8-26 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} k &= 120\ut{N/m}\\ m &= 4.0\ut{kg}\\ \theta &= 40.0\degree\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \LE : \text{Elastic Potential Energy}\\ \end{cases} $$ $$ \begin{aligned} -\Delta h &= \Delta x\sin\theta \end{aligned} $$ $$\Sigma E_i=\Sigma E_f,$$ $$\KE_i+\GE_i+\LE_i=\KE_f+\GE_f+\LE_i$$ $$ \begin{aligned} \GE_i&=\KE_f+\GE_f+\LE_f\\ \KE_f&=-\Delta \GE-\LE_f\\ \frac{1}{2}m{v_f}^2&=-\Delta(mgh)_{i\rarr f}-\frac{1}{2}k{x_f}^2\\ \end{aligned} $$ $$ \begin{aligned} {v_f}^2&=2g(-\Delta h_{i\rarr f})-\frac{k{x_f}^2}{m}\\ &=2g\Delta(x_{i\rarr f}\sin\theta)-\frac{k{x_f}^2}{m}\\ v(x)&=\sqrt{(2g\sin\theta )x-\(\frac{k}{m}\){x}^2}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} v(10\ut{cm})&=v\(\frac{1}{10}\ut{m}\)\\ &=\sqrt{2g\sin\theta \(\frac{1}{10}\)-\(\frac{k}{m}\)\(\frac{1}{10}\)^2}\\ &\approx 0.9801625490226099\ut{m/s}\\ &\approx 0.98\ut{m/s}\\ &\approx 98\ut{cm/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ v(x)=\sqrt{(2g\sin\theta )x-\(\frac{k}{m}\){x}^2}=0,$$ $$ \begin{aligned} x_b&=\frac{2mg}{k}\sin\theta\\ &\approx 0.4202395408355\ut{m}\\ &\approx 0.42\ut{m}\\ &\approx 42\ut{cm}\\ \end{aligned} $$ $$\ab{c}$$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{aligned} a(x)&= \frac{\Sigma F}{m}\\ &=\frac{kx-mg\sin\theta}{m}\\ &=\frac{k}{m}x-g\sin\theta\\ \end{aligned} $$ $$ \begin{aligned} a(x_b)&=\frac{k}{m}\(\frac{2mg}{k}\sin\theta\)-g\sin\theta\\ &=g\sin\theta\\ &\approx 6.3035931125325\ut{m/s^2}\\ &\approx 6.3\ut{m/s^2}\\ \end{aligned} $$ $$\ab{d}$$ $$\text{Up}$$