5-24 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec F_1 &= 5.0\i+4.0\j\ut{N}\\ \vec F_2 &=-2.0\i-5.0\j\ut{N}\\ m&=1\ut{kg} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec F_\text{net}&=\vec F_1+\vec F_2\\ &=3\i-j\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} F_\text{net}&=\sqrt{3^2+(-1)^2}\ut{N}\\ &=\sqrt{10}\ut{N}\\ &\approx 3.16227766016838\ut{N}\\ &\approx 3.2\ut{N}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \theta_F&=-\tan ^{-1}\frac{1}{3}\\ &\approx -0.3217505543966422\ut{rad}\\ &\approx -0.32\ut{rad}\\ \end{aligned} $$ $$\ab{d,e}$$ $$ \begin{aligned} \vec a &=\frac{\vec F}{m}\\ &=\frac{3\i-j}{1}\ut{m/s^2}\\ &=3\i-j\ut{m/s^2}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} a&=\sqrt{3^2+(-1)^2}\ut{N}\\ &=\sqrt{10}\ut{m/s^2}\\ &\approx 3.16227766016838\ut{m/s^2}\\ &\approx 3.2\ut{m/s^2}\\ \end{aligned} $$ $$\ab{e}$$ $$ \begin{aligned} \theta_a&=-\tan ^{-1}\frac{1}{3}\\ &\approx -0.3217505543966422\ut{rad}\\ &\approx -0.32\ut{rad}\\ \end{aligned} $$