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 determine if the block will slide down the inclined plane or remain at rest.
First, let's calculate the gravitational force acting on each block:
Gravity of block 1 = m1 * g
Gravity of block 2 = m2 * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Next, let's calculate the force of static friction between block 1 and the inclined plane:
Fs = μs * (m1 * g * cos(θ))
where θ is the angle of the inclination of the plane.
If the force of static friction is greater than or equal to the gravitational force of block 2, block 2 will remain at rest. Otherwise, block 2 will slide down the inclined plane.
Now, let's calculate the displacement of block 2 after 1.41 seconds using the equations of motion:
For a freely falling object, displacement is given by:
d = 0.5 * a * t^2 + v0 * t
where d is the displacement, a is the acceleration, t is the time, and v0 is the initial velocity.
For block 2 sliding down the inclined plane, the acceleration is given by:
a = g * sin(θ) - μk * g * cos(θ)
If block 2 remains at rest, the displacement will be 0.
Now let's work through the calculations with the given values:
m1 = 21.3 kg (mass of block 1)
m2 = 24.5 kg (mass of block 2)
θ = 30.00 degrees (angle of the inclined plane)
μs = 0.113 (coefficient of static friction)
μk = 0.09 (coefficient of kinetic friction)
t = 1.41 s (time)
First, calculate the gravitational forces:
Gravity of block 1 = m1 * g = 21.3 kg * 9.8 m/s^2
Gravity of block 2 = m2 * g = 24.5 kg * 9.8 m/s^2
Next, calculate the force of static friction:
Fs = μs * (m1 * g * cos(θ)) = 0.113 * (21.3 kg * 9.8 m/s^2 * cos(30.00))
Compare the force of static friction to the gravitational force of block 2. If the force of static friction is greater than or equal to the gravitational force of block 2, block 2 will remain at rest. Otherwise, block 2 will slide down the inclined plane.
If block 2 remains at rest, the displacement will be 0.
If block 2 slides down the inclined plane, we can now calculate the acceleration using:
a = g * sin(θ) - μk * g * cos(θ) = 9.8 m/s^2 * sin(30.00) - 0.09 * 9.8 m/s^2 * cos(30.00)
Finally, we can calculate the displacement:
d = 0.5 * a * t^2 + v0 * t
Since the initial velocity of block 2 is 0 as it starts from rest, the equation simplifies to:
d = 0.5 * a * t^2
Plug in the values for acceleration (a) and time (t) to find the displacement (d). Remember to use positive numbers for upward direction and negative numbers for downward.