Answer
$T=11.25kN$, $F=33.75kN$
Work Step by Step
We can determine the required force as follows:
$v^2=v_{\circ}^2+2a(s-s_{\circ})$
We plug in the known values to obtain:
$0=(15)^2+2a(10-0)$
This simplifies to:
$a=-11.25m/s^2$
According to Newton's second law
$\Sigma F_x=ma_x$
$\implies -T=1000\times (-11.25)$
$\implies T=11250N=11.25kN$
Similarly, $\Sigma F_x=ma_x$
$T-F=ma$
$\implies 11250-F=2000\times (-11.25)$
This simplifies to:
$F=33750N=33.75kN$