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CHAPTER 4 - 6 - 29 30(a sprinter, Fcos...

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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  θ  =  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 0  +  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  α  =  F ay / F ax = (26.0 N)/(20.2 N) = 1.29,  which gives        α  = 52.2 °  below –  x -axis .
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