11판/5. 힘과 운동 I 61

5-61 할리데이 11판 솔루션 일반물리학

{mmax=22300[kg]V0=7682[L] \begin{cases} m_{\max}&=22300\ut{kg}\\ V_0&=7682\ut{L}\\ \end{cases} (a,b)\ab{a,b} kA=1.77[kg/L],k_A=1.77\ut{kg/L}, (a)\ab{a} mA0=kAV0=1.77[kg/L]7682[L]=13597.14[kg] \begin{aligned} m_{A0}&=k_AV_0\\ &=1.77\ut{kg/L}\cdot7682\ut{L}\\ &=13597.14\ut{kg}\\ \end{aligned} (b)\ab{b} $$ \begin{aligned} \Delta V&=\frac{\Delta m}{k_A}\\ &=\frac{m_{\max}-m_{A0}}{k_A}\\ &=\frac{870286}{177}\ut{L}\\ &\approx 4916.870056497175\ut{L}\\ &\approx 49..

5-60 할리데이 11판 솔루션 일반물리학

{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} $$ \begin{aligned} \vec F&=m\vec a\\ &=m\frac{\vec v_1-\vec v_0}{t}..

5-59 할리데이 11판 솔루션 일반물리학

{mg=17.0[kN]a=3.66[m/s2]g=9.80665[m/s2] \begin{cases} mg&=17.0\ut{kN}\\ a&=3.66\ut{m/s^2}\\ g&=9.80665\ut{m/s^2} \end{cases} ΣF=ma=62220g6344.67427714867[N]6.34×103[N]6.34[kN] \begin{aligned} \Sigma F&=ma\\ &=\frac{62220}{g}\\ &\approx 6344.67427714867\ut{N}\\ &\approx 6.34\times10^3\ut{N}\\ &\approx 6.34\ut{kN}\\ \end{aligned}

5-58 할리데이 11판 솔루션 일반물리학

{a=3.0j^[m/s2]m=3.0[kg]F1=8.0i^[N] \begin{cases} \vec a&=3.0\j\ut{m/s^2}\\ m&=3.0\ut{kg}\\ \vec F_1&=8.0\i\ut{N} \end{cases} ΣF=ma,\Sigma \vec F = m\vec a, F1+F2=maF2=maF1=9j^8i^[N] \begin{aligned} \vec F_1+\vec F_2 &= m \vec a\\ \vec F_2&=m \vec a-\vec F_1\\ &=9\j-8\i\ut{N}\\ \end{aligned} F2=92+82=145[N]12.041594578792296[N]12[N] \begin{aligned} F_2&=\sqrt{9^2+8^2}\\ &=\sqrt{145}\ut{N}\\ &\approx 12.041594578792296\ut{N}\\ &\approx 12\ut{N}\\ \end{aligned}

5-57 할리데이 11판 솔루션 일반물리학

{m=9.11×1031[kg]S=1.5[cm]v=6.0×106[m/s]g=9.80665[m/s2] \begin{cases} m&=9.11\times10^{-31}\ut{kg}\\ S&=1.5\ut{cm}\\ v&=6.0\times10^6\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases} (a)\ab{a} 2aS=v2v02,2aS=v^2-{v_0}^2, a=v2022S=v22S \begin{aligned} a&=\frac{v^2-{0}^2}{2S}\\ &=\frac{v^2}{2S} \end{aligned} F=ma=mv22S=1.0932×1015[N]1.1×1015[N]1.1[fN] \begin{aligned} F&=ma\\ &=\frac{mv^2}{2S}\\ &=1.0932\times10^{-15}\ut{N}\\ &\approx 1.1\times10^{-15}\ut{N}\\ &\approx 1.1\ut{fN} \end{aligned} (b)\ab{b}..

5-56 할리데이 11판 솔루션 일반물리학

{mg=2.0[kN]t=6.0[s]v0=0v=88.5[km/h]=29512[m/s]g=9.80665[m/s2] \begin{cases} mg &= 2.0\ut{kN}\\ t&=6.0\ut{s}\\ v_0&=0\\ v&=88.5\ut{km/h}=\frac{295}{12}\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases} (a)\ab{a} v=v0+at,v=v_0+at, a=vv0t=2951206=29572[m/s2]4.097222222222222[m/s2]4.1[m/s2] \begin{aligned} a&=\frac{v-v_0}{t}\\ &=\frac{\frac{295}{12}-0}{6}\\ &=\frac{295}{72}\ut{m/s^2}\\ &\approx 4.097222222222222\ut{m/s^2}\\ &\approx 4.1\ut{m/s^2}\\ \end{aligned} (b)\ab{b} ΣF=ma,\Sigma \vec F = m\vec a, $$ \begin{alig..

5-55 할리데이 11판 솔루션 일반물리학

{m=5.0×104[kg]g=9.80665[m/s2] \begin{cases} m&=5.0\times10^4\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases} (a)\ab{a} ΣF=0Δv=0,\Sigma \vec F = 0 \Harr \Delta \vec v=0, ΣF=Fmg=0F=mg=9806652[N]=490332.5[N]4.9×105[N] \begin{aligned} \Sigma F &= F-mg= 0\\ F&=mg\\ &=\frac{980665}{2}\ut{N}\\ &=490332.5\ut{N}\\ &\approx 4.9\times10^5\ut{N} \end{aligned} (b)\ab{b} ΣF=ma,\Sigma \vec F = m\vec a, $$ \begin{aligned} F-mg &= m a\\ F&=m(a+g)\\ &=\frac{2980665}{2}\ut{N}\\ &=1490332..

5-54 할리데이 11판 솔루션 일반물리학

(풀이자주:F1의 표기에 오타가 있는것으로 보입니다. 임의로 =을 +로 고쳐서 풀도록 하겠습니다.) {m=1[kg]a=4.00[m/s2],θa=160°F1=2.50i^+4.60j^[N] \begin{cases} m&=1\ut{kg}\\ a&=4.00\ut{m/s^2},\theta_a&=160\degree\\ \vec F_1&=2.50\i+4.60\j\ut{N} \end{cases} a=4cos160°i^+4sin160°j^ \begin{aligned} \vec a &=4\cos160\degree\i+4\sin160\degree\j\\ \end{aligned} ΣF=ma,F1+F2=ma \begin{aligned} \Sigma \vec F &= m\vec a,\\ \vec F_1+\vec F_2 &= m \vec a\\ \end{aligned} (a)\ab{a} $$ \begin{aligned} \vec F_2 &= m\ve..

5-53 할리데이 11판 솔루션 일반물리학

{mh=15000[kg]mt=4500[kg]a=1.4[m/s2]g=9.80665[m/s2] \begin{cases} m_h&=15000\ut{kg}\\ m_t&=4500\ut{kg}\\ a&=1.4\ut{m/s^2}\\ g&=9.80665\ut{m/s^2} \end{cases} ΣF=ma,\Sigma \vec F = m\vec a, {ΣFh=mhaΣFt=mta \begin{cases} \Sigma F_h &= m_h a\\ \Sigma F_t &= m_t a\\ \end{cases} {FmhgT=mhaTmtg=mta \begin{cases} F-m_hg-T &= m_h a\\ T-m_tg &= m_t a\\ \end{cases} (a)\ab{a} $$ \begin{aligned} F&=(m_h+m_t)(g+a)\\ &=3900(7+5g)\\ &=218529.675\ut{N}\\ &\approx 2.2\times10^5\ut{..

5-52 할리데이 11판 솔루션 일반물리학

{m=0.400[kg]x(t[s])=(13.00+12.0t+3.00t22.00t3)i^[m]t=1.20[s] \begin{cases} m&=0.400\ut{kg}\\ x(t\ut{s})&=(-13.00+12.0t+3.00t^2-2.00t^3)\i\ut{m}\\ t&=1.20\ut{s} \end{cases} ΣF=ma=m ⁣d2x ⁣dt2=m ⁣d2 ⁣dt2(13+12t+3t22t3)i^=m(612t)i^=125(12t)i^ \begin{aligned} \Sigma F&=ma\\ &=m\dxtt{x}\\ &=m\cdot\dtt(-13+12t+3t^2-2t^3)\i\\ &=m(6-12t)\i\\ &=\frac{12}{5}(1-2t)\i \end{aligned} F(1.2)=8425i^[N]=3.36i^[N] \begin{aligned} F(1.2)&=-\frac{84}{25}\i\ut{N}\\ &=-3.36\i\ut{N}\\ \end{aligned}