$解:设矩形OABC的边CB=a，AB=b$
$∵CE=\frac{1}{3}CB,AF=\frac{1}{3}AB,∴CE=\frac{1}{3}a,AF=\frac{1}{3}b$
$∴△COE的面积为\frac{1}{6}ab,△AOF的面积为\frac{1}{6}ab,矩形OABC的面积为ab$
$∴四边形OEBF的面积为ab-\frac{1}{6}ab-\frac{1}{6}ab=\frac{2}{3}ab$
$∴△AOF的面积: 四边形OEBF的面积=\frac{1}{6}:\frac{2}{3}=1:4$
$∵四边形OEBF的面积为2，∴△AOF的面积=四边形OEBF的面积×\frac{1}{4}=2×\frac{1}{4}=\frac{1}{2}$
$∴\frac{1}{2}×|k|=\frac{1}{2},∴k=±1$
$又∵k＞0,∴k=1$
