question_answer 72)

                  An adult weighing 600 N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging.                

Answer:

                  Energy utilised by adult to raise the centre of gravity of his body during each step                 \[=mgh=600\times 0.25=150\text{ }J\]                 Number of steps, while travelling 1 m = 1                 \ Number of steps, while travelling 6 km                 (= 6000 m) = 6000                 Energy utilised by adult during 6000 steps                 \[=150\text{ }J\times 6000=9\times {{10}^{5}}J\]                 Let E = energy intake in the form of food.                 \ \[10%\] of \[E=9\times {{10}^{5}}J\]                 or \[E=10\times 9\times {{10}^{5}}J\text{ }=9\times {{10}^{6}}J\]