8-64 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} k&=4200\ut{N/m}\\ U_i&=1.10\ut{J}\\ \end{cases} $$ $$U=\frac{1}{2}kx^2,$$ $$ x_i=\frac{1}{10}\sqrt{\frac{11}{210}}\ut{m} $$ $$x_a=x_i+\Delta x$$ $$ \begin{aligned} \Delta U &= U_f-U_i\\ &=\frac{1}{2}k{x_a}^2-\frac{1}{2}k{x_i}^2\\ &=\frac{1}{2}k\Delta x(\Delta x+2x_i)\\ &=2\Delta x(\sqrt{2130}+1050\Delta x) \end{aligned} $$ $$\ab{a}$$ $$\Delta x=+2\ut{cm}$$ $$ \begin{aligned} \Delta U &=2\Delta x(\sqrt{2130}+1050\Delta x)\\ &=\frac{1}{25} \(21+\sqrt{2310}\)\ut{J}\\ &\approx 2.7624983745116665\ut{J}\\ &\approx 2.76\ut{J}\\ \end{aligned} $$ $$\ab{b}$$ $$\Delta x=-2\ut{cm}$$ $$ \begin{aligned} \Delta U &=2\Delta x(\sqrt{2130}+1050\Delta x)\\ &=\frac{1}{25} \(21-\sqrt{2310}\)\ut{J}\\ &\approx -1.0824983745116665\ut{J}\\ &\approx -1.08\ut{J}\\ \end{aligned} $$ $$\ab{c}$$ $$\Delta x=-4\ut{cm}$$ $$ \begin{aligned} \Delta U &=2\Delta x(\sqrt{2130}+1050\Delta x)\\ &=\frac{2}{25} \(42-\sqrt{2310}\)\ut{J}\\ &\approx -0.4849967490233331\ut{J}\\ &\approx -0.485\ut{J}\\ &\approx -485\ut{mJ}\\ \end{aligned} $$