Two blocks connected by a rope of negligible mass are being dragged by a force at a 22° angle above horizontal (see figure below). Suppose F = 89.0 N, m1 = 14.0 kg, m2 = 22.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.128.

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Leonard Mccarthy

Physics

11 September, 04:42

Two blocks connected by a rope of negligible mass are being dragged by a force at a 22° angle above horizontal (see figure below). Suppose F = 89.0 N, m1 = 14.0 kg, m2 = 22.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.128.

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Spud · Answer 1

If the blocks are resting on a horizontal surface and the F force is applied to one of them in an angle, then the effective force has to be x component of the force: Fx = F*cosa=89.9N*cos (22) = 83.4N The normal force for each block will be the same as it weight, as long as they are over a horizontal surface. N1=P1=m1*g=14.0kg*10m/s^2=140N N2=P2=m2*g=22.0kg*10m/s^2=220N The friction force is: Ff=Mu*N Ff1=Mu1*N1=0.128*140N=17.9N Ff2=Mu2*N2=0.128*220N=28.2N So the system has a total mass of: m=m1+m2=14.0kg+22.0kg=36.0kg And a total effective force of: Fr=Fx-Ff1-Ff2=83.4N-17.9N-28.2N=37.3N So, using Newton’s Law: Fr=m*a a=Fr/m=37.3N/36.0kg=1.04m/s^2