8-57 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=6.20\ut{kg}\\ \theta&=30.0\degree\\ v_0&=6.00\ut{m/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \TE : \text{Thermal Energy}\\ \end{cases} $$ $$ \sin\theta=\frac{\Delta h}{\Delta S} $$ $$ \begin{aligned} \Sigma F_y=N-mg\cos\theta=0 \end{aligned} $$ $$\Sigma \Delta E=0,$$ $$\Delta \KE+\Delta \GE+\Delta \TE=0$$ $$ \begin{aligned} -\Delta \KE&=\Delta \GE+\Delta \TE\\ -\Delta \(\frac{1}{2}mv^2\)&=\Delta (mgh)+f\Delta S\\ -\frac{1}{2}m\Delta \(v^2\)&=mg\Delta h+\mu mg\Delta S\cos\theta\\ -\Delta \(v^2\)&=2g\Delta S\sin\theta+2\mu g\Delta S\cos\theta\\ {v_i}^2-{v_f}^2&=2g\Delta S(\sin\theta+\mu \cos\theta)\\ \Delta S&=\frac{{v_i}^2}{2g(\sin\theta+\mu\cos\theta)} \end{aligned} $$ $$\ab{a}$$ $$\mu=0,$$ $$ \begin{aligned} \Delta S&=\frac{{v_i}^2}{2g\sin\theta}\\ &=\frac{36}{g}\\ &\approx 3.670978366720542\ut{m}\\ &\approx 3.67\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$\mu=0.400,$$ $$ \begin{aligned} \Delta S&=\frac{180}{(5+2\sqrt3)g}\\ &\approx 2.1685575939655086\ut{m}\\ &\approx 2.17\ut{m}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \Delta \TE_c&=fS\\ &=\mu mg S \cos\theta\\ &= \frac{\mu m{v_i}^2\cos\theta}{2(\sin\theta+\mu\cos\theta)} \\ &=\frac{1116}{65} \left(5 \sqrt{3}-6\right)\ut{J}\\ &\approx 45.674515480528214\ut{J}\\ &\approx 45.7\ut{J}\\ \end{aligned} $$ $$\ab{d}$$ $$\Sigma \Delta E=0,$$ $$\Delta \KE+\Delta \GE+\Delta \TE=0$$ $$ \begin{aligned} \Delta \KE&=-\Delta TE_d\\ \Delta \(\frac{1}{2}m{v}^2\)&=-2\Delta TE_c\\ \frac{1}{2}m\Delta \({v}^2\)&=\frac{-\mu m{v_i}^2\cos\theta}{\sin\theta+\mu\cos\theta} \\ {v_f}^2-{v_i}^2&=\frac{-2\mu {v_i}^2\cos\theta}{\sin\theta+\mu\cos\theta} \\ \end{aligned} $$ $$ \begin{aligned} v_f&=v_i\sqrt{\frac{ \sin\theta-\mu \cos\theta}{\sin\theta+\mu \cos\theta}}\\ &=6 \sqrt{\frac{1}{13} \(37-20 \sqrt{3}\)}\\ &\approx 2.555889406395302\ut{m/s}\\ &\approx 2.56\ut{m/s}\\ \end{aligned} $$