ch05-p028 - v = 0, the acceleration is found to be ( ) 2 2...

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28. The stopping force G F and the path of the toothpick are horizontal. Our + x axis is in the direction of the toothpick’s motion, so that the toothpick’s acceleration (‘‘deceleration”) is negative-valued and the stopping force is in the – x direction: ˆ i FF =− G . Using Eq. 2-16 with v 0 = 220 m/s and v
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Unformatted text preview: v = 0, the acceleration is found to be ( ) 2 2 2 2 6 2 (220 m/s) 2 1.61 10 m/s . 2 2 0.015 m v v v a x a x = + = = = Thus, the magnitude of the force exerted by the branch on the toothpick is 4 6 2 2 | | (1.3 10 kg)(1.61 10 m/s ) 2.1 10 N. F m a = = =...
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