8-29 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} L&=1.50\ut{m}\\ m&=4.00\ut{kg}\\ \theta_0&=30.0\degree\\ \vec v_0&=0\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \put \begin{cases} 0 : \text{Start}\\ 1 : \text{Lowest Point} \end{cases} $$ $$ \put \begin{cases} \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$ \begin{aligned} -\Delta h_{0\rarr1}&=L-L\cos\theta_0\\ &=L(1-\cos\theta_0)\\ \end{aligned} $$ $$\ab{a}$$ $$W_{i\rarr f}=-\Delta \GE_{i\rarr f},$$ $$ \begin{aligned} W_{0\rarr 1}&=-\Delta \GE_{0\rarr 1}\\ &=-\Delta (mgh)_{0\rarr 1}\\ &=mg(-\Delta h_{0\rarr 1})\\ &=mgL(1-\cos\theta_0)\\ &=3 \(2-\sqrt{3}\) g\\ &\approx 7.883051843864012\ut{J}\\ &\approx 7.88\ut{J}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Delta \GE_{0\rarr 1}&=-W_{0\rarr 1}\\ &=-mgL(1-\cos\theta_0)\\ &=-3 \(2-\sqrt{3}\) g\\ &\approx -7.883051843864012\ut{J}\\ &\approx -7.88\ut{J}\\ \end{aligned} $$ $$\ab{c}$$ $$\GE_0+\Delta \GE_{0\rarr 1}=\GE_1$$ $$ \begin{aligned} \GE_0&=\GE_1-\Delta \GE_{0\rarr 1}\\ &=0-\Delta \GE_{0\rarr 1}\\ &=W_{0\rarr 1}\\ &=mgL(1-\cos\theta_0)\\ &=3 \(2-\sqrt{3}\) g\\ &\approx 7.883051843864012\ut{J}\\ &\approx 7.88\ut{J}\\ \end{aligned} $$ $$\ab{d}$$ $$\Ans(a)= mgL(1-\cos\theta_0),$$ $$\text{Increase}$$ $$\Ans(b)= -mgL(1-\cos\theta_0),$$ $$\text{Decrease}$$ $$\Ans(c)= mgL(1-\cos\theta_0),$$ $$\text{Increase}$$