Answer
$s=5.98m$
Work Step by Step
We can determine the required distance as follows:
$\Sigma F_y=ma_y=0$
$\implies N+500sin45-400sin30-(100\times 9.81)=0$
This simplifies to:
$N=827.45N$
We know that
$\Sigma U_{1\rightarrow 2}=(500cos45+400cos30-(0.2\times 827.45))s=534.47s$
We also know that
$\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$
We plug in the known values to obtain:
$\frac{1}{2}(100)(0)^2+534.47(s)=\frac{1}{2}(100)(8)^2$
This simplifies to:
$s=5.98m$
