Answer
$s=1.35m$
Work Step by Step
The required distance can be determined as follows:
$\Sigma F_y=ma_y=0$
$\implies N+1000(\frac{3}{5})-800sin30-(100)(9.81)=0$
$\implies N=781N$
We know that
$\Sigma U_{1\rightarrow 2}=\Sigma F_x(s_2-s_1)$
$\Sigma U_{1\rightarrow 2}=1000(\frac{4}{5})+800cos 30-(0.2\times 781)(s)$
$\implies \Sigma U_{1\rightarrow 2}=1336.62(s)$
As $\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$
We plug in the known values to obtain:
$\frac{1}{2}(100)(0)^2+1336.62(s)=\frac{1}{2}(100)(6)^2$
This simplifies to:
$s=1.35m$
