Think of this like a car speeding up from 30 km/h to 50 km/h; the extra 'effort' or work the engine must perform to reach that higher speed is represented by the increase in the car's kinetic energy.

Kinetic energy is the energy possessed by a body due to its motion, calculated using the formula $KE = \frac{1}{2}mv^2$, where $m$ is mass and $v$ is velocity. The work-energy theorem states that the net work done on an object equals its change in kinetic energy, which here is calculated as $\Delta KE = \frac{1}{2}m(v_f^2 - v_i^2)$. By substituting the given values $m = 10 \text{ kg}$, $v_i = 3 \text{ m/s}$, and $v_f = 5 \text{ m/s}$, we get $\Delta KE = \frac{1}{2} \times 10 \times (5^2 - 3^2) = 5 \times (25 - 9) = 5 \times 16 = 80 \text{ J}$.

Always remember to square the velocities separately before subtracting them; a common trap is calculating $(v_f - v_i)^2$ instead of $(v_f^2 - v_i^2)$.