Boxes are moved on a conveyor belt from where they are filled to the packing station 12.0 m away. The belt is initially stationary and must finish with zero speed. The most rapid transit is accomplished if the belt accelerates for half the distance, then decelerates for the final half of the trip. If the coefficient of static friction between a box and the belt is 0.58 what is the minimum transit time for each box?

To find the minimum transit time for each box, we need to calculate the acceleration of the belt and then determine the time taken for half the distance during the acceleration phase and the time taken for half the distance during the deceleration phase.

First, let's calculate the acceleration of the belt. We'll assume the maximum static friction force is used to accelerate and decelerate the box.

The static friction force (F) between the box and the belt can be expressed as:

F = μN

Where:
μ is the coefficient of static friction (given as 0.58)
N is the normal force between the box and the belt.

Since the box is on a horizontal surface, the normal force is equal to the weight of the box (mg), where m is the mass of the box and g is the acceleration due to gravity.

Next, let's divide the total distance (12.0 m) into two halves. The distance covered during each phase (acceleration and deceleration) will be 6.0 m.

During the acceleration phase, the maximum static friction force will be applied to accelerate the box. The net force applied to the box is the difference between the maximum static friction force and the force due to the weight of the box:

F_net = F - mg

Using Newton's second law (F_net = ma), we can write:

ma = F - mg

Rearranging the equation, we get:

a = (F - mg) / m

Now, let's substitute the values into the equation:

a = (0.58N - mg) / m

During the deceleration phase, the maximum static friction force will be applied in the opposite direction to decelerate the box. So, the net force applied to the box is:

F_net = -F - mg

Using Newton's second law (F_net = ma), we can write:

ma = -F - mg

Rearranging the equation, we get:

a = (-F - mg) / m

Now, let's substitute the values into the equation:

a = (-0.58N - mg) / m

Since both the acceleration and deceleration phases cover the same distance (6.0 m), we can use the average velocity formula to find the minimum transit time (t_min):

t_min = 2 * (distance / average velocity)

The average velocity (v_avg) can be calculated as:

v_avg = (initial velocity + final velocity) / 2

In this case, the initial velocity is 0 m/s, and the final velocity is also 0 m/s (to finish with zero speed).

Now, let's go step-by-step to calculate the minimum transit time for each box:

Step 1: Calculate the acceleration (a)
- Determine the maximum static friction force (F) using F = μN
- Calculate the normal force (N) as N = mg
- Substitute the values into the equation a = (F - mg) / m

Step 2: Calculate the time taken for the acceleration phase (t_acc)
- Use the equation v = u + at, where v = final velocity, u = initial velocity, a = acceleration, and t = time
- Substitute the values v = 0 m/s, u = 0 m/s, and a from step 1 into the equation
- Solve for t_acc

Step 3: Calculate the time taken for the deceleration phase (t_dec)
- Use the equation v = u + at, where v = final velocity, u = initial velocity, a = acceleration, and t = time
- Substitute the values v = 0 m/s, u = 0 m/s, and a from step 1 into the equation
- Solve for t_dec

Step 4: Calculate the minimum transit time (t_min)
- Substitute the values into the equation t_min = 2 * (distance / average velocity), where distance = 12.0 m and average velocity = (initial velocity + final velocity) / 2

Let's perform these calculations step-by-step:

Step 1: Calculate the acceleration (a)

F = μN
N = mg

a = (F - mg) / m

Step 2: Calculate the time taken for the acceleration phase (t_acc)

v = u + at

u = 0 m/s (initial velocity)
a = acceleration (from Step 1)
v = 0 m/s (final velocity)

t_acc = (v - u) / a

Step 3: Calculate the time taken for the deceleration phase (t_dec)

v = u + at

u = 0 m/s (initial velocity)
a = acceleration (from Step 1)
v = 0 m/s (final velocity)

t_dec = (v - u) / a

Step 4: Calculate the minimum transit time (t_min)

t_min = 2 * (distance / average velocity)

average velocity = (initial velocity + final velocity) / 2

Let's substitute the values and calculate each step.

To determine the minimum transit time for each box, we need to calculate the acceleration required for half the distance and then use it to find the time taken to cover the full distance.

Let's break down the problem step by step:

Step 1: Calculate the maximum acceleration.
Since the belt accelerates for half the distance and decelerates for the other half, the total displacement is 12.0 m. So, the distance covered during acceleration or deceleration is half of that, which is 6.0 m.

The maximum acceleration can be calculated using the formula:
a = (vf^2 - vi^2) / (2 * d)
where a is acceleration, vf is final velocity, vi is initial velocity, and d is distance.

Here, the initial velocity (vi) is 0 m/s.
The final velocity (vf) can be calculated using the equation of motion:
vf^2 = vi^2 + 2ad
Since we want the belt to finish with zero speed, vf is also 0 m/s.

Substituting the values in the formula, we get:
0 = 0^2 + 2 * a * 6
0 = 0 + 12a
12a = 0
a = 0 m/s^2

This means the belt needs to accelerate and decelerate at 0 m/s^2, which effectively means it should move at constant speed.

Step 2: Calculate the transit time.
Since the belt is moving at a constant speed, we can calculate the time taken to cover the full distance using the formula:
t = d / v
where t is time, d is distance, and v is velocity.

Here, the distance (d) is 12.0 m.

To calculate the velocity (v), we need to consider the friction force provided by the belt. The maximum static friction force can be calculated using the formula:
fs_max = μs * N
where fs_max is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force.

In this scenario, the normal force is the weight of the box, which is balanced by the vertical force exerted by the conveyor belt. So, N can be calculated as the weight of the box:
N = mg
where m is the mass of the box and g is the acceleration due to gravity.

Using the maximum static friction force, we can find the maximum tension in the belt, which is equal to the force required to accelerate the box:
T = fs_max

Substituting all the values and rearranging the equation, we can calculate the velocity (v):
T = μs * N
T = μs * mg
v = T / m = (μs * mg) / m = μs * g

Here, the mass of the box (m) and the acceleration due to gravity (g) cancel out, resulting in the velocity (v) being dependent only on the coefficient of static friction (μs).

Finally, substituting the values in the time equation, we get:
t = d / v = 12.0 / (μs * g)

Now, we can calculate the minimum transit time for each box using the given coefficient of static friction (μs = 0.58):
t = 12.0 / (0.58 * 9.8) ≈ 2.27 seconds

Therefore, the minimum transit time for each box is approximately 2.27 seconds.