A runner weighing 60 kg raises the CG of his body by 0.25 m during each step of 1 min length. If he runs for 6 km and there is loss of 10% of the energy due to friction of air etc. What is the total energy required by him from his intake of food ?

A runner weighing 60 kg raises the CG of his body by 0.25 m during eac

1.	A runner weighing 60 kg raises the CG of his body by 0.25 m during each step of 1 min length. If he runs for 6 km and there is loss of 10% of the energy due to friction of air etc. What is the total energy required by him from his intake of food ?
Answer» `9XX10^(5)` J `8xx10^(5)` J `9.9xx10^(5)` J `8.8xx10^(5)` J Solution :Here in taking a RUN of 6000 METRE, there are 6000 steps. For each STEP he raise the C.G. by 0.25 m `:.` P.E. required by the runner `E_P=60xx10xx0.25xx6000` =`9xx10^5J` Total energy required =`(9xx10^5xx110)/(100)=9.9xx10^5 J`

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A runner weighing 60 kg raises the CG of his body by 0.25 m during each step of 1 min length. If he runs for 6 km and there is loss of 10% of the energy due to friction of air etc. What is the total energy required by him from his intake of food ?

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