A 3.00-kg object has a velocity $(6.00\hat{\mathrm{i}}-1.00\hat{\mathrm{j}}) \mathrm{m}/\mathrm{s}$. What is the net work done on the object if its velocity changes to (8.00$\hat{\mathrm{i}}$+ 4.00$\hat{\mathrm{j}}$)m/s ?

$\stackrel{\to }{{\mathrm{v}}_{\mathrm{i}}} = (6.00\hat{\mathrm{i}}-1.00\widehat{\mathrm{j})}\mathrm{m}/{\mathrm{s}}^2$

${\mathrm{v}}_{\mathrm{i}} = \sqrt{{{\mathrm{v}}_{\mathrm{ix}}}^2+{{\mathrm{v}}_{\mathrm{iy}}}^2} = \sqrt{37.0} \mathrm{m}/\mathrm{s}$

$\overrightarrow{{\mathrm{v}}_{\mathrm{f}}} = 8.00\hat{\mathrm{i}}+4.00\hat{\mathrm{j}}$

${{\mathrm{v}}_{\mathrm{f}}}^2 = 64.0 +16.0 = 80.00 {\mathrm{m}}^2/{\mathrm{s}}^2$

$∆\mathrm{K} = {\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}} = \frac{1}{2}\mathrm{m}({{\mathrm{v}}_{\mathrm{f}}}^2-{{\mathrm{v}}^2}_{\mathrm{i}})$

      = $\frac{3.00}{2}(80.0-37)= 64.5\mathrm{J}$