11판/5. 힘과 운동 I

5-60 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 23. 21:09

{m=0.20[kg]v0=2.0i^[m/s]t=0.50[s] \begin{cases} m&=0.20\ut{kg}\\ \vec v_0&=2.0\i\ut{m/s}\\ t&=0.50\ut{s}\\ \end{cases} (a)\ab{a} v1=5.0i^[m/s]\vec v_1=-5.0\i\ut{m/s} F=ma=mv1v0t=(0.2)5i^2i^0.5=145i^F=2.8[N],West \begin{aligned} \vec F&=m\vec a\\ &=m\frac{\vec v_1-\vec v_0}{t}\\ &=(0.2)\frac{-5\i-2\i}{0.5}\\ &=-\frac{14}{5}\i\\ F&=2.8\ut{N},\text{West} \end{aligned} (b)\ab{b} v1=5.0j^[m/s]\vec v_1=-5.0\j\ut{m/s} F=ma=mv1v0t=(0.2)5j^2i^0.5=0.8i^2j^ \begin{aligned} \vec F&=m\vec a\\ &=m\frac{\vec v_1-\vec v_0}{t}\\ &=(0.2)\frac{-5\j-2\i}{0.5}\\ &=-0.8\i-2\j\\ \end{aligned} F=0.82+22=2529[N]2.1540659228538015[N]2.2[N] \begin{aligned} F&=\sqrt{0.8^2+2^2}\\ &=\frac{2}{5}\sqrt{29}\ut{N}\\ &\approx 2.1540659228538015\ut{N}\\ &\approx 2.2\ut{N}\\ \end{aligned} θ=tan120.81.9513027039072615[rad]2.0[rad] \begin{aligned} \theta &=\tan^{-1}\frac{-2}{-0.8}\\ &\approx -1.9513027039072615\ut{rad}\\ &\approx -2.0\ut{rad}\\ \end{aligned}