7-8 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=3.0\ut{kg}\\ \vec F(x\ut{m})&=-6x\i\ut{N}\\ v(3.0\ut{m})&=6.0\ut{m/s}\\ \end{cases} $$ $$W_{i\rarr f}=\int_{i}^{f} \vec F \cdot \dd \vec S,$$ $$ \begin{aligned} W_{3\rarr x}&=\int_{3}^{x} F \dd x\\ &=\int_{3}^{x} (-6x) \dd x\\ &=27-3x^2 \end{aligned} $$ $$K_x=K_3+W_{3\rarr x},$$ $$\frac{1}{2}m{v_x}^2=\frac{1}{2}m{v_3}^2+(27-3x^2)$$ $$ \begin{aligned} {v_x}^2&=\frac{54+m{v_3}^2-6x^2}{m}\\ &=54-2x^2 \end{aligned} $$ $$ \begin{aligned} \therefore v(x)&=\sqrt{54-2x^2} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} v(4.0)&=\sqrt{22}\ut{m/s}\\ &\approx 4.69041575982343\ut{m/s}\\ &\approx 4.7\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$v(x)=\sqrt{54-2x^2}=5,$$ $$ \begin{aligned} x&=\pm\sqrt\frac{29}{2}\ut{m}\\ &\approx \pm 3.8078865529319543\ut{m}\\ &\approx \pm 3.8\ut{m}\\ \end{aligned} $$