8-27 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=0.040\ut{kg}\\ R&=14\ut{cm}=0.14\ut{m}\\ h&=5.0R\\ \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$ \put \begin{cases} P : \text{P Point}\\ L : \text{Lowest Point}\\ Q : \text{Q Point}\\ H : \text{Highest Point}\\ \end{cases} $$ $$ \begin{cases} h_P&=h=5R\\ h_L&=0\\ h_Q&=R\\ h_H&=2R\\ \end{cases} $$ $$W_{i\rarr f}=-\Delta \GE_{i\rarr f},$$ $$ \begin{aligned} W_{i\rarr f}&=-\Delta (mgh)\\ &=mg\Delta (-h_{i\rarr f})\\ &=mg(h_i-h_f)\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} W_{P\rarr Q}&=mg(h_P-h_Q)\\ &=mg(4R)\\ &=\frac{14g}{625}\\ &=0.21966896\ut{J}\\ &\approx 0.22\ut{J} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} W_{P\rarr H}&=mg(h_P-h_H)\\ &=mg(3R)\\ &=\frac{21g}{1250}\\ &=0.16475172\ut{J}\\ &\approx 0.16\ut{J} \end{aligned} $$ $$\ab{c}$$ $$\GE_P+\Delta \GE_{P\rarr L}=\GE_L,$$ $$ \begin{aligned} \GE_P&=\GE_L-\Delta \GE_{P\rarr L}\\ &=0-\Delta(mgh_{P\rarr L})\\ &=-mg(\Delta h_{P\rarr L})\\ &=-mg(-5R)\\ &=5mgR\\ &=\frac{7g}{250}\\ &=0.2745862\ut{J}\\ &\approx 0.27\ut{J}\\ \end{aligned} $$ $$\ab{d}$$ $$\GE_L+\Delta \GE_{L\rarr Q}=\GE_Q,$$ $$ \begin{aligned} \GE_Q&=\GE_L+\Delta \GE_{L\rarr Q}\\ &=0+\Delta(mgh_{L\rarr Q})\\ &=mg(\Delta h_{L\rarr Q})\\ &=mgR\\ &=\frac{7g}{1250}\\ &=0.05491724\ut{J}\\ &\approx 0.055\ut{J}\\ &\approx 55\ut{mJ}\\ \end{aligned} $$ $$\ab{e}$$ $$\GE_L+\Delta \GE_{L\rarr H}=\GE_H,$$ $$ \begin{aligned} \GE_H&=\GE_L+\Delta \GE_{L\rarr H}\\ &=0+\Delta(mgh_{L\rarr H})\\ &=mg(\Delta h_{L\rarr H})\\ &=mg(2R)\\ &=\frac{7g}{625}\\ &=0.10983448\ut{J}\\ &\approx 0.11\ut{J}\\ \end{aligned} $$ $$\ab{f}$$ $$\text{All Answer : Constant about }v_0$$