Answer
$F=19.444KN; T=12.5KN$
Work Step by Step
The required forces $F$ and $T$ can be determined as follows:
$mv_1+\Sigma \int_{t_1}^{t_2} F_x dt=mv_2 $
We plug in the known values to obtain:
$(50000+3\times 30000)(0)+F\Delta t=(50000+3\times 30000)(\frac{40\times 1000}{3600})$
$\implies (80-0)=1.556\times 10^6$
$\implies F=19.444KN$
Similarly,
$mv_1+\Sigma \int_{t_1}^{t_2} F_x dt=mv_2$
We plug in the known values to obtain:
$(3\times 30000)(0)+T\Delta T=(3\times 30000)(\frac{40\times 1000}{3600})$
$\implies T=12.5KN$