8-72 할리데이 11판 솔루션 일반물리학 {m=78[kg]v=64[m/s]g=9.80665[m/s2] \begin{cases} m&= 78\ut{kg}\\ v&=64\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧mvg=78[kg]=64[m/s]=9.80665[m/s2] put {KE:Kinetic EnergyGE:Gravitational Potential EnergyME:Mechanical Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \ME : \text{Mechanical Energy}\\ \end{cases} put ⎩⎨⎧KE:Kinetic EnergyGE:Gravitational Potential EnergyME:Mechanical Energy (a)\ab{a}(a) $$ \begin{aligned} \dyt{\GE}&=\dyt{mgy}\\ &=mg\dyt{y}\\ &=mgv\\ &=4992g\\ &=48954.7968\ut{W}\\ &\approx 4.9\times10^4\ut{W}\\ &\approx 49\ut{kW}\\ \end{alig.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.11
8-71 할리데이 11판 솔루션 일반물리학 {v=0.200[m/s]f=98.0[N] \begin{cases} v&=0.200\ut{m/s}\\ f&=98.0\ut{N}\\ \end{cases} {vf=0.200[m/s]=98.0[N] P=fv=985[W]=19.6[W] \begin{aligned} P&=fv\\ &=\frac{98}{5}\ut{W}\\ &=19.6\ut{W} \end{aligned} P=fv=598[W]=19.6[W] 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.11
8-70 할리데이 11판 솔루션 일반물리학 {m=2.12[kg]v0=2.50[m/s]h=3.80[m]g=9.80665[m/s2] \begin{cases} m&=2.12\ut{kg}\\ v_0&=2.50\ut{m/s}\\ h&=3.80\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧mv0hg=2.12[kg]=2.50[m/s]=3.80[m]=9.80665[m/s2] put {KE:Kinetic EnergyGE:Gravitational Potential Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} put {KE:Kinetic EnergyGE:Gravitational Potential Energy (a,b)\ab{a,b}(a,b) ΣΔE=0,\Sigma \Delta E=0,ΣΔE=0, ΔKE+ΔGE=0\Delta \KE+\Delta \GE=0ΔKE+ΔGE=0 ΔKE=−ΔGEKEf−KEi=−Δ(mgy) \begin{aligned} \Delta \KE&=-\Delta \GE\\ \KE_f-\KE_i&=-\Delta (mgy)\\ \end{aligned} ΔKEKEf−KEi=−ΔGE=−Δ(mgy) $$ \begin{aligned} \KE_f.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.11
8-69 할리데이 11판 솔루션 일반물리학 put {KE:Kinetic EnergyGE:Gravitational Potential EnergyTE:Thermal EnergyME:Mechanical Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \TE : \text{Thermal Energy}\\ \ME : \text{Mechanical Energy}\\ \end{cases} put ⎩⎨⎧KE:Kinetic EnergyGE:Gravitational Potential EnergyTE:Thermal EnergyME:Mechanical Energy {m=8.10[kg]ΔME=−72.0[kJ]=−72000[J]g=9.80665[m/s2] \begin{cases} m&=8.10\ut{kg}\\ \Delta \ME&=-72.0\ut{kJ}=-72000\ut{J}\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧mΔMEg=8.10[kg]=−72.0[kJ]=−72000[J]=9.80665[m/s2] ΔTE=−ΔME,\Delta \TE=-\Delta \ME,ΔTE=−ΔME, ΣΔE=ΔKE1+ΔGE1+ΔTE1=0,\Sigma \Delta E=\Delta \KE_1+\Delta \GE_1+\Delta \TE_1=0,ΣΔE=ΔKE1+ΔGE1+ΔTE1=0, $$ \begin{aligned} \De.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.10
8-68 할리데이 11판 솔루션 일반물리학 (풀이자 주 : 문제의 데이터가 부족하여 풀 수 없습니다. 아마 모터에 공급되는 에너지가 누락된것 같습니다. 풀이자 임의로 두고 풀겠습니다.) {N=212[N]R=20.0[cm]=15[m]f=2.50[rev/s]μ=0.390 \begin{cases} N&=212\ut{N}\\ R&=20.0\ut{cm}=\frac{1}{5}\ut{m}\\ f&=2.50\ut{rev/s}\\ \mu&=0.390\\ \end{cases} ⎩⎨⎧NRfμ=212[N]=20.0[cm]=51[m]=2.50[rev/s]=0.390 put {KE:Kinetic EnergyTE:Thermal EnergyEE:Electric Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \TE : \text{Thermal Energy}\\ \EE : \text{Electric Energy} \end{cases} put ⎩⎨⎧KE:Kinetic EnergyTE:Thermal EnergyEE:Electric Energy $$ \begin{cases} \frac{\KE}{t}&=P_K\\ \frac{\TE}{t}&=P_T\\ \frac{\EE}{t}&=P_E.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.10
8-67 할리데이 11판 솔루션 일반물리학 {m=65.0[kg]v0=0v2=0g= \begin{cases} m&=65.0\ut{kg}\\ v_0&=0\\ v_2&=0\\ g&= \end{cases} ⎩⎨⎧mv0v2g=65.0[kg]=0=0= {y0=0y1=−10.0[m]Δy1→2=−1.20[m] \begin{cases} y_0&=0\\ y_1&=-10.0\ut{m}\\ \Delta y_{1\rarr2}&=-1.20\ut{m}\\ \end{cases} ⎩⎨⎧y0y1Δy1→2=0=−10.0[m]=−1.20[m] put {KE:Kinetic EnergyGE:Gravitational Potential EnergyLE:Elastic Potential Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \LE : \text{Elastic Potential Energy}\\ \end{cases} put ⎩⎨⎧KE:Kinetic EnergyGE:Gravitational Potential EnergyLE:Elastic Potential Energy ΣΔE=0,\Sigma \Delta E=0,ΣΔE=0, ΔKE+ΔGE+ΔLE=0\Delta \KE+\Delta \GE+\Delta \LE=0ΔKE+ΔGE+ΔLE=0 $$\Delta \GE+\Delt.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.10
8-66 할리데이 11판 솔루션 일반물리학 {θ=40°m=52[kg]v=ConstantS=3.0[m]μ=0.35g=9.80665[m/s2] \begin{cases} \theta&=40\degree\\ m&=52\ut{kg}\\ v&=\Cons\\ S&=3.0\ut{m}\\ \mu&=0.35\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧θmvSμg=40°=52[kg]=Constant=3.0[m]=0.35=9.80665[m/s2] put {TE:Thermal Energy \put \begin{cases} \TE : \text{Thermal Energy}\\ \end{cases} put {TE:Thermal Energy {ΣFx=0ΣFy=0 \begin{cases} \Sigma F_x&=0\\ \Sigma F_y&=0\\ \end{cases} {ΣFxΣFy=0=0 {F−f−mgsinθ=0N−mgcosθ=0 \begin{cases} F-f-mg\sin\theta&=0\\ N-mg\cos\theta&=0\\ \end{cases} {F−f−mgsinθN−mgcosθ=0=0 (a)\ab{a}(a) $$ \begin{aligned} W&=FS\\ &=(mg\sin\theta+\mu mg\cos\th.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.10
8-65 할리데이 11판 솔루션 일반물리학 {m=82[kg]vi=16[m/s]Δy=4.2[m]g=9.80665[m/s2] \begin{cases} m&=82\ut{kg}\\ v_i&=16\ut{m/s}\\ \Delta y &=4.2\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧mviΔyg=82[kg]=16[m/s]=4.2[m]=9.80665[m/s2] put {KE:Kinetic EnergyGE:Gravitational Potential Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} put {KE:Kinetic EnergyGE:Gravitational Potential Energy ΣΔE=0,\Sigma \Delta E=0,ΣΔE=0, ΔKE+ΔGE=0\Delta \KE+\Delta \GE=0ΔKE+ΔGE=0 ΔKE=−ΔGEKEf−KEi=−ΔGE \begin{aligned} \Delta \KE&=-\Delta \GE\\ \KE_f-\KE_i&=-\Delta \GE\\ \end{aligned} ΔKEKEf−KEi=−ΔGE=−ΔGE $$ \begin{aligned} \KE_f&=\KE_i-\Del.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.10
8-64 할리데이 11판 솔루션 일반물리학 {k=4200[N/m]Ui=1.10[J] \begin{cases} k&=4200\ut{N/m}\\ U_i&=1.10\ut{J}\\ \end{cases} {kUi=4200[N/m]=1.10[J] U=12kx2,U=\frac{1}{2}kx^2,U=21kx2, xi=11011210[m] x_i=\frac{1}{10}\sqrt{\frac{11}{210}}\ut{m} xi=10121011[m] xa=xi+Δxx_a=x_i+\Delta xxa=xi+Δx ΔU=Uf−Ui=12kxa2−12kxi2=12kΔx(Δx+2xi)=2Δx(2130+1050Δx) \begin{aligned} \Delta U &= U_f-U_i\\ &=\frac{1}{2}k{x_a}^2-\frac{1}{2}k{x_i}^2\\ &=\frac{1}{2}k\Delta x(\Delta x+2x_i)\\ &=2\Delta x(\sqrt{2130}+1050\Delta x) \end{aligned} ΔU=Uf−Ui=21kxa2−21kxi2=21kΔx(Δx+2xi)=2Δx(2130+1050Δx) (a)\ab{a}(a) Δx=+2[cm]\Delta x=+2\ut{cm}Δx=+2[cm] $$ \begin{aligned} \De.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.10
8-63 할리데이 11판 솔루션 일반물리학 {Δy=−14[m]vi=−3.0[m/s]vf=−14[m/s]g=9.80665[m/s2] \begin{cases} \Delta y &= -14\ut{m}\\ v_i&=-3.0\ut{m/s}\\ v_f&=-14\ut{m/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} ⎩⎨⎧Δyvivfg=−14[m]=−3.0[m/s]=−14[m/s]=9.80665[m/s2] put {KE:Kinetic EnergyGE:Gravitational Potential Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} put {KE:Kinetic EnergyGE:Gravitational Potential Energy $$ \begin{aligned} \Ans&=-\frac{\Delta \KE}{\Delta \GE}\\ &=-\frac{\Delta \(\frac{1}{2}m{v}^2\)}{\Delta (mgy)}\\ &=\frac{\Delta \({v}^2\)}{2g(-\Delta y)}\\ &=\frac{{v_.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.10