7-43 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} k&=15.0\ut{N/cm}=1.5\times10^3\ut{N/m}\\ x&=6.20\ut{mm}=6.2\times10^{-3}\ut{m}\\ \Delta x &= 7.60\ut{mm}=7.6\times10^{-3}\ut{m}\\ \end{cases} $$ $$\ab{a}$$ $$W_S=\frac{1}{2}kx^2,$$ $$ \begin{aligned} W_S&=\frac{1}{2}\(1.5\times10^3\)\(6.2\times10^{-3}\)^2\\ &=\frac{2883}{10^5}\ut{J}\\ &=0.02883\ut{J}\\ &\approx 2.88\times10^{-2}\ut{J}\\ &\approx 28.8\ut{mJ}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Delta W &= \frac{1}{2}k(x+\Delta x)^2-\frac{1}{2}k{x}^2\\ &=\frac{1}{2}k\Delta x(2x+\Delta x)\\ &=\frac{57}{500}\ut{J}\\ &=0.114\ut{J}\\ &=114\ut{mJ} \end{aligned} $$