problem04_50

University Physics with Modern Physics with Mastering Physics (11th Edition)

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4.50: a) b) The athlete’s weight is N 882 ) s / m 80 . 9 )( kg 0 . 90 ( 2 = = mg . The acceleration of the barbell is found from s / m 375 . 0 s 6 . 1 / m 60 . 0 av = = v . Its final velocity is thus (2)(0 375m s) 0 750m s . / = . / , and its acceleration is 2 0 s / m 469 . 0 65 . 1 s / m 750 . 0 = = - = t v v a The force needed to lift the barbell is given by: ma w F F = - = barbell lift net The barbell’s mass is 2 (490 N) (9 80m s )
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Unformatted text preview: 50 0kg / . / = . , so N 513 N 23 N 490 ) s / m 469 . )( kg . 50 ( N 490 2 barbell lift = + = + = + = ma w F The athlete is not accelerating, so: athlete lift floor net =--= w F F F N 1395 N 882 N 513 athlete lift floor = + = + = w F F...
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  • NoProfessor
  • Mass, Velocity, m/s, Barbell, final velocity, 0.60 m

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