totaltotal12.0 N 6.0 N(4.0 kg)Ffm aa−=¡which yields the acceleration a= 1.5 m/s2. We have treated Fas 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 mW= 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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