Answer
$0.828m/s^2$
Work Step by Step
We can find the required acceleration as follows:
$F_{net}=m_ca_c+m_sa_s$
This can be rearranged as:
$a_c=\frac{F_{net}-m_sa_s}{m_c}$
We plug in the known values to obtain:
$a_c=\frac{40N-(9.75Kg)(2.32m/s^2)}{21.0Kg}$
$a_c=0.828m/s^2$
