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Boxes are moved on a converyor blet from where they are filled to the packing station 10m away. The belt is initially stationary and must finish with zero speed. The most rapid transit is accomplished if the belt accerlates for half the distance, then decelerates for the final half of the trip. If the coefficent of static friction between a box and the belt is .60, what is the mininum transit time for each box?
I have no idea what it's talking about or how many boxes there are so I just assumed one...
acceleration first half of trip
i got 5.880 s^-2 m
time for first half
i got 4.124 s
velocity at end of first half
i got 7.668 s^-1 m
acceleration for second half
i got - 6.380 s^-2 m
time for second half
i got 1.202 s
total time
i got 5.3 s
This problem was kind of more involved and I got 5.3 s and listed all of the other variables I solved for in order to get that answer... Does it look right???
To solve this problem, you correctly assumed that there is one box being moved on the conveyor belt. Let's go through the steps to find the minimum transit time for each box:
1. Determine the acceleration for the first half of the trip:
- Since the belt starts from rest and finishes with zero speed, the average velocity is half the final velocity.
- The distance covered in the first half is half of the total distance, so it would be 10m/2 = 5m.
- Using the equation v^2 = u^2 + 2as, we can find the acceleration (a) for the first half, where u is the initial velocity (0) and v is the final velocity (unknown):
v^2 = 0^2 + 2a(5)
v^2 = 10a
- Now, we can consider the frictional force acting on the box. The maximum frictional force is given by μN, where μ is the coefficient of static friction and N is the normal force.
- The normal force N is equal to the weight of the box, which can be calculated as mg, where m is the mass of the box and g is the acceleration due to gravity (9.8 m/s^2).
- The maximum frictional force can be written as μmg. Since the box is moving horizontally, the frictional force should be equal and opposite to the force applied to move the box.
- So, we have μmg = ma. Rearranging, we can find the acceleration:
a = μg = 0.6 * 9.8 = 5.88 m/s^2.
2. Find the time taken for the first half:
- We can use the equation v = u + at, where u is the initial velocity (0), a is the acceleration (5.88 m/s^2), and v is the final velocity (unknown):
v = 0 + 5.88t
- Solving for t, we get t = v / 5.88.
- Since the final velocity at the end of the first half is the average velocity for the entire first half, it is half of the final velocity for the whole trip (0 at the end of the trip).
- Thus, the time taken for the first half is t = 0 / 5.88 = 0s.
3. Calculate the velocity at the end of the first half:
- We know that v = u + at, where v is the final velocity (unknown), u is the initial velocity (0), a is the acceleration (5.88 m/s^2), and t is the time (0s).
- Substituting the known values, we get v = 0 + 5.88 * 0 = 0 m/s.
4. Determine the acceleration for the second half of the trip:
- Again, we want the average velocity for the second half to be zero at the end, so the final velocity (0 m/s) is equal to the average velocity.
- The distance covered in the second half is also 5m.
- Using the equation v^2 = u^2 + 2as, we can find the acceleration (a) for the second half, where u is the initial velocity (0 m/s) and v is the final velocity (0 m/s):
0^2 = 0^2 + 2a(5)
0 = 10a
- Solving for a, we get a = 0 m/s^2.
5. Find the time taken for the second half:
- We can use the equation v = u + at, where u is the initial velocity (0 m/s), a is the acceleration (0 m/s^2), and v is the final velocity (0 m/s):
0 = 0 + 0t
- Solving for t, we get t = 0s.
6. Calculate the total time for each box:
- The total time is the sum of the times taken for the first half and the second half.
- Total time = 0s + 0s = 0s.
Based on the calculations, it seems that you made an error somewhere. The minimum transit time for each box is actually 0 seconds, which means there is no time taken to move each box from the filling station to the packing station.