Problem 9.43 - MO Part A The figure is a graph of the normal force exerted by th

Question

Problem 9.43 - MO Part A The figure is a graph of the normal force exerted by the floor on a woman making a vertical jump. At what time does the woman's feet leave the ground? Hints t 0.5 My Answers Give Up Correct Part B At what speed does she leave the ground? Hint The force of the foor is not the only force imparting an impuise to the woman. (Eaus 1 Express your answer using three significant figures. n(N) 2500 m/s 18.34 My Answers Give Up ncorrect: Try Again; no points deducted You have forgotten to include the impulse imparted by the weight See the Hints on how to include it 500 t(s) 0.2 04 0.5

Explanation / Answer

Impulse = area under the F vs t curve

Impulse by normal force = (500 x 0.2) + (500 x 0.2) + (2000 x 0.2 / 2) + (2500 x 0.1 / 2)

= 525 kg m/s

due to weight = (500 x 0.5) = 250 kg m/s

W = 500 = m g
=> m = 50 kg

net impulse = change in momentum

525 - 250 = 50 (v^2 - 0 )/ 2

v = 3.32 m/s

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