Package Dropped On A Conveyor Belt

In summary, the problem involves a package with a mass of 4m being dropped on a conveyor belt with a speed of 3v and a coefficient of kinetic friction of 2k. The questions ask for the time it takes for the package to stop sliding, the displacement of the package during this time, the energy dissipated by friction, and the work done by friction. The attempts at solving these problems involve using the equations F = m*a and W = F*d, and plugging in values for the given variables. However, since the variables are not given with numerical values, the results are left as symbolic expressions.
  • #1
RandyRanderson
1
0

Homework Statement


A package is dropped on a horizontal conveyor belt. The mass of the package is 4m , the speed of the conveyor belt is 3v , and the coefficient of kinetic friction between the package and the belt is 2k .

a) How long does it take for the package to stop sliding on the belt?

b) What is the package’s displacement during this time?

c) What is the energy dissipated by friction?

d) What is the work done by friction on the box?

Homework Equations



F = m*a
W = F*d

The Attempt at a Solution



I got all these problems wrong, and honestly I have no clue what I am doing with this problem. Here are my answers:

a) v/(26.16*k)
basically I calculated this by force of friction = (2k)(4m)(9.81)
then with F = ma
78.48mk = ma
a = 78.48k

So I used v = at
3v = 78.48kt
t = 3v/78.48k
t = v/26.16k

b) (17.44*(v^2))/(k)

basically I calculated this by force of friction = (2k)(4m)(9.81)
then with F = ma
78.48mk = ma
a = 78.48 k
Then d = .5at2
d = .5(78.48k)(v/26.16k)2

c) 1368.7*m*(v^2)

For this I used Ff*d

So Ff = 78.48mk
d = .5(78.48k)(v/26.16k)2

d) 1368.7*m*(v^2)

This is the same as c, right?
 
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  • #2
It seems like all the results are going to be symbolic since you don't have numerical values for m,v,k. Why bother plugging in a numerical value for g? Why not just leave it as 'g'?
 

1. How does a package dropped on a conveyor belt move?

When a package is dropped on a conveyor belt, it moves due to the rotation of the belt. The belt is powered by a motor, causing it to move in a continuous loop. The package is carried along with the belt until it reaches the end or is removed by a worker.

2. How fast does a conveyor belt move?

The speed of a conveyor belt can vary depending on the type of belt and the intended purpose. In general, most conveyor belts have a speed range of 0.5 meters per second to 2.5 meters per second. However, some specialized belts can reach speeds of up to 5 meters per second.

3. Can packages get stuck on a conveyor belt?

Yes, packages can get stuck on a conveyor belt due to various reasons. The package may be too large or oddly shaped, causing it to get caught on the edges of the belt. It can also get stuck if the belt is damaged or if there is debris or objects blocking its path. Regular maintenance and proper operation can help prevent packages from getting stuck on a conveyor belt.

4. How does a conveyor belt handle fragile packages?

Conveyor belts can handle fragile packages by using specific belt materials and features. For example, belts with a non-slip surface can prevent packages from sliding and getting damaged. Some belts also have a cushioning layer or rollers to absorb impact and reduce the chances of breakage. Additionally, conveyor belts can be programmed to move at a slower speed when carrying delicate packages.

5. Are there safety precautions for working with conveyor belts?

Yes, there are safety precautions to follow when working with conveyor belts. Workers should always wear appropriate personal protective equipment, such as gloves and safety glasses, to prevent injuries. It is also essential to follow proper training and guidelines when operating the belt and to regularly check for any malfunctions or hazards. Additionally, workers should never try to remove stuck packages or debris while the belt is running.

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