The formula for kinetic energy is $KE = \frac{1}{2}mv^2$. The constant $\frac{1}{2}$ is exact and does not contribute to the error. The fractional error in a quantity $X = A^p B^q C^r$ is given by $\frac{\Delta X}{X} = p\frac{\Delta A}{A} + q\frac{\Delta B}{B} + r\frac{\Delta C}{C}$. For $KE = mv^2$, the fractional error is $\frac{\Delta KE}{KE} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}$. Given: Mass $m = 5.0 \text{ kg}$ with absolute error $\Delta m = 0.1 \text{ kg}$. Velocity $v = 10.0 \text{ m/s}$ with absolute error $\Delta v = 0.2 \text{ m/s}$. Fractional error in mass: $\frac{\Delta m}{m} = \frac{0.1}{5.0} = 0.02$. Fractional error in velocity: $\frac{\Delta v}{v} = \frac{0.2}{10.0} = 0.02$. Substituting these values into the error formula: $\frac{\Delta KE}{KE} = 0.02 + 2(0.02) = 0.02 + 0.04 = 0.06$. To find the percentage error, multiply by $100\%$: Percentage error in $KE = 0.06 \times 100\% = 6.0\%$.