8-21 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} A:\text{M Box}\\ B:\text{2M Box}\\ C:\text{Spring}\\ \end{cases} $$ $$ \begin{cases} 0:\text{Start}\\ 1:x_1=0.120\ut{m}\\ 2:x_2=\max(x)\\ \end{cases} $$ $$ \begin{cases} m_A&=M=2.80\ut{kg}\\ m_B&=2M\\ k&=230\ut{N/m}\\ \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \LE : \text{Elastic Potential Energy}\\ \end{cases} $$ $$\ab{a}$$ $$\Sigma E_0=\Sigma E_1$$ $$ \begin{aligned} E_{A0}+E_{B0}&=E_{A1}+E_{B1}+E_{C1}\\ \GE_{A0}+\GE_{B0}&=(\KE_{A1}+\GE_{A1})\\ &~~~~~+(\KE_{B1}+\GE_{B1})+LE_{C1}\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=\KE_{A1}+\KE_{B1}\\ &=\GE_{A0}+\GE_{B0}\\ &~~~~~-(\GE_{A1}+\GE_{B1}+\LE_{C1})\\ &=-\Delta \GE_A-\Delta \GE_B-\LE_{C1}\\ &=-m_Ag\Delta (h_A)-m_Bg\Delta (h_B)-\frac{1}{2}k{x_1}^2\\ &=-m_Bg(-x_1)-\frac{1}{2}k{x_1}^2~~~(\because \Delta h_A=0)\\ &=m_Bgx_1-\frac{1}{2}k{x_1}^2\\ &=\frac{3}{125} (28 g-69)\\ &=4.9340688\ut{J}\\ &\approx 4.93 \ut{J}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \KE_{A1}:\KE_{B1}&=\frac{1}{2}m_A{v_1}^2:\frac{1}{2}m_B{v_1}^2,\\ &=m_A:m_B\\ \end{aligned} $$ $$ \begin{aligned} \Ans(a)&=\KE_{A1}+\KE_{B1},\\ \end{aligned} $$ $$ \begin{aligned} \therefore \KE_{B1}&=\frac{m_B}{m_A+m_B}\cdot \Ans(a)\\ &=\frac{2M}{M+2M}\cdot\frac{3}{125} (28 g-69)\\ &=\frac{2}{3}\cdot\frac{3}{125} (28 g-69)\\ &=\frac{2}{125} (28 g-69)\\ &=3.2893792\ut{J}\\ &\approx 3.29\ut{J}\\ \end{aligned} $$ $$\ab{c}$$ $$\Sigma E_0=\Sigma E_2$$ $$ \begin{aligned} E_{A0}+E_{B0}&=E_{A2}+E_{B2}+E_{C2}\\ \GE_{A0}+\GE_{B0}&=(\KE_{A2}+\GE_{A2})\\ &~~~~~+(\KE_{B2}+\GE_{B2})+LE_{C2}\\ &=\GE_{A2}+\GE_{B2}+LE_{C2}\\ \end{aligned} $$ $$ \begin{aligned} 0&=\Delta \GE_A+\Delta \GE_B+LE_{C2}\\ &=m_Ag\Delta h_A+m_Bg\Delta h_B+\frac{1}{2}kx^2\\ &=m_Bg(-x)+\frac{1}{2}kx^2\\ \end{aligned} $$ $$ \begin{aligned} x&=\frac{2m_Bg}{k}\\ &=\frac{28}{575}g\\ &\approx 0.47754121739130434\ut{m}\\ &\approx 0.478\ut{m}\\ &\approx 47.8\ut{cm}\\ \end{aligned} $$