A block of mass m 1 =21.9kg is at rest on a plane inclined at =30.0 degrees abov

Question

A block of mass m1=21.9kg is at rest on a plane inclined at =30.0 degrees above the horizontal. The block is connected via a rope and a massless pulley system to another block of mass m2=25.1 kg. The coefficients of static and kinetic friction between block 1 and the inclined plane are s=0.109 and k=0.086. If the blocks are released from rest, what is the displacement of block 2 in the vertical direction after 1.51s? Use positive numbers for the upward direction and negative numbers for the downward direction.

Explanation / Answer

normal reaction on m1 N= mgcos(30) = 185.87 N static friction = µsN = 20.26 mgsin(30) = 107.31 N weight of m2 = 245.98 weight > mgsin(30) + 20.26 m2 moves in downward direction kinetic friction f= 15.98 N forces on block 2 are Tension T upwards weight donwwards m2g-T = m2a T-f -m1gsin(30) = m1a m2g -(f +m1gsin(30) + m1a) = m2a m2g -f -m1gsin(30) = (m1+m2)a a = -2.61 m/s^2 (doqnwards) displacement of block 2= -.5*2.61*1.51^2 = -2.98 m

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