7-24 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=400\ut{g}\\ k&=2.50\ut{N/cm}=250\ut{N/m}\\ x &= 18.0\ut{cm}=0.18\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} W_{mg}&=(-mg)\cdot(- x)\\ &=\frac{9}{125}g\\ &=0.7060788\ut{J}\\ &\approx 706\ut{mJ} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} W_s&=-\frac{1}{2}k x^2\\ &=-\frac{81}{20}\ut{J}\\ &=-4.05\ut{J} \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} K_1&=K_0+\Sigma W,\\ K_0&=K_1-\Sigma W\\ &=-\Sigma W~~(\because v_1=0)\\ \frac{1}{2}m{v_0}^2&=-\(mgx-\frac{1}{2}kx^2\)\\ m{v_0}^2&=kx^2-2mgx\\ \end{aligned} $$ $$ \begin{aligned} v_0&=\sqrt{\frac{kx^2-2mgx}{m}}\\ &=\frac{3}{10}\sqrt{225-4g}\\ &\approx 4.088961481843525\ut{m/s}\\ &\approx 4.09\ut{m/s}\\ \end{aligned} $$ $$\ab{d}$$ $$kx^2-2mgx-m{v_0}^2=0,$$ $$ \begin{aligned} x&=\frac{ mg\pm\sqrt{m^2g^2+k m{v_0}^2}}{k}\\ &\approx 0.34318365455696676\ut{m}~~(\because x\gt 0)\\ &\approx 34.3\ut{cm}\\ \end{aligned} $$