 4.863798352604151\ut{m/s}\\ &\approx 4.9\ut{m/s}\\ \end{aligned} $$ $$\ab{c}$$ $$v_A=\sqrt{{v_B}^2-2gR(1-\cos\theta)},$$ $$\text{Constant about }mg$$ $$v_B=\sqrt{2 g R (1-\cos \theta)},$$ $$\text{Constant about }mg$$