As shown in the figure below, two masses m1 = 4.50 kg and m2 which has a mass 80

Question

As shown in the figure below, two masses m1 = 4.50 kg and m2 which has a mass 80.0% that of m1, are attached to a cord of negligible mass which passes over a frictionless pulley also of negligible mass. If m1 and m2 start from rest, after they have each traveled a distance h = 1.40 m, use energy content to determine the following. (a) speed v of the masses (b) magnitude of the tension T in the cord

Explanation / Answer

a)For m1: T + m1a = m1g =>T = 4.5*(9.81) - 4.5*a.......(1) For m2: T = m2g + m2a = (0.8*4.5*9.81) + (0.8*4.5*a)......(2) From (1) and (2) we get, 4.5*(9.81) - 4.5*a = (0.8*4.5*9.81) + (0.8*4.5*a) =>(4.5*(9.81))-(0.8*4.5*9.81) = ((4.5)+(0.8*4.5))a =>a = 1.09 m/s^2 S = h = 1.4 m u = initial velocity = 0 m/s We have to find v final velocity. v^2 = u^2 + 2aS =>v = sqrt(2*1.09*1.4) = 1.746997424 m/s....for both m1 and m2 b)T = (0.8*4.5*9.81) + (0.8*4.5*a) = (0.8*4.5*9.81)+(0.8*4.5*1.09) = 39.24 N

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