#### Answer

(a) $F = 32~N$
(b) $T_{1t} = 19.2~N$
(c) $T_{1b} = 16~N$
(d) $T_{2t} = 3.2~N$

#### Work Step by Step

(a) Let $m_b$ be the mass of a block. Let $m_r$ be the mass of a rope. We can find $F$.
$\sum F = (2m_b+2m_r)~a$
$F - (2m_b+2m_r)~g = (2m_b+2m_r)~a$
$F = (2m_b+2m_r)(g+a)$
$F = [(2)(1.0~kg)+(2)(0.25~kg)](9.80~m/s^2+3.0~m/s^2)$
$F = 32~N$
(b) We can find the tension $T_{1t}$ at the top end of rope 1.
$\sum F = (m_b+2m_r)~a$
$T_{1t} - (m_b+2m_r)~g = (m_b+2m_r)~a$
$T_{1t} = (m_b+2m_r)(g+a)$
$T_{1t} = [(1.0~kg)+(2)(0.25~kg)](9.80~m/s^2+3.0~m/s^2)$
$T_{1t} = 19.2~N$
(c) We can find the tension $T_{1b}$ at the bottom end of rope 1.
$\sum F = (m_b+m_r)~a$
$T_{1b} - (m_b+m_r)~g = (m_b+m_r)~a$
$T_{1b} = (m_b+m_r)(g+a)$
$T_{1b} = (1.0~kg+0.25~kg)(9.80~m/s^2+3.0~m/s^2)$
$T_{1b} = 16~N$
(d) We can find the tension $T_{2t}$ at the top end of rope 2.
$\sum F = m_r~a$
$T_{2t} - m_r~g = m_r~a$
$T_{2t} = (m_r)(g+a)$
$T_{2t} = (0.25~kg)(9.80~m/s^2+3.0~m/s^2)$
$T_{2t} = 3.2~N$