CHAPTER 4 - 6 - F 1 F 2 x y ( a ) ( b ) F 2 F 1 x y F a F b...

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Unformatted text preview: F 1 F 2 x y ( a ) ( b ) F 2 F 1 x y F a F b 29. Note that we ignore air resistance. 30. ( a ) We find the horizontal acceleration from the horizontal component of the force exerted on the sprinter, which is the reaction to the force the sprinter exerts on the block: F cos = ma ; (80 N) cos 22 = (57 kg) a , which gives a = 1.3 m/s 2 . ( b ) For the motion of the sprinter we can write v = v + at = 0 + (1.3 m/s 2 )(0.34 s) = 0.44 m/s . 31. ( a ) For the components of the net force we have F ax = F 1 = 20.2 N; F ay = F 2 = 26.0 N. We find the magnitude from F a 2 = F ax 2 + F ay 2 = ( 20.2 N) 2 + ( 26.0 N) 2 , which gives F a = 32.9 N . We find the direction from tan = F ay / F ax = (26.0 N)/(20.2 N) = 1.29, which gives...
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CHAPTER 4 - 6 - F 1 F 2 x y ( a ) ( b ) F 2 F 1 x y F a F b...

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