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} ⎩⎨⎧θmgg=2.5°=69[N]=9.80665[m/s2] ΣF⃗=ma⃗,\Sigma \vec F = m\vec a,ΣF=ma, ma=mgsinθ−ff=mgsinθ−ma=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} maf=mgsinθ−f=mgsinθ−ma=69sin2.5°−1961331380000a (a)\ab{a}(a) v=0,a=0v=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} f=69sin2.5°[N]≈3.009737728208184[N]≈3.0[N] $$\text{D.. 11판/6. 힘과 운동 II 2024.03.09
6-65 할리데이 11판 솔루션 일반물리학 {m=1[kg]R=6.40×106[m]v=465[m/s]F0=9.8[N] \begin{cases} m&=1\ut{kg}\\ R&=6.40\times10^6\ut{m}\\ v&=465\ut{m/s}\\ F_0&=9.8\ut{N}\\ \end{cases} ⎩⎨⎧mRvF0=1[kg]=6.40×106[m]=465[m/s]=9.8[N] (a)\ab{a}(a) ΣFR=mv2R=8649256000[N]=0.03378515625[N]≈33.8[mN] \begin{aligned} \Sigma F_R &= \frac{mv^2}{R}\\ &=\frac{8649}{256000}\ut{N}\\ &=0.03378515625\ut{N}\\ &\approx 33.8\ut{mN}\\ \end{aligned} ΣFR=Rmv2=2560008649[N]=0.03378515625[N]≈33.8[mN] (b)\ab{b}(b) $$ \begin{aligned} \Sigma F_R&= mg-F_N\\ F_N&=mg-\Sigma F_R\\ &=\frac{2500151}{256000}\ut{N}\\ &=9.76621484375\ut{N}\\ &\appro.. 11판/6. 힘과 운동 II 2024.03.09
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} ⎩⎨⎧mgμsμkg=220[N]=0.41=0.32=9.80665[m/s2] (a)\ab{a}(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} fsmax=μsmg=5451[N]=90.2[N]≈90[N] (b)\ab{b}(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} fk=μkmg=5352[N]=70.4[N]≈70[N] (c)\ab{c}(c) ΣF⃗=ma⃗,\Sigma \vec F = m\vec a,ΣF=ma, $$ \begin{a.. 11판/6. 힘과 운동 II 2024.03.08
6-63 할리데이 11판 솔루션 일반물리학 (풀이자주:문제의 의도가 3주기를 원주율로 두라는 경우로 보입니다. 해당 풀이를 별첨합니다.) {m=2.0[g]R=5.0[cm]3T1=3.14[s]g=9.80665[m/s2] \begin{cases} m&=2.0\ut{g}\\ R&=5.0\ut{cm}\\ 3T_1&=3.14\ut{s}\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧mR3T1g=2.0[g]=5.0[cm]=3.14[s]=9.80665[m/s2] (a)\ab{a}(a) v1=2πRT1=1500π157[cm/s]≈30.015216435571272[cm/s]≈30[cm/s] \begin{aligned} v_1&=\frac{2\pi R}{T_1}\\ &=\frac{1500\pi}{157}\ut{cm/s}\\ &\approx 30.015216435571272\ut{cm/s}\\ &\approx 30\ut{cm/s}\\ \end{aligned} v1=T12πR=1571500π[cm/s]≈30.015216435571272[cm/s]≈30[cm/s] (b)\ab{b}(b) aR=v12R,a_R = \frac{{v_1}^2}{R},aR=Rv12, $$ \begin{aligned} a_R&=\frac{4500.. 11판/6. 힘과 운동 II 2024.03.08
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} ⎩⎨⎧θμkmACρ=40.0°=0.0400=85.0[kg]=1.25[m2]=0.140=1.20[kg/m2] ΣF⃗=0⇔Δv⃗=0,\Sigma \vec F = 0 \Harr \Delta \vec v=0,ΣF=0⇔Δv=0, {ΣFx=0ΣFy=0 \begin{cases} \Sigma F_x &= 0\\ \Sigma F_y &= 0\\ \end{cases} {ΣFxΣFy=0=0 D=12CρAv2,D=\frac{1}{2}C\rho Av^2,D=21CρAv2, {0=mgsinθ−μN−12CρAv20=N−mgcosθ \begin{cases} 0 &= mg\sin\theta-\mu N-\frac{1}{2}C\rho Av^2\\ 0 &= N-mg\cos\theta\\ \end{cases} {00=mgsinθ−μN−21CρAv2=N−mgcosθ (a)\ab{a}(a) $$ .. 11판/6. 힘과 운동 II 2024.03.08
6-61 할리데이 11판 솔루션 일반물리학 {m=2.8[kg]θ=30°F=15[N]μ=0.25g=9.80665[m/s2] \begin{cases} m&=2.8\ut{kg}\\ \theta&=30\degree\\ F&=15\ut{N}\\ \mu&=0.25\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧mθFμg=2.8[kg]=30°=15[N]=0.25=9.80665[m/s2] (a)\ab{a}(a) f=μN=μ(Fsinθ+mg)=7g10+158=8.739655[N]≈8.7[N] \begin{aligned} f&=\mu N\\ &=\mu(F\sin\theta+mg)\\ &=\frac{7 g}{10}+\frac{15}{8}\\ &=8.739655\ut{N}\\ &\approx 8.7\ut{N}\\ \end{aligned} f=μN=μ(Fsinθ+mg)=107g+815=8.739655[N]≈8.7[N] (b)\ab{b}(b) ΣF=ma\Sigma F=maΣF=ma $$ \begin{aligned} a&=\frac{F\cos\theta-\mu(F\sin\theta+mg)}{m}\\ &=\frac{1}{112} \(-28 g+300 \sqrt{3}.. 11판/6. 힘과 운동 II 2024.03.08
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} ⎩⎨⎧θμ1μ2t1=30°=μ=0=2t2 S=v0t+12at2,S=v_0t+\frac{1}{2}at^2,S=v0t+21at2, S=(0)t+12at2S=(0)t+\frac{1}{2}a{t}^2S=(0)t+21at2 a=2St2a=\frac{2S}{t^2}a=t22S 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} a2a1=t222S=(2t2)22S=41a2 ΣF⃗=ma⃗,\Sigma \vec F = m\vec a,ΣF=ma, $$ \begin{aligned} \Sigma F_2 &= m a_2\\ m a_2 &= mg\sin\theta\\ a_2&=g\sin\theta=\frac{g}{2}\\ a_1&=\f.. 11판/6. 힘과 운동 II 2024.03.08
6-59 할리데이 11판 솔루션 일반물리학 ΣF⃗=ma⃗,\Sigma \vec F = m\vec a,ΣF=ma, {ma=mgsinθ−μkN0=N−mgcosθ \begin{cases} m a &= mg\sin\theta-\mu_k N\\ 0 &= N-mg\cos\theta\\ \end{cases} {ma0=mgsinθ−μkN=N−mgcosθ a=g(sinθ−μkcosθ) a=g(\sin\theta-\mu_k\cos\theta) a=g(sinθ−μkcosθ) 11판/6. 힘과 운동 II 2024.03.08
6-58 할리데이 11판 솔루션 일반물리학 {v=80.0[km/h]=2009[m/s]R=350[m]m=55.0[kg] \begin{cases} v&=80.0\ut{km/h}=\frac{200}{9}\ut{m/s}\\ R&=350\ut{m}\\ m&=55.0\ut{kg}\\ \end{cases} ⎩⎨⎧vRm=80.0[km/h]=9200[m/s]=350[m]=55.0[kg] (a)\ab{a}(a) ΣFR=mv2R=44000567[N]≈77.60141093474427[N]≈77.6[N] \begin{aligned} \Sigma F_R &= \frac{mv^2}{R}\\ &=\frac{44000}{567}\ut{N}\\ &\approx 77.60141093474427\ut{N}\\ &\approx 77.6\ut{N}\\ \end{aligned} ΣFR=Rmv2=56744000[N]≈77.60141093474427[N]≈77.6[N] (b)\ab{b}(b) 90° 90\degree 90° 11판/6. 힘과 운동 II 2024.03.08
6-57 할리데이 11판 솔루션 일반물리학 {m1=30[kg]m2=15[kg]μs=0.60μk=0.40F=100[N]g=9.80665[m/s2] \begin{cases} m_1&=30\ut{kg}\\ m_2&=15\ut{kg}\\ \mu_s&=0.60\\ \mu_k&=0.40\\ F&=100\ut{N}\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧m1m2μsμkFg=30[kg]=15[kg]=0.60=0.40=100[N]=9.80665[m/s2] ΣF⃗=ma⃗,\Sigma \vec F = m\vec a,ΣF=ma, {ΣF1x=m1a1ΣF1y=0ΣF2x=m2a2ΣF2y=0 \begin{cases} \Sigma F_{1x} &= m_1 a_1\\ \Sigma F_{1y} &= 0\\ \Sigma F_{2x} &= m_2 a_2\\ \Sigma F_{2y} &= 0\\ \end{cases} ⎩⎨⎧ΣF1xΣF1yΣF2xΣF2y=m1a1=0=m2a2=0 {m1a1=−f0=N1−N2−m1gm2a2=−F+f0=N2−m2g \begin{cases} m_1 a_1 &= -f\\ 0 &= N_1-N_2-m_1g\\ m_2 a_2 &= -F+f\\ 0 &= N_2-m_2g\\ \end{cases} ⎩⎨⎧m1a10m2a20=−f=N1−N2−m1g=−F+f=N2−m2g $$ \begin.. 11판/6. 힘과 운동 II 2024.03.08