A block of mass m 1 = 23.0 kg is at rest on a plane inclined at = 30.0° above th

Question

A block of mass m1 = 23.0 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 = 19.5 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 1.61 s? Use positive numbers for the upward direction and negative numbers for the downward direction.

Explanation / Answer

balancing forces on m1 perpendicular to the incline,

N - m1gcos@ = 0

N = m1gcos30

friction force f = uk N = 0.086m1gcos30 = 0.086 x 23 x 9.81 x cos30 = 16.80 N

balancing forces along the incline,

T - m1gsin30 - f = m1a ...... (i)

on hanging block ,

m2g - T = m2a .... (ii)

adding (i) and (ii)

m2g - m1gsin30 - f = (m1 +m2)a

19.5x9.81 - (19.5 x 9.81 x sin30) - 16.80 = (23 + 19.5)a

a = 1.86 m/s^2

using d = ut + at^2 /2

d = 0 - 1.86 x 1.61^2 /2

d = - 2.40 m

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