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Two blocks connected by a rope of negligible mass are being dragged across a tab

ID: 2142674 • Letter: T

Question

Two blocks connected by a rope of negligible mass are being dragged across a table. Suppose F = 179N, m1 = 8.10 kg, m2 = 15.4 kg,  the acceleration of the system is 4.25m/s2. If the coefficient of kinetic friction between block 1 and the table is equal to the coefficient of friction between block 2 and the table.

what is  the coefficient of kinetic friction between the blocks and the table?

what is the tension in the rope

Two blocks connected by a rope of negligible mass are being dragged across a table. Suppose F = 179N, m1 = 8.10 kg, m2 = 15.4 kg, the acceleration of the system is 4.25m/s2. If the coefficient of kinetic friction between block 1 and the table is equal to the coefficient of friction between block 2 and the table. what is the coefficient of kinetic friction between the blocks and the table? what is the tension in the rope

Explanation / Answer

F - friction = (m1 + m2 )a


F - u (m1 + m2 ) g = (m1 + m2) a


179-u*(8.1+15.4)*9.81 = (8.1+15.4)*4.25

u=0.343


sum forces for block 1

T - friction = m1 a

T = m1a + friction = 8.1*4.25 + 0.343*8.1*9.81=61.68 N

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