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Solução Do Halliday 8.ed.vol.1. Cap.5 - Ch05 - P078

A solução do Halliday da 8ª edição volume 1. do Cap.5

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    December 2018
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78. With SI units understood, the net force on the box is G Fnet = ( 3.0 + 14 cos 30° − 11) ˆi + (14 sin30° + 5.0 − 17 ) ˆj G which yields Fnet = (4.1 N) ˆi − ( 5.0 N) ˆj . (a) Newton’s second law applied to the m = 4.0 kg box leads to G G Fnet a= = (1.0 m/s 2 )iˆ − (1.3m/s 2 )ˆj . m 2 G (b) The magnitude of a is a = (1.0 m/s 2 ) 2 + ( −1.3 m/s 2 ) = 1.6 m s 2 . (c) Its angle is tan–1 [(–1.3 m/s2)/(1.0 m/s2)] = –50° (that is, 50° measured clockwise from the rightward axis).