## College Physics (4th Edition)

The weight of block 1 pulls down to the left while the weight of block 2 pulls down to the right. We can find the acceleration of the system: $\sum F = (m_1+m_2)~a$ $m_2~g-m_1~g = (m_1+m_2)~a$ $a = \frac{(m_2~-m_1)~g}{m_1+m_2}$ $a = \frac{(9.2~kg-3.6~kg)(9.80~m/s^2)}{3.6~kg+9.2~kg}$ $a = 4.3~m/s^2$ We can find the time it takes block 2 to reach the floor: $\Delta y = \frac{1}{2}at^2$ $t = \sqrt{\frac{2\Delta y}{a}}$ $t = \sqrt{\frac{(2)(1.40~m)}{4.3~m/s^2}}$ $t = 0.81~s$ It takes 0.81 seconds for block 2 to reach the floor.