At the highest position; $\;h=5 \text{m} ; m=10 \text{kg} ; g=10 \text{m} / \text{s}^2$ $\; \therefore \;\; \;\text{Potential Energy}\; (U_1) = m g h$ $\;\;=10 \times 10 \times 5$ $\;\;=500 \text{J}$Kinetic Energy $(K_1)=0$ $( \because \;\text{velocity}\;=0)$ $\therefore$ Total mechanical Energy $\;=U_1+K_1$ $\;=500+0$ $\;=500 \text{J}$ At the height of $2 \text{m}$, the total mechanical energy will remain same as it was at the highest position. ( $\because U+K=$ constant, for a freely falling body)$\therefore$ Total mechanical energy $=500 \text{J}$