A block of mass m 1 = 23.6 kg is at rest on a plane inclined at = 30.0° above th

Question

A block of mass m1 = 23.6 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

Lets calculate the weight along the rope
For m1,it is m1*g*sin 30 = 23.6*9.8*sin 30 = 115.64 N

for m2, it is m2*g =19.5*9.8 = 191.1 N

Since force along string is more for m2, m2 has tendency move down

m1 has tendency move up, so friction on m1 will be acting downward parallel to m1*g*sin 30

static frictional force,fs = mius*m1*g*cos thetha
=0.109*23.6*9.8*cos 30
= 21.83 N

So, net force along the riope for m1 = 115.64 + 21.83 = 137.47 N

Since this is less than 191.1 N, objects will move.

Kinetic frictional, fk = miul*m1*g*cos thetha
=0.086*23.6*9.8*cos 30
= 17.23 N

Net force along rope = 191.1 - 115.64 - 17.23
= 58.23 N

Accelration magnitude a = Net force /net mass
= 58.23 / (m1+m2)
= 58.23 / (23.6 + 19.5)
= 1.35 m/s^2

for m2 this accelration is in downward direction.
so,a = -1.35 m/s^2
t = 1.6 s
use:
d = Vi*t + 0.5*a*t^2
= 0 + 0.5*(-1.35)*(1.6)^2
= -1.73 m
Answer: -1.73 m

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