Two masses of 3.00 kg and 5.00 kg are connected by a light stringthat passes over a frictionless pulley, and was shown in Figure5.15a.Figure 5.15a(a) Determine the tension in the string. N(b) Determine the acceleration of each mass. m/s2 upwards (3.00 kg mass) m/s2 downwards (5.00 kg mass)(c) Determine the distance each mass will move in the firstsecond of motion if they start from rest. m (3.00 kg mass) m (5.00 kg mass)i don't understand how to do like any of this problem, someone please help me!!!!!

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Two masses of 3.00 kg and 5.00 kg are connected by a light stringthat passes over a frictionless pulley, and was shown in Figure5.15a.Figure 5.15a(a) Determine the tension in the string. N(b) Determine the acceleration of each mass. m/s2 upwards (3.00 kg mass) m/s2 downwards (5.00 kg mass)(c) Determine the distance each mass will move in the firstsecond of motion if they start from rest. m (3.00 kg mass) m (5.00 kg mass)i don&#039;t understand how to do like any of this problem, someone please help me!!!!!

casincal · Accepted Answer

Total forces acting on both masses together mustbe:    forward, Mg     backward,mg    The masses move together, so they have the same acceleration. So   total force = total mass×a    Mg - mg = ( M + m ) a    a=g(M−m)(M+m)=9.8×2.008.00=2.45ms  this is the acc for both Tension? consider just the smaller mass.  total force = ma    T - mg = ma    T = mg + ma = m ( g + a) = 3.00×(9.8+2.45) =   36.75 Newtons    And the distance? They move the same distance. This can befound by using dist = 12at2=12×9.8×12=4.9meters    they bothmove this distance in the first second

Two masses of 3.00 kg and 5.00 kg are connected by a light stringthat passes over a frictionless pulley, and was shown in Figure5.15a. Figure 5.15a (a

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