솔루션피아

$$ \begin{cases} F_1&=4.00\ut{N}\\ F_2&=4.00\ut{N},\theta_2&=50.0\degree\\ F_3&=10.0\ut{N},\theta_3&=35.0\degree\\ S&=2.50\ut{m} \end{cases} $$ $$ \begin{cases} \vec F_1&=-4\i\ut{N}\\ \vec F_2&=-4\sin50\degree\i-4\cos50\degree\j\ut{N}\\ \vec F_3&=10\cos35\degree\i+10\sin35\degree\j\ut{N}\\ \end{cases} $$ $$ \begin{aligned} \theta_{F} &= \vec F_1+\vec F_2+\vec F_3\\ =&(-4-4\sin50\degree+10\cos35\..

$$ \begin{cases} F_a&=20.0\ut{N}\\ m&=2.00\ut{kg}\\ \theta&=25.0\degree\\ d&=0.700\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$W=\vec F \cdot \vec S,$$ $$ \begin{aligned} W_{F_a}&=F_a d \cos\theta\\ &\approx 12.688309018513099\ut{J}\\ &\approx 12.7\ut{J}\\ \\ W_{mg}&=mg d \cos(90\degree+\theta)\\ &\approx -5.802257127099202\ut{J}\\ &\approx -5.80\ut{J}\\ \\ W_{F_N}&=F_N d \cos90\deg..

$$ \begin{cases} k&=130\ut{N/m}\\ \Delta x&=5.00\ut{m} \end{cases} $$ $$ F=kx, $$ $$W_{i\rarr f}=\int_{i}^{f} \vec F \cdot \dd \vec S,$$ $$ \begin{aligned} W&=\int_{0}^{x} kx \cdot \dd x\\ &=\frac{1}{2}kx^2\\ &=1625\ut{J}\\ &\approx 1.63\times10^3\ut{J}\\ &\approx 1.63\ut{kJ}\\ \end{aligned} $$

$$ \begin{cases} K(0\ut{m})&=80.0\ut{J}\\ K(2\ut{m})&=0\\ v_0&=4.00\ut{m/s}\\ \end{cases} $$ $$K=\frac{1}{2}mv^2,$$ $$K(0\ut{m})=\frac{1}{2}m{v_0}^2=80.0\ut{J}$$ $$\therefore m=10.0\ut{kg}$$ $$ \begin{aligned} K_0&=W=F\cdot S\\ &=mg\sin\theta\cdot\frac{\Delta x}{\cos\theta}\\ &=mg\Delta x \tan\theta\\ \end{aligned} $$ $$ \begin{aligned} \tan\theta&=\frac{4}{g},\\ \cos\theta&=\frac{1}{\sqrt{\frac..

풀이자 주 : 일을하는 평균비율이라는 표현이 애매합니다. 정황상 일률을 구하라는 것으로 두고 풀겠습니다. $$ \begin{cases} m&= 4.1\times 10^3\ut{kg}\\ t&=20\ut{s}\\ h&=210\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} P&=F\cdot v\\ &=mg \cdot \frac{h}{t}\\ &=43050g\\ &=422176.2825\ut{W}\\ &\approx 4.2\times10^5\ut{W}\\ &\approx 0.42\ut{MW}\\ \end{aligned} $$

$$ \begin{cases} m_A&=1400\ut{kg}\\ t&=3.0\ut{min}\\ h&=60\ut{m}\\ m_B&=900\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} P&=F\cdot v\\ &=(m_A-m_B)g\cdot \frac{h}{t}\\ &=\frac{500}{3}g\\ &\approx 1634.4416666666666\ut{W}\\ &\approx 1.6\ut{kW}\\ \end{aligned} $$

풀이자주: 변하는힘F 라고만 주어져있어 F를 이대로는 특정할 수가 없습니다. 물체의 속도가 일정하다고 임의로 두겠습니다. 문제상황을 어떻게 두냐에 따라 답은 전혀 달라질 수 있습니다. $$ \begin{cases} L&=12.0\ut{m}\\ m&=170\ut{kg}\\ d &=3.50\ut{m}\\ v_1&=0\\ g&=9.80665\ut{m/s^2}\\ \Delta \vec v&=0 \end{cases} $$ $$\Sigma \vec F = ma=0,$$ $$ \begin{cases} \Sigma F_x &= F-F_T\sin\theta=0\\ \Sigma F_y &= F_T\cos\theta-mg=0 \end{cases} $$ $$ \begin{aligned} F&=mg\tan\theta\\ \en..

$$ \begin{cases} m&=3.0\ut{kg}\\ \vec F(x\ut{m})&=-6x\i\ut{N}\\ v(3.0\ut{m})&=6.0\ut{m/s}\\ \end{cases} $$ $$W_{i\rarr f}=\int_{i}^{f} \vec F \cdot \dd \vec S,$$ $$ \begin{aligned} W_{3\rarr x}&=\int_{3}^{x} F \dd x\\ &=\int_{3}^{x} (-6x) \dd x\\ &=27-3x^2 \end{aligned} $$ $$K_x=K_3+W_{3\rarr x},$$ $$\frac{1}{2}m{v_x}^2=\frac{1}{2}m{v_3}^2+(27-3x^2)$$ $$ \begin{aligned} {v_x}^2&=\frac{54+m{v_3}^..

$$ \begin{cases} v_0&=0\\ m&=1.5\ut{kg}\\ \vec F&=(2.5-x^2)\i\ut{N}\\ x_0&=0 \end{cases} $$ $$ \begin{aligned} W(x)&=\int_{0}^{x} F \cdot \dd x\\ &=\int_{0}^{x}(2.5-x^2)\dd x\\ &=\frac{5x}{2}-\frac{x^3}{3}\\ \end{aligned} $$ $$\ab{a}$$ $$x=2.8\ut{m}$$ $$ \begin{aligned} W(2.8)&=-\frac{119}{375}\ut{J}\\ \abs{K}&=\abs{K_0+W}\\ &=\frac{119}{375}\ut{J}\\ &\approx 0.31733333333333336\ut{J}\\ &\approx..

$$ \begin{cases} v_0&=0\\ m&=11.0\ut{kg}\\ F&=7.00\ut{N} \end{cases} $$ $$\Sigma \vec F = m\vec a,$$ $$a=\frac{F}{m}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} d&=(0)t+\frac{1}{2}\(\frac{F}{m}\)t^2\\ &=\frac{Ft^2}{2m} \end{aligned} $$ $$ \begin{aligned} W(t)&=\vec F \cdot \vec d,\\ &=\frac{F^2t^2}{2m}\\ &=\frac{49}{22}t^2\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} W(1)&=\frac{49}{22}..