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mrs_skeptik [129]
3 years ago
9

A Each time he does one pushup, Jose, who has a mass of 89kg, raises his center of mass by 25 cm. He completes an impressive set

of 150 pushups in 5 minutes, exercising at a steady rate If we assume that lowering his body has no energetic cost, what is his metabolic power during this workout? Express your answer with the appropriate units.Part B In fact, it costs Jose a certain amount of energy to lower his body-about half of what it costs to raise it. If you include this in your calculation, what is his metabolic power?
Physics
1 answer:
k0ka [10]3 years ago
5 0

The Metabolic power,If we assume that lowering his body has no energetic cost is  667 J/min

The Metabolic power, If lowering of his body needs half of the energy is 1000J/min

<u>Explanation:</u>

<u>1.If we assume that lowering his body has no energetic cost:</u>

<u>Metabolic energy in 5 minutes</u>

        = Increase in potential energy for 150 pushups

        = (89 kg) × (25/100 m) × 150

        = 3337.5 J

<u>Metabolic power</u>

        = (3375.5 J) / (5 min)

        = 667 J/min

The Metabolic power,If we assume that lowering his body has no energetic cost is  667 J/min

<u>2.If lowering of his body needs half of the energy what it costs to raise it.</u>

<u>Metabolic power</u>

         = (667 J/min) × [1 + (1/2)]

         = 1000 J/min

The Metabolic power, If lowering of his body needs half of the energy is 1000J/min

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w_f = 1.0345 rad/s

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                                     L_i = L_f

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                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

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                                   L_(c,f) = I_c*w_f

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                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

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-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

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                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

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