11판/8. 퍼텐셜에너지와 에너지 보존

8-30 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 4. 2. 19:43
$$ \begin{cases} L&=1.50\ut{m}\\ m&=4.00\ut{kg}\\ \theta_0&=30.0\degree\\ \vec v_0&=0\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \put \begin{cases} 0 : \text{Start}\\ 1 : \text{Lowest Point} \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$ \begin{aligned} -\Delta h_{0\rarr1}&=L-L\cos\theta_0\\ &=L(1-\cos\theta_0)\\ \end{aligned} $$ $$\ab{a}$$ $$\Sigma E_i=\Sigma E_f,$$ $$ \begin{aligned} \KE_0+\GE_0&=\KE_1+\GE_1\\ \GE_0&=\KE_1+\GE_1\\ \KE_1&=-\Delta \GE_{0\rarr1}\\ \frac{1}{2}m{v_1}^2&=-\Delta (mgh)_{0\rarr1}\\ \end{aligned} $$ $$ \begin{aligned} v_1&=\sqrt{2g(-\Delta h_{0\rarr1})}\\ &=\sqrt{2gL(1-\cos\theta_0)}\\ &=\sqrt{\frac{3}{2} \(2-\sqrt{3}\)g}\\ &\approx 1.9853276611008082\ut{m/s}\\ &\approx 1.99\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$v_1=\sqrt{2gL(1-\cos\theta_0)},$$ $$\text{Constant about }m$$