A block of mass m1 = 24.2 kg is at rest on a plane inclined at = 30.0° above the

Question

A block of mass m1 = 24.2 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 = 18.9 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 1.47 s? Use positive numbers for the upward direction and negative numbers for the downward direction.

Explanation / Answer

component of m1 acting along incline = -m1*g*sin thetha
= -24.2*9.8*sin 30
= -118.58 N
Component of m2 acting along rope = -m2*g = -18.9*9.8 = -185.22 N

Since component of m2 is more than that of m1, m1 will have tendency to move down

so frictional force acting on m1 will be al0ng the incline in donward direction
fs = -mius*m1*g* cos thetha = -0.109*24.2*9.8*cos 30 = -22.39 N

Since f + component of m1 acting along incline < Component of m2 acting along rope
m1 will move upward

kinetic friction will act on it
fs = -miuk*m1*g* cos thetha = -0.086*24.2*9.8*cos 30 = -17.66 N

a= net force along the rope / net mass
= - (185.22 - 17.66 - 118.58) / (24.2 + 18.9)
= -1.14 m/s^2

use:
d= vi*t +0.5*a*t^2
= 0 + 0.5*(-1.14)*(1.47)^2
= - 1.23 m
Answer: - 1.23 m

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