The kinetic energy of a moving body is given by K = (1)/(2)mv^2 ....

Given:
$m = (2.0 \pm 0.1)$ kg
$v = (5.0 \pm 0.2)$ m/s

The percentage error in mass is:
$\dfrac{\Delta m}{m} \times 100 = \dfrac{0.1}{2.0} \times 100 = 5\%$

The percentage error in velocity is:
$\dfrac{\Delta v}{v} \times 100 = \dfrac{0.2}{5.0} \times 100 = 4\%$

The formula for kinetic energy is $K = \dfrac{1}{2}mv^2$. The constant $\dfrac{1}{2}$ does not contribute to the error.
The maximum percentage error in $K$ is:
$\dfrac{\Delta K}{K} \times 100 = \left( \dfrac{\Delta m}{m} \times 100 \right) + 2 \left( \dfrac{\Delta v}{v} \times 100 \right)$

Substituting the values:
$\dfrac{\Delta K}{K} \times 100 = 5\% + 2(4\%) = 5\% + 8\% = 13\%$

Answer: $13\%$