Answer
(a) $0.76m/s^2$
(b) $6.0N$
Work Step by Step
(a) We can find the required acceleration as follows:
$a=\frac{F}{m_1+m_2}-\mu_kg$
We plug in the known values to obtain:
$a=\frac{9.4}{3.02}-(0.24)(9.81)$
$a=0.76m/s^2$
(b) We can find the required tension as follows:
$T=m_2a +\mu_k m_2g$
$\implies T=m_2(a+\mu_kg)$
We plug in the known values to obtain:
$T=(1.92)[0.76+(0.24)(9.81)]$
$T=6.0N$
