8-18 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec F_{0\rarr2.00\ut{m}}&=+3.00\i\ut{N}\\ \vec F_{2.00\rarr3.00\ut{m}}&=+5.00\i\ut{N}\\ \vec F_{3.00\rarr8.00\ut{m}}&=0\\ \vec F_{8.00\rarr11.0\ut{m}}&=-4.00\i\ut{N}\\ \vec F_{11.0\rarr12.0\ut{m}}&=-1.00\i\ut{N}\\ \vec F_{12.0\rarr15.0\ut{m}}&=0\\ \end{cases} $$

$$ \begin{cases} \int_0^2F\dd x&=3x\\ \int_2^3F\dd x&=5(x-2)\\ \int_3^8F\dd x&=0\\ \int_8^{11}F\dd x&=-4(x-8)\\ \int_{11}^{12}F\dd x&=-(x-11)\\ \int_{12}^{15}F\dd x&=0\\ \end{cases} $$ $$ \put \begin{cases} K : \text{Kinetic Energy}\\ U:\text{Potential Energy}\\ \end{cases} $$ $$ U(x)=U(i)-\int_i^x F\dd x, $$ $$ \begin{cases} U_{0\rarr2}&=U(0)-3x\\ U_{2\rarr3}&=U(2)-5(x-2)\\ U_{3\rarr8}&=U(3)-0\\ U_{8\rarr11}&=U(8)+4(x-8)\\ U_{11\rarr12}&=U(11)+(x-11)\\ U_{12\rarr15}&=U(12)-0\\ \end{cases} $$ $$ \begin{cases} U_{0\rarr2}&=11-3x\\ U_{2\rarr3}&=5-5(x-2)\\ U_{3\rarr8}&=0-0\\ U_{8\rarr11}&=0+4(x-8)\\ U_{11\rarr12}&=12+(x-11)\\ U_{12\rarr15}&=13-0\\ \end{cases} $$ $$ \begin{cases} U_{0\rarr2}&=11-3x\\ U_{2\rarr3}&=15-5x\\ U_{3\rarr8}&=0\\ U_{8\rarr11}&=4x-32\\ U_{11\rarr12}&=x+1\\ U_{12\rarr15}&=13\\ \end{cases} $$

$$\ab{a}$$ $$ \begin{cases} m&=2.00\ut{kg}\\ v_5&=-3.45\ut{m/s} \end{cases} $$ $$K=\frac{1}{2}mv^2,$$ $$ \begin{aligned} K_5&=\frac{1}{2} \cdot 2\cdot (-3.45)^2\\ &=\frac{4761}{400}\ut{J}\\ &=11.9025\ut{J}\\ \end{aligned} $$ $$K_5\gt U_0,$$ $$\text{Available for Arrival }$$ $$ \begin{aligned} E_5&=E_0\\ K_5&=U_0+K_0\\ K_0&=K_5-U_0\\ \frac{1}{2}m{v_0}^2&=K_5-U_0\\ \end{aligned} $$ $$ \begin{aligned} v_0&=\sqrt\frac{2(K_5-U_0)}{m}\\ &=\sqrt\frac{2\(\frac{4761}{400}-11\)}{2}\\ &=\frac{19}{20}\ut{m/s}\\ &=0.95\ut{m/s}\\ &=0.950\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{cases} m&=2.00\ut{kg}\\ v_5&=+3.45\ut{m/s} \end{cases} $$ $$ \begin{aligned} K_5&=\frac{1}{2} \cdot 2\cdot (+3.45)^2\\ &=\frac{4761}{400}\ut{J}\\ &=11.9025\ut{J}\\ \end{aligned} $$ $$K_5\lt U_{13},$$ $$\text{Impossible to Arrive}$$ $$\text{Stop : } 8\lt x \lt 11$$ $$ \begin{aligned} E_x&=E_5\\ U_x&=K_5\\ 4x-32&=\frac{4761}{400}\\ (\because U_{8\rarr11}&=4x-32) \end{aligned} $$ $$ \begin{aligned} x&=\frac{17561}{1600}\ut{m}\\ &=10.975625\ut{m}\\ &\approx 11.0\ut{m}\\ \end{aligned} $$