A block of mass m1 = 21.2 kg is at rest on a plane inclined at ? = 30.0° above t

Question

A block of mass m1 = 21.2 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 = 28.0 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

The static force impeding the movement of the block on the ramp is
21.2*g*0.109*cos30 = 19.6N
The component of the block's weight impeding it's movement on the ramp is
21.2*g*sin30 = 103.9N
Total force impeding the block's movement on ramp = 123.5N
The force pulling the block up the ramp is 28*9.8 = 274.4N
Fnet = 274.4-123.5=150.9N
The total mass mt being accelerated is 21.2+28=49.2kg
The acceleration at t=0 is then Fnet/mt=3.067m/s^2
Immediately after breaking the block loose from the static friction, the kinetic friction force decreases to 21.2*9.8*0.086*cos30=15.5N
The new acceleration will be 3.15m/s^2
d=0.5*a*t^2=0.5*3.15*1.89^2=5.626m at t=1.89s

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