Answer
(a) $a = 2.7 ~m/s^2$
(b) t = 0.96 seconds
(c) $m_A = 99 ~kg$
Work Step by Step
(a) Let's consider the force equation for $m_A$.
$\sum ~F = ma$
$F_T = (m_A) ~a$
Let's consider the force equation for $m_B$.
$\sum ~F = ma$
$(m_B) ~g - F_T = (m_B) ~a$
$(m_B) ~g - (m_A) ~a = (m_B) ~a$
$a = \frac{(m_B) ~g}{m_A + m_B}$
$a = \frac{(5.0)(9.80 ~m/s^2)}{(13.0 ~kg) + (5.0 ~kg)} = 2.7 ~m/s^2$
(b) $x = \frac{1}{2}at^2$
$t = \sqrt{\frac{2x}{a}} = \sqrt{\frac{(2)(1.250 ~m)}{2.7 ~m/s^2}} = 0.96 ~s$
(c) $a = \frac{(m_B) ~g}{m_A + m_B}$
$\frac{1}{100}~g = \frac{(m_B) ~g}{m_A + m_B}$
${m_A + m_B} = 100~(m_B)$
$m_A = 99 ~(m_B) = (99)(1.0 ~kg) = 99 ~kg$
