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M m 8 Consider 0 - 41 degrees. IfM -2.6m, what is the acceleration of the blocks? The inclines is frictionless. Consider DOW
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given 0=410 M = 2.6m free body diagaam of m AT f Finet .ma mg T= tension mg = weight Net force ma=t-mg - body diagram of M aMat 2.6ma = 13 у solving at -mg +1.706 mg 3.6 mi a = 0.706 mg acceleration = 1.92 a = 1.92 m/s2 (Answed) as at base of inclin

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