8-82 할리데이 11판 솔루션 일반물리학 {F=750[N]k=2.5×105[N/m] \begin{cases} F&=750\ut{N}\\ k&=2.5\times10^5\ut{N/m}\\ \end{cases} {Fk=750[N]=2.5×105[N/m] (a)\ab{a}(a) F=kx,F=kx,F=kx, x=Fk=3.0×10−3[m]=3.0[mm] \begin{aligned} x&=\frac{F}{k}\\ &=3.0\times10^{-3}\ut{m}\\ &=3.0\ut{mm}\\ \end{aligned} x=kF=3.0×10−3[m]=3.0[mm] (b)\ab{b}(b) W=12kx2=12k(Fk)2=F22k=98[J]=1.125[J]≈1.1[J] \begin{aligned} W&=\frac{1}{2}kx^2\\ &=\frac{1}{2}k\(\frac{F}{k}\)^2\\ &=\frac{F^2}{2k}\\ &=\frac{9}{8}\ut{J}\\ &=1.125\ut{J}\\ &\approx 1.1\ut{J} \end{aligned} W=21kx2=21k(kF)2=2kF2=89[J]=1.125[J]≈1.1[J] (c)\ab{c}(c) F=kx,F=kx,F=kx, $$\ab{d}$.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.12
8-81 할리데이 11판 솔루션 일반물리학 {vA1=2.00[m/s]vB1=2.60[m/s]vA2=6.00[m/s] \begin{cases} v_{A1}&=2.00\ut{m/s}\\ v_{B1}&=2.60\ut{m/s}\\ v_{A2}&=6.00\ut{m/s}\\ \end{cases} ⎩⎨⎧vA1vB1vA2=2.00[m/s]=2.60[m/s]=6.00[m/s] ΣΔE=0,\Sigma \Delta E=0,ΣΔE=0, ΔKE+ΔGE=0\Delta \KE+\Delta \GE=0ΔKE+ΔGE=0 Δ(12mv2)=−Δ(mgy)Δ(v2)=2g(−Δy)vf2−vi2=2g(−Δy)=Constant \begin{aligned} \Delta \(\frac{1}{2}m{v}^2\)&=-\Delta (mgy)\\ \Delta \({v}^2\)&=2g(-\Delta y)\\ {v_f}^2-{v_i}^2&=2g(-\Delta y)\\ &=\Cons\\ \end{aligned} Δ(21mv2)Δ(v2)vf2−vi2=−Δ(mgy)=2g(−Δy)=2g(−Δy)=Constant vB22−vA22=vB12−vA12{v_{B2}}^2-{v_{A2}}^2={v_{B1}}^2-{v_{A1}}^2vB22−vA22=vB12−vA12 $$ \begin{aligned} v_{B2}&=\sqrt{{v_{A2.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.12
8-80 할리데이 11판 솔루션 일반물리학 {L=230[cm]g=9.80665[m/s2] \begin{cases} L&=230\ut{cm}\\ g&=9.80665\ut{m/s^2} \end{cases} {Lg=230[cm]=9.80665[m/s2] put {0:Start1:Lowest Point2:Over P Highest Point \put \begin{cases} 0:\text{Start}\\ 1:\text{Lowest Point}\\ 2:\text{Over P Highest Point} \end{cases} put ⎩⎨⎧0:Start1:Lowest Point2:Over P Highest Point {y0=Ly1=0y2=2r \begin{cases} y_0&=L\\ y_1&=0\\ y_2&=2r\\ \end{cases} ⎩⎨⎧y0y1y2=L=0=2r ΣΔE=0,\Sigma \Delta E=0,ΣΔE=0, ΔKE+ΔGE=0\Delta \KE+\Delta \GE=0ΔKE+ΔGE=0 $$ \begin{aligned} \Delta \KE&=-\Delta \GE\\ \Delta \(\frac{1}{2}m{v}^2\)&=-\Delta (mgy)\\ \Delta \({v}^2\)&=2.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.12
8-79 할리데이 11판 솔루션 일반물리학 {m=3.6[kg]y=3.8[m]g=9.80665[m/s2] \begin{cases} m&=3.6\ut{kg}\\ y&=3.8\ut{m}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} ⎩⎨⎧myg=3.6[kg]=3.8[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)\ab{a}(a) GE=mgy=342g25=134.154972[J]≈1.3×102[J] \begin{aligned} \GE&=mgy\\ &=\frac{342g}{25}\\ &=134.154972\ut{J}\\ &\approx 1.3\times10^2\ut{J}\\ \end{aligned} GE=mgy=25342g=134.154972[J]≈1.3×102[J] (b)\ab{b}(b) ΣΔE=0,\Sigma \Delta E=0,ΣΔE=0, $$ \begin{aligned} \KE&=\GE\\ &=134.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.12
8-78 할리데이 11판 솔루션 일반물리학 {vi=7.0[m/s]h=2.0[m]μ=0.50 \begin{cases} v_i&=7.0\ut{m/s}\\ h&=2.0\ut{m}\\ \mu&=0.50\\ \end{cases} ⎩⎨⎧vihμ=7.0[m/s]=2.0[m]=0.50 put {KE:Kinetic EnergyGE:Gravitational Potential EnergyTE:Thermal Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \TE : \text{Thermal Energy}\\ \end{cases} put ⎩⎨⎧KE:Kinetic EnergyGE:Gravitational Potential EnergyTE:Thermal Energy ΣΔE=0,\Sigma \Delta E=0,ΣΔE=0, ΔKE+ΔGE+ΔTE=0\Delta \KE+\Delta \GE+\Delta \TE=0ΔKE+ΔGE+ΔTE=0 $$ \begin{aligned} \Delta \TE&=-\Delta \KE-\Delta \GE\\ fd&=-\Delta\(\frac{1}{2}mv^2\)-\Delta (mgy)\\ \mu m g d&=-\.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.11
8-77 할리데이 11판 솔루션 일반물리학 {m=5.0[kg]k=425[N/m]x=6.0[cm]=0.06[m]μ=0.25g=9.80665[m/s2] \begin{cases} m&=5.0\ut{kg}\\ k&=425\ut{N/m}\\ x&=6.0\ut{cm}=0.06\ut{m}\\ \mu&=0.25\\ g&=9.80665\ut{m/s^2} \end{cases} ⎩⎨⎧mkxμg=5.0[kg]=425[N/m]=6.0[cm]=0.06[m]=0.25=9.80665[m/s2] put {KE:Kinetic EnergyLE:Elastic Potential EnergyTE:Thermal Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \LE : \text{Elastic Potential Energy}\\ \TE : \text{Thermal Energy}\\ \end{cases} put ⎩⎨⎧KE:Kinetic EnergyLE:Elastic Potential EnergyTE:Thermal Energy (a)\ab{a}(a) $$ \begin{aligned} W_s&=-\Delta \LE\\ &=\LE_i-\LE_f\\ &=-\LE_f\\ &=-\frac{1}{2}kx^2\\ &=-\frac{153}{200}\ut{J}\\ &=-0.765\ut{.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.11
8-76 할리데이 11판 솔루션 일반물리학 put {KE:Kinetic EnergyGE:Gravitational Potential EnergyTE:Thermal Energy \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \TE : \text{Thermal Energy}\\ \end{cases} put ⎩⎨⎧KE:Kinetic EnergyGE:Gravitational Potential EnergyTE:Thermal Energy {S=6.00[m]F=2.00[N]f=10.0[N]ΔKE=+40.0[J] \begin{cases} S&=6.00\ut{m}\\ F&=2.00\ut{N}\\ f&=10.0\ut{N}\\ \Delta \KE&=+40.0\ut{J} \end{cases} ⎩⎨⎧SFfΔKE=6.00[m]=2.00[N]=10.0[N]=+40.0[J] ΣΔE=WF,\Sigma \Delta E=W_F,ΣΔE=WF, ΔKE+ΔGE+ΔTE=WF\Delta \KE+\Delta \GE+\Delta \TE=W_FΔKE+ΔGE+ΔTE=WF $$ \begin{aligned} W_G&=-\Delta \GE\\ &=\Delta \KE+\Delta \TE- W_F\\ &=\Delta \KE+fS-.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.11
8-75 할리데이 11판 솔루션 일반물리학 {R=18.0[m]θi=0yi=Rvi=0 \begin{cases} R&=18.0\ut{m}\\ \theta_i&=0\\ y_i&=R\\ v_i&=0\\ \end{cases} ⎩⎨⎧Rθiyivi=18.0[m]=0=R=0 Δy=−(R−Rcosθf) \begin{aligned} \Delta y &= -(R-R\cos\theta_f)\\ \end{aligned} Δy=−(R−Rcosθf) ΣΔE=0,\Sigma \Delta E=0,ΣΔE=0, ΔKE+ΔGE=0\Delta \KE+\Delta \GE=0ΔKE+ΔGE=0 ΔKE=−ΔGEΔ(12mv2)=−Δ(mgy)Δ(v2)=2g(−Δy) \begin{aligned} \Delta \KE&=-\Delta \GE\\ \Delta \(\frac{1}{2}m{v}^2\)&=-\Delta (mgy)\\ \Delta \({v}^2\)&=2g(-\Delta y)\\ \end{aligned} ΔKEΔ(21mv2)Δ(v2)=−ΔGE=−Δ(mgy)=2g(−Δy) $$ \begin{aligned} {v_f}^2&={v_i}^2+2gh\\ &=2g(-\Delta y.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.11
8-74 할리데이 11판 솔루션 일반물리학 put {K:Kinetic EnergyU:Potential Energy \put \begin{cases} K : \text{Kinetic Energy}\\ U : \text{Potential Energy}\\ \end{cases} put {K:Kinetic EnergyU:Potential Energy {UA=15.0[J]UB=35.0[J]UC=45.0[J] \begin{cases} U_A&=15.0\ut{J}\\ U_B&=35.0\ut{J}\\ U_C&=45.0\ut{J}\\ \end{cases} ⎩⎨⎧UAUBUC=15.0[J]=35.0[J]=45.0[J] {U0→2=35U2→4=35−10(x−2)=55−10xU4→5=15U5→6=15+30(x−5)=30x−135U6→∞=45 \begin{cases} U_{0\rarr2}&=35\\ U_{2\rarr4}&=35-10(x-2)=55-10x\\ U_{4\rarr5}&=15\\ U_{5\rarr6}&=15+30(x-5)=30 x-135 \\ U_{6\rarr\infin}&=45\\ \end{cases} ⎩⎨⎧U0→2U2→4U4→5U5→6U6→∞=35=35−10(x−2)=55−10x=15=15+30(x−5)=30x−135=45 $$ \begin{aligned} \Delta U&=-\int F\dd x,\\ F&=-\dyx{U} .. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.11
8-73 할리데이 11판 솔루션 일반물리학 {m=12.0[kg]x1=10.0[cm]=0.1[m]Δx=30.0[cm]=0.3[m]g=9.80665[m/s2] \begin{cases} m&=12.0\ut{kg}\\ x_1&=10.0\ut{cm}=0.1\ut{m}\\ \Delta x&=30.0\ut{cm}=0.3\ut{m}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} ⎩⎨⎧mx1Δxg=12.0[kg]=10.0[cm]=0.1[m]=30.0[cm]=0.3[m]=9.80665[m/s2] x2=x1+Δx=0.4[m]x_2=x_1+\Delta x=0.4\ut{m}x2=x1+Δx=0.4[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 (a)\ab{a}(a) F=kx,F=kx,F=kx, $$ \begin{aligned} k&=\frac{F}{x}\\ &=\frac{mg}{x_1}\\ &=1.. 11판/8. 퍼텐셜에너지와 에너지 보존 2024.04.11