Answer
6.63 m/s.
Work Step by Step
According to the work-energy theorem,
$K_{f}=K_{i}+W$
$=\frac{1}{2}mv_{i}^{2}+\vec{F}\cdot \vec{d}$
$\vec{F}\cdot \vec{d}=F_{x}d_{x}+F_{y}d_{y}+F_{z}d_{z}$$=(-5.00\times2.00)+(5.00\times2.00)+(4.00\times7.00)=28.0\,J$
Therefore,
$K_{f}=\frac{1}{2}\times2.00\,kg(4.00\,m/s)^{2}+28.0\,J=44.0\,J$
That is, $\frac{1}{2}mv_{f}^{2}=44.0\,J$
$\implies 1.0\,kg\times v_{f}^{2}=44.0\,J$
Or $v_{f}=\sqrt {44.0\,m^{2}/s^{2}}=6.63\,m/s$
