솔루션피아

$$ \begin{cases} F&=25\ut{N}\\ \vec d&=2.00\i-4.00\j+3.00\k\ut{m}\\ \end{cases} $$ $$ \begin{aligned} d&=\sqrt{2^2+(-4)^2+3^2}\\ &=\sqrt{29}\ut{m} \end{aligned} $$ $$\Delta K = W = Fd\cos\theta$$ $$ \begin{aligned} \theta &= \cos^{-1}\frac{\Delta K}{Fd}\\ &=\cos^{-1}\frac{\Delta K}{25\sqrt{29}} \end{aligned} $$ $$\ab{a}$$ $$\Delta K = +30.0\ut{J}$$ $$ \begin{aligned} \theta &= \cos^{-1}\frac{30}..

$$ \begin{cases} m&=10\ut{kg}\\ x_i&=0\\ x_f&=8.0\ut{m} \end{cases} $$ $$ \begin{cases} a(0\ut{m})&=0\\ a(8\ut{m})&=40.0\ut{m/s^2}\\ \end{cases} $$ $$ \begin{aligned} a(x\ut{m})&=\frac{40}{8}x\\ &=5x\ut{m/s^2} \end{aligned} $$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{aligned} F&=10\cdot5x\\ &=50x \end{aligned} $$ $$W_{i\rarr f}=\int_{i}^{f} \vec F \cdot \dd \vec S,$$ $$ \begin{aligned} W&=\int_{i}..

(풀이자주:풀이에 테일러급수가 필요합니다. 관련한 내용은 별도의 링크로 분리했습니다. 테일러 급수 자체에 대한 이해는 현재과정에서의 필수는 아니니 결론만 보고 건너뛰어도 무방합니다.) https://solutionpia.tistory.com/624 테일러 급수 $$\title{About Taylor Series}$$ $$ \put \begin{cases} f(x)&=\text{Any Function}\\ g(x)&=c_0+c_1x+c_2x^2+\cdots+c_nx^n+\cdots \end{cases} $$ $$ \put \begin{cases} f(0)&=g(0)\\ f'(0)&=g'(0)\\ f''(0)&=g''(0)\\ &\vdots\\ f^{(n)}(0)&=g^{(n)}(0)\\ &\vdots\\ ..

$$ \begin{cases} m&=2.00\ut{kg}\\ x&=9.0\ut{m}\\ a_s&=12\ut{m/s^2} \end{cases} $$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{aligned} F_s&=24\ut{N} \end{aligned} $$ $$W_{i\rarr f}=\int_{i}^{f} \vec F \cdot \dd \vec S,$$ $$\ab{a}$$ $$ \begin{aligned} W_{0\rarr4}&=\int_{0}^{4} F \cdot \dd x\\ &=\frac{1}{2}(3+4)24\\ &=84\ut{J} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} W_{0\rarr7}&=\int_{0}^{7} F \..

$$ \begin{cases} m&=60\ut{kg}\\ \theta&=20\degree\\ F&=230\ut{N}\\ S&= 4.20\ut{m} \end{cases} $$ $$\ab{a}$$ $$W=\vec F \cdot \vec S,$$ $$ \begin{aligned} W_F&=FS\cos\theta\\ &=966\cos20\degree\ut{J}\\ &\approx 907.7430716791876\ut{J}\\ &\approx 9.1\times10^2\ut{J}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} W_{mg}&=mgS\cos(90\degree+\theta)\\ &=-60g\sin20\degree\ut{J}\\ &\approx -201.244310..

$$ \begin{cases} \vec F&=3x\i+6\j\ut{N}\\ \vec r_i&=2\i+3\j\ut{m}\\ \vec r_f&=-4\i-3\j\ut{m}\\ \end{cases} $$ $$ \begin{aligned} \Delta r &= (-4\i-3\j)-(2\i+3\j)\\ &=-6\i-6\j\ut{m} \end{aligned} $$ $$W=\vec F \cdot \vec S,$$ $$ \begin{aligned} W&=(3x\i+6\j)\cdot(-6\i-6\j)\\ &=-18x-36\ut{J} \end{aligned} $$

$$ \begin{cases} \vec F&=5.00\i+c\j\ut{N}\\ \vec d&=2.00\i-3.00\j\ut{m}\\ \end{cases} $$ $$W=\vec F \cdot \vec S,$$ $$ \begin{aligned} W&=(5\i+c\j)\cdot(2\i-3\j)\\ &=10-3c\ut{J} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} 0&=10-3c\\ c&=\frac{10}{3}\\ &\approx 3.333333333333333\\ &\approx 3.33\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} 17&=10-3c\\ c&=-\frac{7}{3}\\ &\approx -2.3333333333..

$$ \begin{cases} m&=2050\ut{kg}\\ v&=110\ut{km/h}=\frac{275}{9}\ut{m/s}\\ \end{cases} $$ $$K=\frac{1}{2}mv^2,$$ $$ \begin{aligned} K&=\frac{1}{2}(2050)\(\frac{275}{9}\)^2\\ &=\frac{77515625}{81}\ut{J}\\ &\approx 956983.024691358\ut{J}\\ &\approx 9.6\times10^5\ut{J}\\ \end{aligned} $$

$$ \begin{cases} \vec d&=9.0\i+c\j\ut{m}\\ \vec F&=5.0\i-8.0\j\ut{N\\} \end{cases} $$ $$W=\vec F \cdot \vec S,$$ $$ \begin{aligned} W&=(5.0\i-8.0\j)\cdot(9.0\i+c\j)\\ &=45-8c \end{aligned} $$ $$\ab{a}$$ $$W=0,$$ $$45-8c=0$$ $$c=\frac{45}{8}=5.625\approx 5.6$$ $$\ab{b}$$ $$W\gt 0,$$ $$45-8c\gt0$$ $$c\lt\frac{45}{8}=5.625\approx 5.6$$ $$\ab{c}$$ $$W\lt 0,$$ $$45-8c\lt0$$ $$c\gt\frac{45}{8}=5.625\a..

$$ \begin{cases} F_{ax}&=9x-3x^2\\ x_i&=0\\ v_0&=0\\ v_f&=0\\ \end{cases} $$ $$\ab{a}$$ $$W_{i\rarr f}=\int_{i}^{f} \vec F \cdot \dd \vec S,$$ $$ \begin{aligned} W(x)&=\int_{0}^{x} (9x-3x^2) \cdot \dd x\\ &=\frac{9}{2}{x}^2-x^3 \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \dyx{W}&=\dx\int_{0}^{x} F \cdot \dd x\\ &=F\\ &=9x-3x^2=0,\\ \end{aligned} $$ $$\therefore x_{W \max}=3\ut{m}$$ $$\ab{c}$$..