9-41 할리데이 11판 솔루션 일반물리학

$$\put \begin{cases} \KE : \text{Kinetic Energy}\\ \end{cases}$$ $$\begin{cases} m&=150\ut{g}\\ v_i&=5.2\ut{m/s}\\ \KE_f&=\frac{1}{2}\KE_i \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} \KE_f&=\frac{1}{2}\KE_i\\ \frac{1}{2}m{v_f}^2&=\frac{1}{2}\cdot\frac{1}{2}m{v_i}^2\\ \end{aligned}$$ $$\begin{aligned} v_f&=\sqrt\frac{1}{2}\cdot{v_i}\\ &=\frac{v_i}{\sqrt2}\\ &=\frac{13\sqrt2}{5}\ut{m/s}\\ &\approx 3.6769552621700474\ut{m/s}\\ &\approx 3.68\ut{m/s}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} J&=\abs{\Delta \vec p}\\ &=m\abs{\Delta \vec v}\\ &=m\abs{v_f-v_i}\\ &=m\abs{-\frac{v_i}{\sqrt2}-v_i}\\ &=m\(\frac{v_i}{\sqrt2}+v_i\)\\ &=\frac{39}{100} \(2+\sqrt{2}\)\ut{N\cdot s}\\ &\approx 1.331543289325507\ut{N\cdot s}\\ &\approx 1.3\ut{N\cdot s}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} F&=\frac{J}{t}\\ &=\frac{\frac{39}{100} \(2+\sqrt{2}\)}{7.6 \times 10^{-3}}\ut{N}\\ &=\frac{975}{19}\(2+\sqrt{2}\)\ut{N}\\ &\approx 175.20306438493515\ut{N}\\ &\approx 1.8\times10^2\ut{N}\\ \end{aligned}$$