11판/5. 힘과 운동 I 61

5-31 할리데이 11판 솔루션 일반물리학

{m=6.3[kg]F1=11i^[N]F2=25j^[N]F3=6.0i^[N]F4=14[N],θ4=30°F5=5.0j^[N] \begin{cases} m&=6.3\ut{kg}\\ \vec F_1&=-11\i\ut{N}\\ \vec F_2&=-25\j\ut{N}\\ \vec F_3&=6.0\i\ut{N}\\ F_4&=14\ut{N},\theta_4&=30\degree\\ \vec F_5&=5.0\j\ut{N}\\ \end{cases} (a)\ab{a} $$ \begin{aligned} \vec a&=\frac{1}{m}\Sigma \vec F\\ &=\frac{1}{6.3}\bra{\(7\sqrt3-5\)\i-13\j}\ut{m/s^2}\\ &=\frac{10\(7 \sqrt{3}-5\)}{63} \i-\frac{130}{63}\j\ut{m/s^2}\\ &\approx 1.1\i-2.1\j\ut{m/s^2}\\ \e..

5-30 할리데이 11판 솔루션 일반물리학

{v=17i^23j^[m/s]F1=4.0i^+5.0j^6.0k^[N]F2=6.0i^+7.0j^3.5k^[N] \begin{cases} \vec v&=17\i-23\j\ut{m/s}\\ \vec F_1&=4.0\i+5.0\j-6.0\k\ut{N}\\ \vec F_2&=-6.0\i+7.0\j-3.5\k\ut{N}\\ \end{cases} ΣF=0Δv=0\Sigma \vec F=0 \Harr \Delta \vec v=0 F3=F1F2=2.0i^12j^+9.5k^[N] \begin{aligned} \vec F_3 &= -\vec F_1-\vec F_2\\ &=2.0\i-12\j+9.5\k\ut{N} \end{aligned}

5-29 할리데이 11판 솔루션 일반물리학

{mA=60[kg]mB=5.0[kg]a=2.8[m/s2] \begin{cases} m_A&=60\ut{kg}\\ m_B&=5.0\ut{kg}\\ a&=2.8\ut{m/s^2}\\ \end{cases} 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=mAaFTmBg=mBa \begin{cases} T-m_Ag &= -m_A a\\ F-T-m_Bg &= -m_B a\\ \end{cases} (a)\ab{a} F=13(5g14)=455.43225[N]4.6×102[N] \begin{aligned} F&=13 (5 g-14)\\ &=455.43225\ut{N}\\ &\approx 4.6\times10^2\ut{N}\\ \end{aligned} (b)\ab{b} $$ \begin..

5-28 할리데이 11판 솔루션 일반물리학

{m1=2.00[kg]m2=2.70[kg]a=4.50[m/s2]F=7.00[N]g=9.80665[m/s2] \begin{cases} m_1&=2.00\ut{kg}\\ m_2&=2.70\ut{kg}\\ a&=4.50\ut{m/s^2}\\ F&=7.00\ut{N}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} ΣF=ma,\Sigma \vec F = m\vec a, {ΣF1=m1aΣF2=m2a \begin{cases} \Sigma F_1 &= m_1 a\\ \Sigma F_2 &= m_2 a\\ \end{cases} {T+m1gsinβ=m1am2gTF=m2a \begin{cases} T+m_1g\sin\beta &= m_1 a\\ m_2g-T-F&= m_2 a\\ \end{cases} (a)\ab{a} $$ \begin{aligned} T&=\frac{1}{20} (54 g-383)\\ &\approx 7.327954999999998\..

5-27 할리데이 11판 솔루션 일반물리학

{θ=30.0°F=510[N]f=90.0[N]g=9.80665[m/s2] \begin{cases} \theta &= 30.0\degree\\ F&=510\ut{N}\\ f&=90.0\ut{N}\\ g&=9.80665\ut{m/s^2} \end{cases} a=ΣFm=Fcos30°fm=255390m \begin{aligned} a&=\frac{\Sigma F}{m}\\ &=\frac{F\cos30\degree-f}{m}\\ &=\frac{255\sqrt3-90}{m}\\ \end{aligned} (a)\ab{a} m1=282[kg]a1=2553902821.247067219610155[m/s2]1.25[m/s2] \begin{aligned} m_1&=282\ut{kg}\\ a_1&=\frac{255\sqrt3-90}{282}\\ &\approx 1.247067219610155\ut{m/s^2}\\ &\approx 1.25\ut{m/s^2} \end{aligned} (b)\ab{b} $..

5-26 할리데이 11판 솔루션 일반물리학

{mg=2.50×104[N]v0=55.0[km/h]=27518[m/s]S=24.0[m]g=9.80665[m/s2] \begin{cases} mg&=2.50\times10^4\ut{N}\\ v_0&=55.0\ut{km/h}=\frac{275}{18}\ut{m/s}\\ S&=24.0\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} (a)\ab{a} 2aS=v2v02,2aS=v^2-{v_0}^2, 2(Fm)S=(0)2(v0)22(F2.5×104g)(24)=(27518)2 \begin{aligned} 2\(-\frac{F}{m}\)S&=(0)^2-\(v_0\)^2\\ 2\(-\frac{F}{2.5\times10^4}g\)(24)&=-\(\frac{275}{18}\)^2\\ \end{aligned} $$ \begin{aligned} F&\approx 12396.48254347605\ut{N}\\ &\approx 12.4\ut{kN}\\ \end{aligned} ..

5-25 할리데이 11판 솔루션 일반물리학

{m=85[kg]gE=9.80665[m/s2]gM=3.7[m/s2]gS=0 \begin{cases} m&=85\ut{kg}\\ g_E&=9.80665\ut{m/s^2}\\ g_M&=3.7\ut{m/s^2}\\ g_S&=0\\ \end{cases} (a)\ab{a} mgE=833.56525[N]8.3×102[N] \begin{aligned} mg_E&=833.56525\ut{N}\\ &\approx 8.3\times10^2\ut{N} \end{aligned} (b)\ab{b} mgM=314.5[N]3.1×102[N] \begin{aligned} mg_M&=314.5\ut{N}\\ &\approx 3.1\times10^2\ut{N} \end{aligned} (c)\ab{c} mgS=0 \begin{aligned} mg_S&=0\\ \end{aligned} (d)\ab{d} m=mE=mM=mS=85[kg] m=m_E=m_M=m_S=85\ut{kg}

5-24 할리데이 11판 솔루션 일반물리학

{F1=5.0i^+4.0j^[N]F2=2.0i^5.0j^[N]m=1[kg] \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} (a)\ab{a} Fnet=F1+F2=3i^j[N] \begin{aligned} \vec F_\text{net}&=\vec F_1+\vec F_2\\ &=3\i-j\ut{N}\\ \end{aligned} (b)\ab{b} Fnet=32+(1)2[N]=10[N]3.16227766016838[N]3.2[N] \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} (c)\ab{c} $$ \begin{align..

5-23 할리데이 11판 솔루션 일반물리학

{m=3.50×106[kg]t1=34.0[day]v=0.100cc=3.0×108[m/s]g=9.80665[m/s2] \begin{cases} m&=3.50\times10^6\ut{kg}\\ t_1&=34.0\ut{day}\\ v&=0.100c\\ c&=3.0\times10^8\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases} (a)\ab{a} v=v0+at,v=v_0+at, $$ \begin{aligned} a&=\frac{v-v_0}{t}\\ &=\frac{0.1c-0}{34\ut{day}}\\ &=\frac{3\times10^7\ut{m/s}}{34\ut{day}}\cdot\frac{1\ut{day}}{24\ut{h}}\cdot\frac{1\ut{h}}{3600\ut{s}}\\ &=\frac{3125}{306}\ut{m/s^2}\\ &\approx 10.21241830065359\u..

5-22 할리데이 11판 솔루션 일반물리학

{m=72.0[kg]θ=5.00°a=3.5[m/s2]g=9.80665[m/s2] \begin{cases} m&=72.0\ut{kg}\\ \theta&=5.00\degree\\ a&=3.5\ut{m/s^2}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} (a)\ab{a} ΣF=ma=723.5=252[N]2.5×102[N] \begin{aligned} \Sigma F&=ma\\ &=72\cdot3.5\\ &=252\ut{N}\\ &\approx 2.5\times10^2\ut{N}\\ \end{aligned} (b)\ab{b} ΣFy=FNmgcosθ=0FN=mgcosθ, \begin{aligned} \Sigma F_y&=F_N-mg\cos\theta=0\\ F_N&=mg\cos\theta, \end{aligned} $$ \begin{aligned} \Sigma \vec F&=\vec F+m\vec g=m\vec a\\ \end{..