6-31 할리데이 11판 솔루션 일반물리학

(풀이자주:문제에 주어진 데이터가 부족해 운동마찰계수는 정확히 구할 수 없습니다. 경사로의 각도를 임의로 두고 풀었습니다.) $$ \begin{cases} S&=250\ut{m}\\ t_A&=62\ut{s}\\ t_B&=45\ut{s}\\ \end{cases} $$ $$\ab{a,b}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} S&=(0)t+\frac{1}{2}at^2\\ a&=\frac{2S}{t^2}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} a_A&=\frac{2S}{{t_A}^2}\\ &=\frac{125}{961}\ut{m/s^2}\\ &\approx 0.13007284079084286\ut{m/s^2}\\ &\approx 0.13\ut{m/s^2}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} a_B&=\frac{2S}{{t_B}^2}\\ &=\frac{20}{81}\ut{m/s^2}\\ &\approx 0.24691358024691357\ut{m/s^2}\\ &\approx 0.25\ut{m/s^2}\\ \end{aligned} $$ $$\ab{c,d}$$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{cases} \Sigma F_x &= m a_x\\ \Sigma F_y &= 0\\ \end{cases} $$ $$ \begin{cases} ma &= mg\sin\theta-\mu_kF_N\\ 0 &= F_N-mg\cos\theta\\ \end{cases} $$ $$ \begin{aligned} a&=g\sin\theta-\mu_kg\cos\theta\\ \mu_k&=\frac{g\sin\theta-a}{g\cos\theta}\\ &=\tan\theta-\frac{a}{g\cos\theta}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \mu_{kA}&=\tan\theta-\frac{a_A}{g\cos\theta}\\ &=\tan\theta-\frac{125}{961g\cos\theta}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \mu_{kB}&=\tan\theta-\frac{a_B}{g\cos\theta}\\ &=\tan\theta-\frac{20}{81g\cos\theta}\\ \end{aligned} $$