8-28 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=0.040\ut{kg}\\ R&=14\ut{cm}=0.14\ut{m}\\ h&=5.0R\\ \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$ \put \begin{cases} P : \text{P Point}\\ L : \text{Lowest Point}\\ Q : \text{Q Point}\\ H : \text{Highest Point}\\ \end{cases} $$ $$ \begin{cases} h_P&=h=5R\\ h_L&=0\\ h_Q&=R\\ h_H&=2R\\ \end{cases} $$ $$\ab{a}$$ $$\Sigma E_P=\Sigma E_Q,$$ $$ \begin{aligned} \GE_P&=\GE_Q+\KE_Q\\ \KE_Q&=-\Delta \GE_{P\rarr Q}\\ \frac{1}{2}m{v_Q}^2&=-\Delta (mgh)_{P\rarr Q}\\ \end{aligned} $$ $$ \begin{aligned} {v_Q}^2&=-2g\Delta h_{P\rarr Q}\\ &=-2g(-4R)\\ &=8gR\\ \end{aligned} $$ $$\Sigma F_R = \frac{mv^2}{R},$$ $$ \begin{aligned} \Sigma F_x &= \Sigma F_R\\ &= \frac{mv^2}{R}\\ &= \frac{m(8gR)}{R}\\ &= 8mg\\ &= \frac{8g}{25}\\ &=3.138128\ut{N}\\ &\approx 3.1\ut{N} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Sigma F_y &= mg\\ &= \frac{g}{25}\\ &=0.392266\ut{N}\\ &\approx 0.39\ut{N} \end{aligned} $$ $$\ab{c}$$ $$\Sigma E_i=\Sigma E_H,$$ $$ \begin{aligned} \GE_i&=\GE_H+\KE_H\\ \KE_H&=-\Delta \GE_{i\rarr H}\\ \frac{1}{2}m{v_H}^2&=-\Delta (mgh)_{i\rarr H}\\ {v_H}^2&=2g(-\Delta h_{i\rarr H})\\ &=2g(h_i-h_H)\\ &=2g(h_i-2R)\\ \end{aligned} $$ $$ \begin{aligned} \Sigma F_{Hy}&=\Sigma F_{HR}\\ mg+N&=\frac{m{v_H}^2}{R}\\ mg&=\frac{m{v_H}^2}{R}\\ \end{aligned} $$ $$ \begin{aligned} gR&={v_H}^2\\ &=2g(h_i-2R)\\ R&=2(h_i-2R)\\ \end{aligned} $$ $$ \begin{aligned} 2h_i&=5R\\ \end{aligned} $$ $$ \begin{aligned} h_i&=\frac{5}{2}R\\ &=\frac{7}{20}\ut{m}\\ &=0.35\ut{m}\\ &=35\ut{cm} \end{aligned} $$