8-17 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_A&=2.00\ut{kg}\\ m_B&=4.00\ut{kg}\\ \theta&=30.0\degree\\ v_0&=0\\ \Delta y_B&=-32.0\ut{cm}=-0.32\ut{m} \end{cases} $$ $$ \begin{aligned} \sin\theta &= \frac{\Delta y_A}{\Delta d_A}\\ &= \frac{\Delta y_A}{-\Delta y_B}\\ \therefore \Delta y_A&=-\Delta y_B\sin\theta \end{aligned} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$ \begin{aligned} \Sigma E_i&=\Sigma E_f,\\ &=E_{Af}+E_{Bf}\\ &=(\KE_{Af}+\GE_{Af})+(\KE_{Bf}+\GE_{Bf})\\ &=(\KE_{Af}+\KE_{Bf})+(\GE_{Af}+\GE_{Bf})\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=\KE_{Af}+\KE_{Bf}\\ &=\Sigma E_i-(\GE_{Af}+\GE_{Bf})\\ &=(\GE_{Ai}+\GE_{Bi})-(\GE_{Af}+\GE_{Bf})~~~(\because v_0=0)\\ &=m_Agy_{Ai}+m_Bgy_{Bi}-m_Agy_{Af}-m_Bgy_{Bf}\\ &=m_Ag(y_{Ai}-y_{Af})+m_Bg(y_{Bi}-y_{Bf})\\ &=-m_Ag\Delta y_A-m_Bg\Delta y_B\\ &=-m_Ag(-\Delta y_B\sin\theta)-m_Bg\Delta y_B\\ &=g\Delta y_B(m_A\sin\theta-m_B)\\ &=\frac{24}{25}g\\ &=9.414384\ut{J}\\ &\approx 9.41\ut{J}\\ \end{aligned} $$