1. **Problem statement:** Calculate the work done on a body of mass 10 kg when its velocity increases from 2 m/s to 8 m/s. 2. **Formula used:** Work done on a body is equal to the change in its kinetic energy. $$ W = \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) $$ where $m$ is mass, $v_i$ is initial velocity, and $v_f$ is final velocity. 3. **Given values:** - Mass, $m = 10$ kg - Initial velocity, $v_i = 2$ m/s - Final velocity, $v_f = 8$ m/s 4. **Calculate the change in kinetic energy:** $$ W = \frac{1}{2} \times 10 \times (8^2 - 2^2) $$ $$ = 5 \times (64 - 4) $$ $$ = 5 \times 60 $$ $$ = 300 $$ 5. **Interpretation:** The work done on the body to increase its velocity from 2 m/s to 8 m/s is 300 joules. **Final answer:** 300 J