Two crates rest on top of one another.There is a crate of mass 4.62 kg on top of another crate of mass 2.19 kg. The coefficient of friction between the lower crate and the floor is μk = 0.440 and the coefficient of static friction between the two crates is μs = 0.870. If the crates start at rest and a force is applied to the right so that both move a distance of 8.70 m, what is the minimal amount of time required in which this can be accomplished without the top crate sliding on the lower crate? (force applies upper crate) THANK YOU

To find the minimal amount of time required for the crates to move without the top crate sliding on the lower crate, we need to determine the maximum acceleration that can be applied to the crates without exceeding the static friction force between them.

1. First, let's find the maximum static friction force between the crates.
The maximum static friction force (Fs) can be calculated using the formula:
Fs = μs * N
where μs is the coefficient of static friction and N is the normal force between the crates.

2. The normal force (N) between the crates can be calculated as the weight of the lower crate, which is equal to its mass times the acceleration due to gravity (g).
N = m * g
where m is the mass of the lower crate and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Given:
Mass of lower crate (m) = 2.19 kg
Acceleration due to gravity (g) = 9.8 m/s^2

N = 2.19 kg * 9.8 m/s^2
N ≈ 21.462 N

3. Now, let's calculate the maximum static friction force (Fs) between the crates.
Fs = μs * N
Given:
Coefficient of static friction (μs) = 0.870

Fs = 0.870 * 21.462 N
Fs ≈ 18.665 N

4. The maximum static friction force (Fs) also represents the maximum force that can be applied to the crates without causing sliding. Therefore, we need to find the value of force (F) that is equal to or less than the maximum static friction force.
Given:
Distance to move (d) = 8.70 m

Force-distance relationship:
F * d = work done
work done = change in kinetic energy
Since the crates start at rest and end at rest, the change in kinetic energy is zero.

F * 8.70 m = 0

Therefore, F must be zero or any value less than the maximum static friction force (Fs) to ensure the top crate does not slide on the lower crate.

5. Since the force (F) is zero, the crates will not experience any external force acting on them. Therefore, they will stay at rest with no motion.

Therefore, the minimal amount of time required to move the crates without the top crate sliding on the lower crate is zero seconds, as they will not move due to no external force being applied.

To find the minimal amount of time required for the two crates to move a distance of 8.70 m without the top crate sliding on the lower crate, we need to consider the forces acting on both crates.

Since the force is applied to the upper crate, let's analyze the forces acting on it. There are two forces acting on the upper crate: the force applied to the right and the force of friction between the two crates.

1. Force applied to the right (F_applied): This force is the driving force that causes the crates to move. The magnitude of this force is not given in the question, so we'll assume it's the minimum force required to move the crates without the top crate sliding on the lower crate.

2. Force of friction between the two crates (F_friction): The force of friction between the two crates can be static or kinetic, depending on whether the top crate is sliding or not. Since we want to prevent sliding, we need to consider static friction.

Now let's do a force analysis on the upper crate:

1. Net horizontal force (F_net) on the upper crate:
F_net = F_applied - F_friction
The net force should be positive to ensure the upper crate moves to the right without sliding.

2. Maximum static friction force (F_static_max):
F_static_max = μs * normal force
The normal force is equal to the weight of the top crate since there is no vertical acceleration.

3. Equation relating F_net and F_static_max:
F_net = F_applied - μs * (mass of upper crate) * (gravitational acceleration)

To find the minimum amount of time required, we need to determine the acceleration of the upper crate. We can use Newton's second law of motion:

F_net = (mass of upper crate) * acceleration

Now let's solve for the acceleration:

F_applied - μs * (mass of upper crate) * (gravitational acceleration) = (mass of upper crate) * acceleration
acceleration = (F_applied - μs * (mass of upper crate) * (gravitational acceleration)) / (mass of upper crate)

Once we have the acceleration, we can use the kinematic equation to find the time required:

distance = initial velocity * time + (1/2) * acceleration * time^2

Since the crates start at rest, the initial velocity is zero, and the equation simplifies to:

distance = (1/2) * acceleration * time^2

Now we can rearrange the equation to solve for time:

time = sqrt((2 * distance) / acceleration)

Substituting the given values, you can calculate the minimal amount of time required for the crates to move without sliding.