1. Write the formula for kinetic energy

   $K = \frac{1}{2}mv^2$

   Given: $m = (5 \pm 0.5)\,\text{kg}, \qquad v = (20 \pm 0.4)\,\text{m/s}$

2. Calculate the nominal value of kinetic energy

   $K = \frac{1}{2}\times 5 \times (20)^2$ $K = \frac{1}{2}\times 5 \times 400 = 1000\,\text{J}$

3. Find the fractional errors

   For $K = \frac{1}{2}mv^2$ the constant $\frac{1}{2}$ has no error, so

   $\frac{\Delta K}{K} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}$

   Now, $\frac{\Delta m}{m} = \frac{0.5}{5} = 0.1$ $\frac{\Delta v}{v} = \frac{0.4}{20} = 0.02$

   Therefore, $\frac{\Delta K}{K} = 0.1 + 2(0.02) = 0.1 + 0.04 = 0.14$

4. Calculate absolute error in kinetic energy

   $\Delta K = 0.14 \times 1000 = 140\,\text{J}$

5. Final result

   $K = (1000 \pm 140)\,\text{J}$

6. Check options

   • A: $(1000 \pm 140)\,\text{J}$ ✅
   • B: $(500 \pm 0.14)\,\text{J}$ ❌
   • C: $(1000 \pm 0.14)\,\text{J}$ ❌
   • D: $(500 \pm 140)\,\text{J}$ ❌

Therefore, the correct option is A.