ch05-p082 - 82 We take x uphill for the m2 = 1.0 kg box and...

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22 1 sin FTm g m a Tm a θ −− = = which can be added to obtain F m 2 g sin = ( m 1 + m 2 ) a . This yields the acceleration 2 2 12 N (1.0 kg)(9.8 m/s )sin 37 1.53 m/s . 1.0 kg 3.0 kg a −° == + Thus, the tension is T = m 1 a = (3.0 kg)(1.53 m/s 2 ) = 4.6 N. 82. We take + x uphill for the m 2 = 1.0 kg box and + x rightward for the m 1 = 3.0 kg box (so the accelerations of the two boxes have the same magnitude and the same sign). The
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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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