ch06-p064 - 64. Note that since no static friction...

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2 12 sin k mg f a mm θ = + For f k = μ k F N = k m 1 g , we obtain 22 2 (3.0 kg)(9.8 m/s )sin30 (0.25)(2.0 kg)(9.8 m/s ) 1.96 m/s 3.0 kg 2.0 kg a °− == + . Returning this value to either of the above two equations, we find T = 8.8 N. 64. Note that since no static friction coefficient is mentioned, we assume f s is not relevant
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