8-23 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} R&=L\\ v_A&=v_0\\ v_D&=0 \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \ME : \text{Mechanical Energy} \end{cases} $$ $$\ab{a}$$ $$\Sigma E_A=\Sigma E_D,$$ $$ \begin{aligned} \KE_A+\GE_A&=\KE_D+\GE_D\\ \KE_A&=\Delta \GE_{A\rarr D}~~~(\because v_D=0)\\ \end{aligned} $$ $$ \begin{aligned} \frac{1}{2}m{v_A}^2&=mg\Delta h_{A\rarr D}\\ {v_0}^2&=2gL\\ v_0&=\sqrt{2gL} \end{aligned} $$ $$\ab{b}$$ $$\Sigma E_A=\Sigma E_B,$$ $$ \begin{aligned} \KE_A+\GE_A&=\KE_B+\GE_B\\ \KE_B&=\KE_A-\Delta \GE_{A\rarr B}\\ \frac{1}{2}m{v_B}^2&=\frac{1}{2}m{v_A}^2-mg\Delta h_{A\rarr B}\\ \end{aligned} $$ $$ \begin{aligned} {v_B}^2&={v_A}^2-2g(-L)\\ &={v_0}^2+2gL\\ &=2gL+2gL\\ &=4gL \end{aligned} $$ $$ \Sigma F_R=\frac{mv^2}{R},$$ $$ \begin{aligned} T-mg&=\frac{m{v_B}^2}{L}\\ \end{aligned} $$ $$ \begin{aligned} T&=\frac{m{v_B}^2}{L}+mg\\ &=\frac{m(4gL)}{L}+mg\\ &=5mg\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \Ans&=-\Delta \ME_{A\rarr C}\\ &=-\Delta (\KE+\GE)_{A\rarr C}\\ &=-(\Delta \KE_{A\rarr C}+\Delta \GE_{A\rarr C})\\ &=-\Delta \KE_{A\rarr C}\\ &=-(\KE_C-\KE_A)\\ &=\KE_A\\ &=\frac{1}{2}m{v_0}^2\\ &=\frac{1}{2}m(2gL)\\ &=mgL \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \Ans&=-\Delta \ME_{A\rarr B}\\ &=-\Delta (\KE+\GE)_{A\rarr B}\\ &=-(\Delta \KE_{A\rarr B}+\Delta \GE_{A\rarr B})\\ &=-(\KE_B-\KE_A)-\Delta \GE_{A\rarr B}\\ &=\KE_A-\Delta \GE_{A\rarr B}\\ &=mgL-mg\Delta h_{A\rarr B}\\ &=mgL-mg(-L)\\ &=2mgL\\ \end{aligned} $$