The elastic energy stored in your tendons can contribute up to 35 of your energy

Question

The elastic energy stored in your tendons can contribute up to 35 of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 , while nonathletes' stretch only 33 . The spring constant for the tendon is the same for both groups, 31 . What is the difference in maximum stored energy between the sprinters and the nonathletes? The elastic energy stored in your tendons can contribute up to 35 of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 , while nonathletes' stretch only 33 . The spring constant for the tendon is the same for both groups, 31 . What is the difference in maximum stored energy between the sprinters and the nonathletes? The spring constant for the tendon is the same for both groups, 31 . What is the difference in maximum stored energy between the sprinters and the nonathletes?

Explanation / Answer

      The sprinters tendons stretch, x2 = 43 mm                                                            = (43 mm)(10-3 m/1 mm)                                                            = 0.043 m       The sprinters nonathletes' stretch , x1 = 33 mm                                                                   = (33 mm)(10-3 m/1 mm)                                                                   = 0.033 m       The spring constant, k = 31 N/mm                                          = 31000 N/m _____________________________________________________________       The energy stored is              E = (1/2)k(x22-x12)                 = (1/2)(31000 N/m)[(0.043 m)2-(0.033 m)2]                 = 11.78 J ____________________________________________________________       If efficiency is 35%, then          E = (35%)(11.78 J)              = (35/100)(11.78 J)             = 4.12 J                                                                   = (33 mm)(10-3 m/1 mm)                                                                   = 0.033 m       The spring constant, k = 31 N/mm                                          = 31000 N/m _____________________________________________________________       The energy stored is              E = (1/2)k(x22-x12)                 = (1/2)(31000 N/m)[(0.043 m)2-(0.033 m)2]                 = 11.78 J ____________________________________________________________       If efficiency is 35%, then          E = (35%)(11.78 J)              = (35/100)(11.78 J)             = 4.12 J

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