A block of mass m1 = 21.3 kg is at rest on a plane inclined by θ = 30.00 degrees. It is connected via a rope and pulley system to another mass m2 = 24.5 kg, as shown. The coefficients of static and kinetic friction between block 1 and the inclined plane are μs = 0.113 and μk = 0.09, respectively. If the system is released from rest, what is the displacement of block 2 in vertical direction after 1.41 s? Use positive numbers for upward direction and negative numbers for downward.

To find the displacement of block 2 in the vertical direction after 1.41 seconds, we need to analyze the forces acting on the system and calculate the net force.

First, let's draw a free-body diagram for each object:

For block 1:
- Weight downward (mg1)
- Normal force perpendicular to the inclined plane (N1)
- Friction force opposing motion up the incline (f1)
- Tension force in the rope (T)

For block 2:
- Weight downward (mg2)
- Tension force in the rope (T)

Now, let's break down the forces along the x-axis and y-axis:

Along the x-axis:
For block 1: Tension force (T) down the incline and friction force (f1) up the incline.
For block 2: No forces along the x-axis.

Along the y-axis:
For block 1: Weight (mg1) downward, normal force (N1) perpendicular to the incline.
For block 2: Weight (mg2) downward.

In this case, we are interested in the vertical displacement of block 2, which means we need to calculate the net force acting on it along the y-axis.

The net force on block 2 is given by:

Net force on block 2 = Weight of block 2 - Tension force

The weight of block 2 is calculated by:

Weight of block 2 = m2 * g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Next, we need to calculate the tension force.

Since the two blocks are connected by a rope, they must have the same tension force. Therefore, we can use the tension force acting on block 1 to calculate the tension force acting on block 2.

The tension force is given by:

Tension force = T

To find the tension force, we need to consider the forces acting on block 1.

The net force on block 1 is given by:

Net force on block 1 = Tension force - Friction force - Weight of block 1 * sin(θ)

The friction force is given by:

Friction force = μs * N1

where μs is the coefficient of static friction and N1 is the normal force.

The normal force is given by:

Normal force = Weight of block 1 * cos(θ)

Finally, we can substitute these values into the net force equation for block 1 and solve for the tension force:

Net force on block 1 = Tension force - μs * N1 - Weight of block 1 * sin(θ) = 0 (since block 1 is at rest)

Now, we can substitute the tension force value into the net force equation for block 2:

Net force on block 2 = Weight of block 2 - Tension force

Finally, using Newton's second law (F = ma), we can calculate the acceleration of block 2 by dividing the net force by its mass:

Acceleration of block 2 = Net force on block 2 / m2

Now, we can calculate the displacement of block 2 in the vertical direction after 1.41 seconds using the kinematic equation:

Displacement of block 2 = initial velocity * time + (1 / 2) * acceleration * time^2

Since the system is released from rest, the initial velocity is 0. Therefore, the displacement simplifies to:

Displacement of block 2 = (1 / 2) * acceleration * time^2

Substitute the values into this equation and solve for the displacement of block 2.