11판/6. 힘과 운동 II 66

6-66 할리데이 11판 솔루션 일반물리학

{θ=2.5°mg=69[N]g=9.80665[m/s2] \begin{cases} \theta&=2.5\degree\\ mg&=69\ut{N}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} ΣF=ma,\Sigma \vec F = m\vec a, ma=mgsinθff=mgsinθma=69sin2.5°1380000196133a \begin{aligned} m a &= mg\sin\theta-f\\ f&=mg\sin\theta-ma\\ &=69 \sin2.5\degree-\frac{1380000}{196133}a \end{aligned} (a)\ab{a} v=0,a=0v=0,a=0 f=69sin2.5°[N]3.009737728208184[N]3.0[N] \begin{aligned} f&=69 \sin2.5\degree\ut{N}\\ &\approx 3.009737728208184\ut{N}\\ &\approx 3.0\ut{N}\\ \end{aligned} $$\text{D..

6-65 할리데이 11판 솔루션 일반물리학

{m=1[kg]R=6.40×106[m]v=465[m/s]F0=9.8[N] \begin{cases} m&=1\ut{kg}\\ R&=6.40\times10^6\ut{m}\\ v&=465\ut{m/s}\\ F_0&=9.8\ut{N}\\ \end{cases} (a)\ab{a} ΣFR=mv2R=8649256000[N]=0.03378515625[N]33.8[mN] \begin{aligned} \Sigma F_R &= \frac{mv^2}{R}\\ &=\frac{8649}{256000}\ut{N}\\ &=0.03378515625\ut{N}\\ &\approx 33.8\ut{mN}\\ \end{aligned} (b)\ab{b} $$ \begin{aligned} \Sigma F_R&= mg-F_N\\ F_N&=mg-\Sigma F_R\\ &=\frac{2500151}{256000}\ut{N}\\ &=9.76621484375\ut{N}\\ &\appro..

6-64 할리데이 11판 솔루션 일반물리학

{mg=220[N]μs=0.41μk=0.32g=9.80665[m/s2] \begin{cases} mg&=220\ut{N}\\ \mu_s&=0.41\\ \mu_k&=0.32\\ g&=9.80665\ut{m/s^2} \end{cases} (a)\ab{a} fsmax=μsmg=4515[N]=90.2[N]90[N] \begin{aligned} f_{s\max}&=\mu_s mg\\ &=\frac{451}{5}\ut{N}\\ &=90.2\ut{N}\\ &\approx 90\ut{N}\\ \end{aligned} (b)\ab{b} fk=μkmg=3525[N]=70.4[N]70[N] \begin{aligned} f_k&=\mu_k mg\\ &=\frac{352}{5}\ut{N}\\ &=70.4\ut{N}\\ &\approx 70\ut{N}\\ \end{aligned} (c)\ab{c} ΣF=ma,\Sigma \vec F = m\vec a, $$ \begin{a..

6-63 할리데이 11판 솔루션 일반물리학

(풀이자주:문제의 의도가 3주기를 원주율로 두라는 경우로 보입니다. 해당 풀이를 별첨합니다.) {m=2.0[g]R=5.0[cm]3T1=3.14[s]g=9.80665[m/s2] \begin{cases} m&=2.0\ut{g}\\ R&=5.0\ut{cm}\\ 3T_1&=3.14\ut{s}\\ g&=9.80665\ut{m/s^2} \end{cases} (a)\ab{a} v1=2πRT1=1500π157[cm/s]30.015216435571272[cm/s]30[cm/s] \begin{aligned} v_1&=\frac{2\pi R}{T_1}\\ &=\frac{1500\pi}{157}\ut{cm/s}\\ &\approx 30.015216435571272\ut{cm/s}\\ &\approx 30\ut{cm/s}\\ \end{aligned} (b)\ab{b} aR=v12R,a_R = \frac{{v_1}^2}{R}, $$ \begin{aligned} a_R&=\frac{4500..

6-62 할리데이 11판 솔루션 일반물리학

{θ=40.0°μk=0.0400m=85.0[kg]A=1.25[m2]C=0.140ρ=1.20[kg/m2] \begin{cases} \theta&=40.0\degree\\ \mu_k&=0.0400\\ m&=85.0\ut{kg}\\ A&=1.25\ut{m^2}\\ C&=0.140\\ \rho&=1.20\ut{kg/m^2}\\ \end{cases} ΣF=0Δv=0,\Sigma \vec F = 0 \Harr \Delta \vec v=0, {ΣFx=0ΣFy=0 \begin{cases} \Sigma F_x &= 0\\ \Sigma F_y &= 0\\ \end{cases} D=12CρAv2,D=\frac{1}{2}C\rho Av^2, {0=mgsinθμN12CρAv20=Nmgcosθ \begin{cases} 0 &= mg\sin\theta-\mu N-\frac{1}{2}C\rho Av^2\\ 0 &= N-mg\cos\theta\\ \end{cases} (a)\ab{a} $$ ..

6-61 할리데이 11판 솔루션 일반물리학

{m=2.8[kg]θ=30°F=15[N]μ=0.25g=9.80665[m/s2] \begin{cases} m&=2.8\ut{kg}\\ \theta&=30\degree\\ F&=15\ut{N}\\ \mu&=0.25\\ g&=9.80665\ut{m/s^2} \end{cases} (a)\ab{a} f=μN=μ(Fsinθ+mg)=7g10+158=8.739655[N]8.7[N] \begin{aligned} f&=\mu N\\ &=\mu(F\sin\theta+mg)\\ &=\frac{7 g}{10}+\frac{15}{8}\\ &=8.739655\ut{N}\\ &\approx 8.7\ut{N}\\ \end{aligned} (b)\ab{b} ΣF=ma\Sigma F=ma $$ \begin{aligned} a&=\frac{F\cos\theta-\mu(F\sin\theta+mg)}{m}\\ &=\frac{1}{112} \(-28 g+300 \sqrt{3}..

6-60 할리데이 11판 솔루션 일반물리학

{θ=30°μ1=μμ2=0t1=2t2 \begin{cases} \theta&=30\degree\\ \mu_1&=\mu\\ \mu_2&=0\\ t_1&=2t_2 \end{cases} S=v0t+12at2,S=v_0t+\frac{1}{2}at^2, S=(0)t+12at2S=(0)t+\frac{1}{2}a{t}^2 a=2St2a=\frac{2S}{t^2} a2=2St22a1=2S(2t2)2=14a2 \begin{aligned} a_2&=\frac{2S}{{t_2}^2}\\ a_1&=\frac{2S}{(2t_2)^2}=\frac{1}{4}a_2 \end{aligned} ΣF=ma,\Sigma \vec F = m\vec a, $$ \begin{aligned} \Sigma F_2 &= m a_2\\ m a_2 &= mg\sin\theta\\ a_2&=g\sin\theta=\frac{g}{2}\\ a_1&=\f..

6-58 할리데이 11판 솔루션 일반물리학

{v=80.0[km/h]=2009[m/s]R=350[m]m=55.0[kg] \begin{cases} v&=80.0\ut{km/h}=\frac{200}{9}\ut{m/s}\\ R&=350\ut{m}\\ m&=55.0\ut{kg}\\ \end{cases} (a)\ab{a} ΣFR=mv2R=44000567[N]77.60141093474427[N]77.6[N] \begin{aligned} \Sigma F_R &= \frac{mv^2}{R}\\ &=\frac{44000}{567}\ut{N}\\ &\approx 77.60141093474427\ut{N}\\ &\approx 77.6\ut{N}\\ \end{aligned} (b)\ab{b} 90° 90\degree

6-57 할리데이 11판 솔루션 일반물리학

{m1=30[kg]m2=15[kg]μs=0.60μk=0.40F=100[N]g=9.80665[m/s2] \begin{cases} m_1&=30\ut{kg}\\ m_2&=15\ut{kg}\\ \mu_s&=0.60\\ \mu_k&=0.40\\ F&=100\ut{N}\\ g&=9.80665\ut{m/s^2} \end{cases} ΣF=ma,\Sigma \vec F = m\vec a, {ΣF1x=m1a1ΣF1y=0ΣF2x=m2a2ΣF2y=0 \begin{cases} \Sigma F_{1x} &= m_1 a_1\\ \Sigma F_{1y} &= 0\\ \Sigma F_{2x} &= m_2 a_2\\ \Sigma F_{2y} &= 0\\ \end{cases} {m1a1=f0=N1N2m1gm2a2=F+f0=N2m2g \begin{cases} m_1 a_1 &= -f\\ 0 &= N_1-N_2-m_1g\\ m_2 a_2 &= -F+f\\ 0 &= N_2-m_2g\\ \end{cases} $$ \begin..