11판/5. 힘과 운동 I

5-50 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 21. 21:54
{mA=13[kg]mB=20[kg] \begin{cases} m_A&=13\ut{kg}\\ m_B&=20\ut{kg} \end{cases} (a)\ab{a} ΣF=ma,\Sigma \vec F = m\vec a, {ΣFA=mAaAΣFB0 \begin{cases} \Sigma F_A &= m_A a_A\\ \Sigma F_B &\ge0\\ \end{cases} {TmAg=mAaAT>mBg \begin{cases} T-m_Ag &= m_A a_A\\ T &\gt m_Bg\\ \end{cases} a713g    5.280503846153846[m/s2]    5.3[m/s2] \begin{aligned} a&\ge\frac{7}{13}g\\ &~~~~\approx5.280503846153846\ut{m/s^2}\\ &~~~~\approx5.3\ut{m/s^2}\\ \end{aligned} (b)\ab{b} ΣF=ma,\Sigma \vec F = m\vec a, {ΣFA=mAaΣFB=mBa \begin{cases} \Sigma F_A &= m_A a\\ \Sigma F_B &= m_B a\\ \end{cases} {TmAg=mAamBgT=mBa \begin{cases} T-m_Ag &= m_A a\\ m_Bg-T &= m_B a\\ \end{cases} (b)\ab{b} a=733g2.0801984848484847[m/s2]2.1[m/s2] \begin{aligned} a&=\frac{7}{33}g\\ &\approx 2.0801984848484847\ut{m/s^2}\\ &\approx 2.1\ut{m/s^2}\\ \end{aligned} (c)\ab{c} y+ y+ (d)\ab{d} T=52033g154.5290303030303[N]1.5×102[N] \begin{aligned} T&=\frac{520}{33}g\\ &\approx 154.5290303030303\ut{N}\\ &\approx 1.5\times10^2\ut{N}\\ \end{aligned}