8-70 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=2.12\ut{kg}\\ v_0&=2.50\ut{m/s}\\ h&=3.80\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$\ab{a,b}$$ $$\Sigma \Delta E=0,$$ $$\Delta \KE+\Delta \GE=0$$ $$ \begin{aligned} \Delta \KE&=-\Delta \GE\\ \KE_f-\KE_i&=-\Delta (mgy)\\ \end{aligned} $$ $$ \begin{aligned} \KE_f&=\KE_i-mg\Delta y\\ &=\frac{1}{2}m{v_0}^2-mg(y_f-y_i)\\ \end{aligned} $$ $$ \begin{aligned} \KE(y)&=\frac{1}{2}m{v_0}^2+mg(h-y)\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \KE\(0\)&=\frac{1}{2}m{v_0}^2+mg\(h-0\)\\ &=\frac{53}{8}+\frac{1007 g}{125}\\ &=85.6273724\ut{J}\\ &\approx 85.6\ut{J}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \KE\(\frac{h}{2}\)&=\frac{1}{2}m{v_0}^2+mg\(h-\frac{h}{2}\)\\ &=\frac{53}{8}+\frac{1007g}{250}\\ &=46.1261862\ut{J}\\ &\approx 46.1\ut{J}\\ \end{aligned} $$ $$\ab{c,d}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$-h=-v_0t+\frac{1}{2}(-g)t^2$$ $$ \begin{aligned} t_f&=\frac{-v_0+\sqrt{{v_0}^2+2 g h}}{g}\\ \end{aligned} $$ $$\Delta t = 0.180\ut{s},$$ $$ \begin{aligned} t_1&=t_f-\Delta t\\ &=\frac{-v_0+\sqrt{{v_0}^2+2 g h}}{g}-\Delta t\\ \end{aligned} $$ $$\ab{c}$$ $$$$ $$ \begin{aligned} v&=v_0+at,\\ -v_1&=-v_0-gt_1 \end{aligned} $$ $$ \begin{aligned} \KE(t_1)&=\frac{1}{2}m{v_1}^2\\ &=\frac{1}{2}m(v_0+gt_1)^2\\ &=\frac{1}{2} m \bra{v_0+g \(\frac{\sqrt{2 g h+v_0^2}-v_0}{g}-\Delta t\)}^2\\ &=\frac{1}{2} m \(\sqrt{2 g h+{v_0}^2}-g\Delta t\)^2\\ &=\frac{53}{50} \(\sqrt{\frac{38 g}{5}+\frac{25}{4}}-\frac{9 g}{50}\)^2\\ &\approx 55.29594621128494\ut{J}\\ &\approx 55.3\ut{J}\\ \end{aligned} $$ $$\ab{d}$$ $$$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ -S_1&=-v_0t_1+\frac{1}{2}(-g){t_1}^2 \end{aligned} $$ $$ \begin{aligned} y_1&=h-S_1\\ &=h-v_0t_1-\frac{1}{2}g{t_1}^2\\ \end{aligned} $$ $$ \begin{aligned} \GE(t_1)&=mgy_1\\ &=mg\(h-v_0t_1-\frac{1}{2}g{t_1}^2\)\\ &=\frac{1}{2} m g\Delta t\(2 \sqrt{2 g h+v^2}-g\Delta t\)\\ &=\frac{477 g \(10 \sqrt{760 g+625}-9 g\)}{125000}\\ &\approx 30.331426188715064\ut{J}\\ &\approx 30.3\ut{J}\\ \end{aligned} $$