7-17 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=1.67\times10^{-27}\ut{kg}\\ a&=1.7\times10^{15}\ut{m/s^2}\\ v_0&=2.4\times10^6\ut{m/s}\\ S&=2.8\ut{cm}=0.028\ut{m} \end{cases} $$ $$\ab{a}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} v_1&=\sqrt{2aS+{v_0}^2}\\ &=4\sqrt{631}\times10^5\ut{m/s}\\ &\approx 1.0047885349664377\times10^7\ut{m/s}\\ &\approx 1.0\times10^7\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$K=\frac{1}{2}mv^2,$$ $$ \begin{aligned} \Delta K &= K_1-K_0\\ &=\frac{1}{2}m{v_1}^2-\frac{1}{2}m{v_0}^2\\ &=\frac{1}{2}m\({v_1}^2-{v_0}^2\)\\ &=\frac{19873}{25}\times10^{-16}\ut{J}\\ &=7.9492\times10^{-14}\ut{J}\\ &\approx 7.9\times10^{-14}\ut{J}\\ \end{aligned} $$ $$\title{Other Answer}$$ $$\ab{b}$$ $$ \begin{aligned} \Delta K &= W\\ &=F\cdot S\\ &= maS\\ &=\frac{19873}{25}\times10^{-16}\ut{J}\\ &=7.9492\times10^{-14}\ut{J}\\ &\approx 7.9\times10^{-14}\ut{J}\\ \end{aligned} $$