솔루션피아

$$ \begin{cases} v_0&=0\\ \vec F&=315\i\ut{N}\\ \theta&=20\degree\\ S&=9.2\ut{m} \end{cases} $$ $$ \begin{aligned} K&=K_0+W\\ &=W~~(\because v_0=0)\\ &=\vec F \cdot \vec S\\ &= FS\cos\theta\\ &=2898\cos20\degree\ut{J}\\ &\approx 2723.2292150375624\ut{J}\\ &\approx 2.7\times10^3\ut{J}\\ &\approx 2.7\ut{kJ}\\ \end{aligned} $$

$$ \begin{cases} m&=400\ut{g}\\ k&=2.50\ut{N/cm}=250\ut{N/m}\\ x &= 18.0\ut{cm}=0.18\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} W_{mg}&=(-mg)\cdot(- x)\\ &=\frac{9}{125}g\\ &=0.7060788\ut{J}\\ &\approx 706\ut{mJ} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} W_s&=-\frac{1}{2}k x^2\\ &=-\frac{81}{20}\ut{J}\\ &=-4.05\ut{J} \end{aligned} $$ $$\ab{c}$$ $$ \begin{align..

$$ \begin{cases} F&=12.0\ut{N}\\ \theta &= 150\degree\\ m&=2.0\ut{kg}\\ \vec r&=5.00\i-6.00\j\ut{m}\\ \end{cases} $$ $$W=\vec F \cdot \vec S,$$ $$ \begin{aligned} W&=(F\cos\theta\i+F\sin\theta\j) \cdot (5\i-6\j)\\ &=12\cos150\degree\cdot5+12\sin150\degree\cdot(-6)\\ &=-36-30\sqrt3\ut{J}\\ &\approx -87.96152422706632\ut{J}\\ &\approx -88\ut{J}\\ \end{aligned} $$

$$ \begin{cases} F_1&=110\ut{N},& x_1=40\ut{mm}\\ F_2&=240\ut{N},& x_2=60\ut{mm}\\ F_3&=W,& x_3=30\ut{mm}\\ \end{cases} $$ $$\ab{a}$$ $$ F=kx, $$ $$ \begin{cases} F_1&=k(x_1-x_0)\\ F_2&=k(x_2-x_0)\\ \end{cases} $$ $$k=\frac{13}{2}\ut{N/mm}$$ $$ \begin{aligned} x_0&=\frac{300}{13}\\ &\approx 23.076923076923077\ut{mm}\\ &\approx 23\ut{mm}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} W&=k(x_3-x..

$$ \begin{cases} m&=5.0\ut{kg}\\ F(x)&=\text{in Graph}\\ \end{cases} $$ $$W_{i\rarr f}=\int_{i}^{f} \vec F \cdot \dd \vec S,$$ $$ \begin{aligned} W_{0\rarr 8}&=\int_{0}^{8} F \cdot \dd x\\ &=\bra{\frac{1}{2}\cdot(2\ut{m}+4\ut{m})\cdot20\ut{N}}-\(\frac{1}{2}\cdot2\ut{m}\cdot10\ut{N}\)\\ &=50\ut{J}\\ &=5.0\times10\ut{J} \end{aligned} $$

$$ \begin{cases} \vec r_1&=2.70\i-2.90\j+5.50\k\ut{m}\\ \vec r_2&=-2.10\i+3.30\j+5.40\k\ut{m}\\ m&=2.90\ut{kg}\\ \vec F&=3.00\i+7.00\j+5.30\k\ut{N}\\ t&=3.00\ut{s} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} W&=\vec F\cdot \vec S\\ &=\vec F\cdot (\vec r_2-\vec r_1)\\ &=(3\i+7\j+5.3\k)\cdot(-4.8\i+6.2\j-0.1\k)\\ &=3\cdot(-4.8)+7\cdot6.2+5.3\cdot(-0.1)\\ &=28.47\ut{J}\\ &\approx 28.5\ut{J}\\ \end..

$$ \begin{cases} \theta&=30\degree\\ F&=38\ut{lbf}\\ v&=3.0\ut{mi/h}\\ t&=8.0\ut{min}\\ 1\ut{mi}&=5280\ut{ft}\\ \end{cases} $$ $$ \begin{aligned} S&=v t\\ &=3.0\ut{mi/h}\cdot 8.0\ut{min}\cdot\frac{1\ut{h}}{60\ut{min}}\cdot\frac{5280\ut{ft}}{1\ut{mi}}\\ &=2112\ut{ft} \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} W&=\vec F\cdot \vec S\\ &= F S \cos \theta\\ &=40128\sqrt3 \ut{ft\cdot lbf}\\ &\appr..

$$ \begin{cases} \theta &= 50\degree\\ F_r&=70\ut{N}\\ d&=0.60\ut{m}\\ \Delta K_1 &=80\ut{J} \end{cases} $$ $$ \begin{cases} \Delta K_1 = \Sigma W_1 = W_{mg}+W_F\\ \Delta K_2 = \Sigma W_2 = W_{mg} \end{cases} $$ $$ \begin{aligned} \Ans &= \Delta K_2-\Delta K_1\\ &=W_{mg}-(W_{mg}+W_F)\\ &=-W_F\\ &=-\vec F \cdot \vec d\\ &=-(-42\ut{J})\\ &=42\ut{J} \end{aligned} $$

$$ \begin{cases} m&=1.67\times10^{-27}\ut{kg}\\ a&=1.7\times10^{15}\ut{m/s^2}\\ v_0&=2.4\times10^6\ut{m/s}\\ S&=2.8\ut{cm}=0.028\ut{m} \end{cases} $$ $$\ab{a}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} v_1&=\sqrt{2aS+{v_0}^2}\\ &=4\sqrt{631}\times10^5\ut{m/s}\\ &\approx 1.0047885349664377\times10^7\ut{m/s}\\ &\approx 1.0\times10^7\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$K=\frac{1}{2}mv^2,$$ $$ \b..

풀이자주:유효숫자가 애매하여 다양하게 표현했습니다. $$ \begin{cases} \vec F&=290\i-230\j\ut{N}\\ \vec d&=15\i-12\j\ut{m}\\ \end{cases} $$ $$ \begin{aligned} W&=\vec F \cdot \vec d\\ &=290\cdot15+(-230)\cdot(-12)\\ &=7110\ut{J}\\ &=7.11\ut{kJ}\\ &\approx 7.1\ut{kJ} \end{aligned} $$