#### Answer

$F=227N$

#### Work Step by Step

We can determine the required force as follows:
According to the conservation of energy equation
$\frac{1}{2}mv_1^2+mgh_1=\frac{1}{2}mv_2^2+mgh_2$
$\implies \frac{1}{2}(40)(0)^2+(40\times 9.81-2cos60)=\frac{1}{2}(40)v_2^2+(40\times 9.81(-2))$
This simplifies to:
$v_2=4.429m/s$
We know that
$T-mg=m\frac{v^2}{\rho}$
$\implies T-(40)(9.81)=40\times (\frac{(4.429)^2}{2})=784.8N$
We also know that
$\Sigma F_y=0$
$\implies 2\times 2Fcos30-T=0$
$\implies 4Fcos30=784.8$
This simplifies to:
$F=227N$