Question

Two blocks are connected by a string that is wrapped around a pulley wheel. The light block with mass 5.3 kg sits a flat table; the heavier block, which is twice as massive, is suspended below the pulley. There is friction acting between the lighter block and the table with a coefficient of friction 0.79. The system is released from rest. What is the speed of the blocks after the bottom block has fallen 4.2 meters in units of meters/second? You can ignore the mass of the string and the pulley wheel.

Answer 1

o a T my % AT a Along vertical! al mig mi=5.3kg M2=2m. m₂ =2x5.3 kg M2 =10.6 kg m la Tis tension in the string. a is acceleration for me an Bu N = mig normal force coefficient of kinetic friction. f = Man ha=0.79 x mig fic=0.79x5.349:8N he 41.0326M> Kinetic friction Along horizontali T-fx=m, a Mer=0.79

41:0326 = 5:30 a For Mr. AT Im La Mag a = M g-T= mza > (10.6x9.8) – T=10.69 103-88-T= 10.69 from equation 6 and ② 7-41.0326 +103.88 -X= 5.39 +0.69 7. 62.8476= 15.9 a 162.8476 16] m/s² 15.g 19= 3.95268 m/s2 Initial speed of blocks is fu=om's Velocity of battom block is when et fall by ay=4.2m. So, w cel +29 44 = 0+2x3.95268x42 > V? 33.202512 > V=533:202512 m/s V~5.762 m/s