A block of mass m1 = 21.8 kg is at rest on a plane inclined at = 30.0° above the

Question

A block of mass m1 = 21.8 kg is at rest on a plane inclined at = 30.0° above the horizontal. The block is connected via a rope and massless pulley system to another block of mass m2 = 27.1 kg, as shown in the figure. The coefficients of static and kinetic friction between block 1 and the inclined plane are s = 0.109 and k = 0.086, respectively. If the blocks are released from rest, what is the displacement of block 2 in the vertical direction after 1.89 s? Use positive numbers for the upward direction and negative numbers for the downward direction.

Explanation / Answer

When the system is moving the net (unbalanced) force acting (Fr) is ..
Fr = (Weight of m2) - (weight component of m1 down incline) - (kinetic friction)
Fr = (m2.g) - (m1.g x sin30) - (k.m1.g x cos30)
Fr = (27.1*9.80) - (21.8*9.80*sin30) - (0.086*21.8*9.80*cos30) .. ..

         = (265.58)-106.82-15.91

            = 142.85          

Fr = 142.85 N (up incline)

Acceleration of joined mass, a = Fr / (m1 + m2)
a = 142.85N / 48.90kg = 2.92 m/s²

Distance moved .. d = ut + ½.at²

d = 0 + (½ x 2.92 x 1.89²) .. .. ..

d = 5.21 m (down)

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