8-25 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} L&=140\ut{cm}=1.40\ut{m}\\ d&=80.0\ut{cm}=0.800\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$\Sigma E_i=\Sigma E_f,$$ $$ \begin{aligned} \GE_i&=\KE_f+\GE_f\\ \KE_f&=-\Delta \GE\\ \frac{1}{2}m{v_f}^2&=-\Delta (mg h)_{i\rarr f}\\ {v_f}^2&=2g(-\Delta h_{i\rarr f})\\ {v_f}&=\sqrt{2g(-\Delta h_{i\rarr f})}\\ \end{aligned} $$ $$\put \begin{cases} 0:\text{Start}\\ 1:\text{Lowest Point}\\ 2:\text{Highest Point}\\ \end{cases} $$  $$\ab{a}$$ $$ \begin{aligned} v_1&=\sqrt{2g(-\Delta h_{0\rarr 1})}\\ &=\sqrt{2gL}\\ &\approx 5.2400973273404\ut{m/s}\\ &\approx 5.24\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} -\Delta h_{0\rarr2}&=L-2r,\\ &=L-2(L-d)\\ &=2d-L\\ &=\frac{1}{5}\ut{m} \end{aligned} $$ $$ \begin{aligned} v_2&=\sqrt{2g(-\Delta h_{0\rarr 2})}\\ &=\sqrt{\frac{2g}{5}}\\ &\approx 1.9805706248452744\ut{m/s}\\ &\approx 1.98\ut{m/s}\\ \end{aligned} $$