6-11 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=5.0\ut{kg}\\ \vec F&=40\i\ut{N}\\ \theta&=37\degree\\ \mu_k&=0.30\\ v_0&=4.0\ut{m/s}\\ \end{cases} $$ $$\ab{a}$$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{cases} \Sigma F_x &= m a_x\\ \Sigma F_y &= 0\\ \end{cases} $$ $$ \begin{cases} F\cos\theta -mg\sin\theta -\mu_kF_N &= m a\\ F_N-mg\cos\theta-F\sin\theta &= 0\\ \end{cases} $$ $$ \begin{aligned} a&=\frac{\cos\theta (F-\mu_k mg)-\sin\theta (F \mu_k +mg)}{m}\\ &\approx -3.3066429493621925\ut{m/s^2}\\ &\approx -3.3\ut{m/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$\text{Down}$$ $$\ab{c}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} S&=\frac{v^2-{v_0}^2}{2a}\\ &=\frac{0^2-{v_0}^2}{2a}\\ &\approx 2.419372191830719\ut{m}\\ &\approx 2.4\ut{m}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \Sigma F_x &=F\cos\theta -mg\sin\theta\\ &\approx 2.4364739179215427\ut{N} \end{aligned} $$ $$ \begin{aligned} f&=\mu_k(mg\cos\theta-F\sin\theta)\\ &\approx 18.969688664732505\ut{N} \end{aligned} $$ $$\text{Stop}$$