#### Answer

$s=0.0735ft$

#### Work Step by Step

According to the work-energy principle
$T_1+\Sigma U_{1\rightarrow}=T_2$
$\implies \frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$
$\implies \frac{1}{2}mv_1^2+W(3+s)+\frac{1}{2}k(x_1^2-x_2^2)=\frac{1}{2}mv_2^2$
We plug in the known values to obtain:
$\implies \frac{1}{2}(\frac{5}{100})(10)^2+5(3+s)+\frac{1}{2}(4)(0.75)^2-(0.75+s)^2=0$
This simplifies to:
$s=0.0735ft$