11판/5. 힘과 운동 I

5-39 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 19. 19:05
{m=4.0[kg]F1=3.0i^+4.0j^[N] \begin{cases} m&=4.0\ut{kg}\\ \vec F_1 &= 3.0\i+4.0\j\ut{N} \end{cases} a=ΣFm=F1+F2m \begin{aligned} \vec a&=\frac{\Sigma \vec F}{m}\\ &=\frac{\vec F_1+\vec F_2}{m}\\ \end{aligned} (a)\ab{a} F2=3.0i^4.0j^[N],\vec F_2 = -3.0\i-4.0\j\ut{N}, aa=F1+F2m=0 \begin{aligned} \vec a_a&=\frac{\vec F_1+\vec F_2}{m}\\ &=0 \end{aligned} (b)\ab{b} F2=3.0i^+4.0j^[N],\vec F_2 = -3.0\i+4.0\j\ut{N}, ab=F1+F2m=84j^[m/s2]=2j^[m/s2] \begin{aligned} \vec a_b&=\frac{\vec F_1+\vec F_2}{m}\\ &=\frac{8}{4}\j\ut{m/s^2}\\ &=2\j\ut{m/s^2}\\ \end{aligned} (c)\ab{c} F2=3.0i^4.0j^[N],\vec F_2 = 3.0\i-4.0\j\ut{N}, ab=F1+F2m=64i^[m/s2]=1.5j^[m/s2] \begin{aligned} \vec a_b&=\frac{\vec F_1+\vec F_2}{m}\\ &=\frac{6}{4}\i\ut{m/s^2}\\ &=1.5\j\ut{m/s^2}\\ \end{aligned}