5-26 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} mg&=2.50\times10^4\ut{N}\\ v_0&=55.0\ut{km/h}=\frac{275}{18}\ut{m/s}\\ S&=24.0\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} 2\(-\frac{F}{m}\)S&=(0)^2-\(v_0\)^2\\ 2\(-\frac{F}{2.5\times10^4}g\)(24)&=-\(\frac{275}{18}\)^2\\ \end{aligned} $$ $$ \begin{aligned} F&\approx 12396.48254347605\ut{N}\\ &\approx 12.4\ut{kN}\\ \end{aligned} $$ $$\ab{b}$$ $$S=\frac{1}{2}\(v+v_0\)t,$$ $$ \begin{aligned} t&=\frac{2S}{(0)+v_0}\\ &=\frac{2\cdot 24}{\frac{275}{18}}\ut{s}\\ &=\frac{864}{275}\ut{s}\\ &\approx 3.141818181818182\ut{s}\\ &\approx 3.14\ut{s}\\ \end{aligned} $$ $$\ab{c}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{cases} 2\(\frac{F}{m}\)S_B&=(0)^2-(2v_0)^2\\ 2\(\frac{F}{m}\)S_A&=(0)^2-{v_0}^2\\ \end{cases} $$ $$ \frac{S_B}{S_A}=2^2=4 $$ $$\ab{d}$$ $$v=v_0+at,$$ $$ \begin{cases} 0&=2v_0+\(\frac{F}{m}\)t_B\\ 0&=v_0+\(\frac{F}{m}\)t_A\\ \end{cases} $$ $$ \begin{aligned} \frac{t_B}{t_A}&=\frac{-2v_0}{-v_0}\\ &=2 \end{aligned} $$