determine the acceleration of the system and the time it takes the heavier mass to reach the table\nenter…

Explanation:

Step1: Set up force - equations

Let $m_1 = 288\ g=0.288\ kg$ and $m_2 = 199\ g = 0.199\ kg$. For the heavier mass $m_1$, $m_1g - T=m_1a$. For the lighter mass $m_2$, $T - m_2g=m_2a$. Adding these two equations gives $(m_1 - m_2)g=(m_1 + m_2)a$. So, $a=\frac{(m_1 - m_2)g}{m_1 + m_2}$, where $g = 9.8\ m/s^2$.

Step2: Calculate the acceleration

Substitute $m_1 = 0.288\ kg$ and $m_2 = 0.199\ kg$ into the acceleration formula: [ \begin{align*} a&=\frac{(0.288 - 0.199)\times9.8}{0.288+ 0.199}\ &=\frac{0.089\times9.8}{0.487}\ &=\frac{0.8722}{0.487}\ &\approx1.8\ m/s^2 \end{align*} ]

Step3: Calculate the time

The heavier mass $m_1$ is initially at a height $h = 90 - 10=80\ cm = 0.8\ m$. Using the kinematic - equation $h=v_0t+\frac{1}{2}at^2$. Since $v_0 = 0$, we have $h=\frac{1}{2}at^2$. Then $t=\sqrt{\frac{2h}{a}}$.

Step4: Substitute values to find time

Substitute $h = 0.8\ m$ and $a\approx1.8\ m/s^2$ into the time formula: [ \begin{align*} t&=\sqrt{\frac{2\times0.8}{1.8}}\ &=\sqrt{\frac{1.6}{1.8}}\ &=\sqrt{\frac{8}{9}}\ &\approx0.94\ s \end{align*} ]

Answer:

Acceleration: 1.8 Time: 0.94