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Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 68.0 N, m1 = 12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100.
Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 68.0 N, m1 = 12.0 kg, m2 = 18.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.100.(a) Draw a free-body diagram for each block.
(b) Determine the tension T and the magnitude of the acceleration of the system.
ThetensionT is 27.2N and the magnitude of the acceleration of the system is 1.29m/s^2 when twoblocksconnected by a rope of negligible mass are being dragged by a horizontal force.Givenforceexerted on block F = 68.0 NMass of first block m1 = 12.0 kgMassof second block m2 = 18.0 kgThe coefficient ofkineticfrictionis 0.100.(a) Block 1's tension force on block 2 is equal to object 2's tension force on object 1, in terms of size. A light string'stension, which remains constant along its length, indicates how firmly the string is pulling on the objects at both ends.(b) The second law of Newton is applied using the free-body diagrams.On m1 we have net force acting = T-f = maFnet = N1-m1g = 0 then N= m1gthecoefficientof friction gives f1 = μN1f1 = 0.100x12x9.8 = 11.8NOn m2 we have net force acting = F1 - T- f = maWe have from the y component N2 = m2gthen f2 = 0.100x18x9.8 = 17.6NF−f1 −f2 =m2a+m1a68-11.8-17.6 = a(12+18)a = 1.29m/s^2Theaccelerationof the system is 1.29m/s^2we have tension T = m1a+f1 = 12x1.29+11.8 = 27.2NTo learn more abouttensionclick herebrainly.com/question/11348644#SPJ4...