## College Physics (4th Edition)

The correct answer is: (b) less than $m_2~g$
The correct answer is: (b) less than $m_2~g$ We can consider the motion of block $m_1$: $T = m_1~a$ $a = \frac{T}{m_1}$ We can consider the motion of block $m_2$: $m_2~g-T = m_2~a$ $m_2~g-T = m_2~(\frac{T}{m_1})$ $m_1~m_2~g-m_1~T = m_2~T$ $T = \frac{m_1~m_2~g}{m_1+m_2}$ $T = (\frac{m_1}{m_1+m_2})~m_2~g \lt m_2~g$