55. (a) We use constant acceleration kinematics. If down is taken to be positive and ais the acceleration of the heavier block m2, then its coordinate is given by yat=122, so ayt===×−22 0 750500600 1022(.))..msm/sBlock 1 has an acceleration of 6.00 ×10–2m/s2upward. (b) Newton’s second law for block 2 is 2mg Tma−=, where m2is its mass and T2is the tension force on the block. Thus, ()2()(0.500 kg) 9.8 m/s6.00 10m/s4.87 N.Tmga−=−=−×=(c) Newton’s second law for block 1 is 111,−where T1is the tension force on the block. Thus, 2()(0.460 kg) 9.8 m/s6.00 10m/s4.54 N.ga−=+
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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.