Answer
Please see the work below.
Work Step by Step
We know that
$F_2=Kx$
We plug in the known values to obtain:
$F_2=(8100)(0.051)=413.1N$
Now $a_2=\frac{F_2}{m_2}$
$a_2=a=\frac{413.1}{4900}=0.84306\frac{m}{s^2}$
Hence, the net force is given as
$F=(m_1+m_2)a$
We plug in the known values to obtain:
$F=(640+490)(0.84306)=950N$
