ch06-p022 - 22 Treating the two boxes as a single system of...

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total total 12.0 N 6.0 N (4.0 kg) Ff m a a −= ¡ which yields the acceleration a = 1.5 m/s 2 . We have treated F as if it were known to the nearest tenth of a Newton so that our acceleration is “good” to two significant figures. Turning our attention to the larger box (the Wheaties box of mass m W = 3.0 kg) we apply Newton’s second law to find the contact force F' exerted by the Cheerios box on it.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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