Answer
$v_A=28.3m/s$
Work Step by Step
We can determine the required speed as follows:
According to the work-energy principle
$T_1+\Sigma U_{1\rightarrow 2}=T_2$
$\implies 0+\frac{1}{2}F(2S_B)=\frac{1}{2}mv_A^2$
We plug in the known values to obtain:
$0+\frac{1}{2}(20000)(2)(0.2)=\frac{1}{2}(10)v_A^2$
This simplifies to:
$v_A=28.3m/s$
