솔루션피아

$$\begin{cases} 0:\text{Start}\\ 1:h_{\max}\\ 2:\Delta h = -3.0\ut{m} \end{cases}$$ $$\begin{cases} m&=88\ut{g}\\ \theta&=30\degree\\ v_0&=8.0\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\begin{aligned} \vec v_0 &= v_0\cos\theta\i+v_0\sin\theta\j,\\ \end{aligned}$$ $$\put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases}$$ $$\ab{a}$$..

$$\begin{cases} A:\text{M Box}\\ B:\text{2M Box}\\ C:\text{Spring}\\ \end{cases}$$ $$\begin{cases} 0:\text{Start}\\ 1:x_1=0.120\ut{m}\\ 2:x_2=\max(x)\\ \end{cases}$$ $$\begin{cases} m_A&=M=2.80\ut{kg}\\ m_B&=2M\\ k&=230\ut{N/m}\\ \end{cases}$$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \LE : \text{Elastic Potential Energy}\\ \end{cases..

$$\begin{cases} m&=265\ut{kg}\\ d&=6.2\ut{m}\\ \theta&=39\degree\\ \mu &= 0.28\\ g&= 9.80665\ut{m/s^2} \end{cases}$$ $$h = d\sin\theta$$ $$\Sigma F_y=0,$$ $$N_{0\rarr1}-mg\cos\theta=0$$ $$N_{0\rarr1}=mg\cos\theta$$ $$\therefore f_{0\rarr1}=\mu mg\cos\theta$$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \TE : \text{Thermal Energy}\\ \end{case..

$$\begin{cases} m&=1750\ut{kg}\\ \theta&=5.0\degree\\ S&=65\ut{m}\\ v_i&=32\ut{km/h}=\frac{80}{9}\ut{m/s}\\ v_f&=40\ut{km/h}=\frac{100}{9}\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \ME : \text{Mechanical Energy} \end{cases}$$ $$ \begin{aligned} \Delta y &= -S\sin\theta\\ &=-65\sin5\degree\\ \e..

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

$$\begin{cases} m_A&=2.00\ut{kg}\\ m_B&=4.00\ut{kg}\\ \theta&=30.0\degree\\ v_0&=0\\ \Delta y_B&=-32.0\ut{cm}=-0.32\ut{m} \end{cases}$$ $$\begin{aligned} \sin\theta &= \frac{\Delta y_A}{\Delta d_A}\\ &= \frac{\Delta y_A}{-\Delta y_B}\\ \therefore \Delta y_A&=-\Delta y_B\sin\theta \end{aligned}$$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\..

$$\begin{cases} m&=0.30\ut{kg}\\ v&=4.00\ut{m/s}\\ h_{\max}&=0.80\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases}$$ $$ \begin{aligned} \Delta E &=E_2-E_1\\ &=\GE_2-\KE_1\\ &=mgh_2-\frac{1}{2}m{v_1}^2\\ &=\frac{6 g}{25}-\frac{12}{5}\\ &=-0.046404\ut{J}\\ &=-4.6404\times10^{-2}\ut{J}\\ &\app..

$$\begin{cases} m&=425\ut{g}\\ L&=2.4\ut{m}\\ \theta_i&=30.0\degree\\ v_i&=0\\ g&=9.80665\ut{m/s^2}\\ \end{cases}$$ $$\put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases}$$ $$\Sigma E_i=\Sigma E_f,$$ $$\begin{aligned} \KE_i+\GE_i&=\KE_f+\GE_f\\ mgy_i&=\frac{1}{2}m{v_f}^2+mgy_f\\ 2gy_i&={v_f}^2+2gy_f\\ \end{aligned}$$ $$y=L-L\cos\theta=L..

(풀이자 주 : 일반적인 디젤기관차의 출력은 3000마력, 2MW 수준입니다. 그리고 2mW는 엄청나게 작은 출력으로, 선풍기를 꺼뒀을때의 대기전력 조차 200mW 수준입니다. 현실적으로 오타로 간주하고 기관차의 출력을 2MW로 두고 풀겠습니다. 만일 기관차의 출력을 2mW로 두고 푼다면 답은 달라집니다.) $$\begin{cases} P&=2.0\ut{MW}=2.0\times10^{3}\ut{W}\\ t&=6.0\ut{min}=360\ut{s}\\ v_i&=10\ut{m/s}\\ v_f&=25\ut{m/s}\\ \end{cases}$$ $$\ab{a}$$ $$ \begin{aligned} P&=\frac{W}{t}\\ &=\frac{\Delta \KE}{t}\\ &=\frac{\KE_f-\KE_i}..

$$\begin{cases} m&=1700\ut{kg}\\ t&=30\ut{s}\\ v&=82\ut{km/h}=\frac{205}{9}\ut{m/s} \end{cases}$$ $$\ab{a}$$ $$\KE=\frac{1}{2}mv^2,$$ $$\begin{aligned} \KE&=\frac{1}{2}\cdot 1700 \cdot \(\frac{205}{9}\)^2\\ &=\frac{35721250}{81}\ut{J}\\ &\approx 441003.0864197531\ut{J}\\ &\approx 4.4 \times 10^5 \ut{J}\\ &\approx 0.44 \ut{MJ}\\ \end{aligned}$$ $$\ab{b}$$ $$ \begin{aligned} P&=\frac{W}{t}\\ &..