How Is Work Calculated on a Conveyor Belt with Friction?

AI Thread Summary
The discussion centers on calculating work done on a suitcase on a conveyor belt with friction. It highlights that work is defined as the product of force and distance, and emphasizes the importance of net force and changes in kinetic energy (KE) when determining work done. The participants clarify that if the suitcase moves at a constant speed, there is no net force acting on it, resulting in zero net work. Static friction and air resistance are acknowledged but considered negligible in this idealized scenario. Ultimately, the conversation reinforces that understanding the relationship between force, motion, and energy is crucial for solving such physics problems.
hello478
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Homework Statement
correct answer is c
Relevant Equations
w=f*d
my answer was c
but i didn't understand why the work done in 16m or 12m wont be the greatest?
how would we calculate work done in 12 m and 16 m??
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hello478 said:
Homework Statement: correct answer is c
Relevant Equations: w=f*d

my answer was c
but i didn't understand why the work done in 16m or 12m wont be the greatest?
how would we calculate work done in 12 m and 16 m??
View attachment 342511
Let’s look at the 16 m section. How do you calculate the work done by the conveyor on the suit case? In other words…is there a force that is carrying it along? Think carefully with respect to the assumptions of the problem (there are no forces like "air resistance" pushing it the opposite direction) and the fact that suitcase moves at constant velocity. It can be a tricky thing to catch, because it is an ideal scenario...something we never experience first hand. However, if you apply Newtons Second Law, you will see what the answer must be.
 
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I take exception to the asinine statement "Resistive force opposing the motion of the suitcase are negligible". We can not neglect static friction which is a "resistive force" in the sense that it adjusts itself to provide the observed acceleration. Specifically, when the suitcase travels down the 12 m incline, static friction is clearly opposite to the direction of motion and, therefore, should qualify as negligible. This problem would not work if the conveyor belt were frictionless.

That was my rant for the day.
 
kuruman said:
I take exception to the asinine statement "Resistive force opposing the motion of the suitcase are negligible". We can not neglect static friction which is a "resistive force" in the sense that it adjusts itself to provide the observed acceleration. Specifically, when the suitcase travels down the 12 m incline, static friction is clearly opposite to the direction of motion and, therefore, should qualify as negligible. This problem would not work if the conveyor belt were frictionless.

That was my rant for the day.
The 12 m ramp is a frictionless slide. However,I completely agree ( but didn’t initially see) that the 8m inclined conveyor belt without friction is problematic!
 
erobz said:
The 12 m ramp is a frictionless slide.
Of course. I confess hasty reading.
hello478 said:
how would we calculate work done in 12 m and 16 m??
Note that the problem is asking for the net work, i.e. the sum of the works done on the suitcase by all the forces acting on it. There is an equation that relates ##~W_{net}~## to ##\dots##
 
hello478 said:
but i didn't understand why the work done in 16m or 12m wont be the greatest?
You have to distinguish between work done on and work done by.
The work done on an object is always the increase in its KE, so in all those stages of constant speed no work is done on it.
This is the same as the work done by the net force, but individual forces may do various amounts of work. E.g. in the first stage, the belt does some amount of work W, but gravity does work -W, for zero net work.

One niggle: on the first ramp, in practice, the case will start from rest, so a bit of work is done on the case.
 
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for the 16m part of the slide, how do i know which force to use, how is the force acting towards the right?
and what about the 8m?
and i dont understand where the newtons second law is it to be used...
 
hello478 said:
for the 16m part of the slide, how do i know which force to use,
As I wrote in post #6, you don't care about the individual forces. For finding the work done on a body you only have to consider its change in KE.
hello478 said:
how is the force acting towards the right?
Who said it is? Looks to me that no force is needed in that stage.
hello478 said:
and i dont understand where the newtons second law is it to be used...
Why do you think that law is to be used?
 
haruspex said:
Why do you think that law is to be used?
Because I told them. I was going to have them figure out why no force would be required in the direction of motion on the 16m part.
 
  • #10
erobz said:
Because I told them. I was going to have them figure out why no force would be required in the direction of motion on the 16m part.
Strange. I did a search for newton in the thread and it didn’t find it; now it does.
 
  • #11
erobz said:
Because I told them. I was going to have them figure out why no force would be required in the direction of motion on the 16m part.
thank you!!!

haruspex said:
As I wrote in post #6, you don't care about the individual forces. For finding the work done on a body you only have to consider its change in KE.

Who said it is? Looks to me that no force is needed in that stage.
ok ill try to think more on it... 🫡
 
  • #12
so, now i know that there is no force acting at 16m
but how will i know about it?
using the second law of motion or the kinetic energy??
and why is there no force acting...?
 
  • #13
hello478 said:
so, now i know that there is no force acting at 16m
You know there is no net force because if there were the case would accelerate.
The instruction to ignore resistive forces is redundant. In practice, the belt would exert a horizontal force in order to overcome air resistance, but since those balance this does not lead to any work done on the case.
 
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  • #14
haruspex said:
if there were the case would accelerate.
how do we know its not accelerating?

haruspex said:
but since those balance this does not lead to any work done on the case.
how can i know if they balance?

if im ignoring resistive forces wouldnt it accelerate?
 
  • #15
hello478 said:
how do we know its not accelerating?
Read options of the test.
 
  • #16
MatinSAR said:
Read options of the test.
but like there is no other hint of it in the question, right?
 
  • #17
hello478 said:
but like there is no other hint of it in the question, right?
Yes. These conditions (constant or changing speed) are stated in options so you can compare them.
 
  • #18
MatinSAR said:
Yes. These conditions (constant or changing speed) are stated in options so you can compare them.
so, in 16m, its moving at a constant speed... which means no change in speed, and no KE
but what about work done?
is it 0? because no net force so 0 work?
 
  • #19
haruspex said:
In practice, the belt would exert a horizontal force in order to overcome air resistance, but since those balance this does not lead to any [net] work done on the case.
I know you meant net work.
 
  • #20
hello478 said:
but like there is no other hint of it in the question, right?
Is it’s velocity changing?
 
  • #21
erobz said:
Is it’s velocity changing?
the question doesnt say it... but the options do...
 
  • #22
hello478 said:
the question doesnt say it... but the options do...
You aren’t reading B carefully.
 
  • #23
erobz said:
You aren’t reading B carefully
yeah it does say constant speed, like i said above the options tell me that... but not the question
so okay, ill consider it now
 
  • #24
haruspex said:
In practice, the belt would exert a horizontal force in order to overcome air resistance, but since those balance this does not lead to any work done on the case.
what does this mean?
please explain in simple words as im no expert in physics... 😅
 
  • #25
hello478 said:
what about work done?
Do you know about Work and Energy theorem? Use it.
 
  • #26
MatinSAR said:
Do you know about Work and Energy theorem? Use it.
that energy = work?
but first tell me if this is correct...
work done is 0 because there is no net force...?

also if there is no net force how can the box move??
 
  • #27
hello478 said:
yeah it does say constant speed, like i said above the options tell me that... but not the question
so okay, ill consider it now
Those give you the conditions to analyze each section. They are part of the problem statement. It’s unanswerable without them.
 
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  • #28
erobz said:
Those give you the conditions to analyze each section. They are part of the problem statement. It’s unanswerable without them.
ok thank you for making it sense to me
so like it sets some limitations so i can solve the problem and its sort of part of the question...
 
  • #29
hello478 said:
also if there is no net force how can the box move??
Because it’s not accelerating. Notice when you ride in the car, do you feel a force pulling you along when you are moving at a constant velocity? It’s not until you accelerate forward that you feel the normal force from the seat pushing you along. Likewise when the brakes are jammed you are thrown forward.
 
  • #30
erobz said:
Because it’s not accelerating. Notice when you ride in the car, do you feel a force pulling you along when you are moving at a constant velocity? It’s not until you accelerate forward that you feel the normal force from the seat pushing you along. Likewise when the brakes are jammed you are thrown forward.
ok and can you please answer this?
hello478 said:
but first tell me if this is correct...
work done is 0 because there is no net force...?
or is it because there is no change in KE?
 
  • #31
hello478 said:
ok and can you please answer this?

or is it because there is no change in KE?
The net work is 0 because there is no change in KE( or visa-versa)
 
  • #32
erobz said:
The net work is 0 because there is no change in KE.
work done is 0 because there is no net force so w=16*0 ... so this explanation is wrong?
 
  • #33
hello478 said:
work done is 0 because there is no net force so w=16*0 ... so this explanation is wrong?
Conceptually It’s important that you say there is no net work done. If there are other forces like air resistance and static friction from the belt that are in balance the there is no net work done on the box, but the air resistance would be doing some work, and the belt( through friction) would be doing work also.

But otherwise yeah, no net force means No acceleration, no acceleration means no net force.
 
  • #34
erobz said:
Conceptually It’s important that you say there is no net work done. If there are other forces like air resistance and static friction from the belt that are in balance the there is no net work done on the box, but the air resistance would be doing some work, and the belt( through friction) would be doing work also.

But otherwise yeah, no net force means No acceleration, no acceleration means no net force.
sorry for bothering too much but the question says to ignore them... so it would be correct, right?
 
  • #35
hello478 said:
sorry for bothering too much but the question says to ignore them... so it would be correct, right?
I just didn’t want you to be confused about this later on.
 
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  • #36
erobz said:
I just didn’t want you to be confused about this later on.
ok thank you soo much! ill keep that in mind

can you also help me with the other options?
for a
work is done because it gains GPE
for c, it gains GPE too
for d it looses gpe and gains KE
 
  • #37
hello478 said:
that energy = work?
but first tell me if this is correct...
work done is 0 because there is no net force...?

also if there is no net force how can the box move??
@erobz Answered them clearly.
 
  • #38
hello478 said:
ok thank you soo much! ill keep that in mind

can you also help me with the other options?
for a
work is done because it gains GPE
for c, it gains GPE too
for d it looses gpe and gains KE
You gotta try to apply the definition we just explained to each case…and tell us what you get. Did you apply the definition or just wing it to get those results…
 
  • #39
hello478 said:
ok thank you soo much! ill keep that in mind

can you also help me with the other options?
for a
work is done because it gains GPE
for c, it gains GPE too
for d it looses gpe and gains KE
For a particle total work is related to change in KE. Not PE.
 
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  • #40
erobz said:
You gotta try to apply the definition we just explained to each case…and tell us what you get. Did you apply the definition or just wing it to get those results…
will get back to you in a while 😶
(might take a few hours)
 
  • #41
MatinSAR said:
For a particle total work is related to change in KE. Not PE.
but isnt PE also energy?
and so energy = work?
 
  • #42
hello478 said:
but isnt PE also energy?
and so energy = work?
Read option a again. Think of work and energy theorem. Do you think work done on the object is nonezero?
 
  • #43
MatinSAR said:
Read option a again. Think of work and energy theorem. Do you think work done on the object is nonezero?
will do and let you know... :)
 
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  • #44
hello478 said:
will do an let you know... :)
Are you working out of a textbook?
 
  • #45
hello478 said:
but isnt PE also energy?
and so energy = work?
Work tells you how much energy changes:$$\Delta E = F_\text{net} \times D$$where ##D## is the distance moved in the direction of the net force.

You can count gravitational potential energy on the energy side. But then you need to remove gravity from consideration on the force side:$$\Delta (\text{GPE} + \text{KE}) = F_\text{non-g} \times D$$
Or you can treat gravity as just another force. In this case you need to remove gravitational potential energy from the energy side:$$\Delta \text{KE} = F_\text{net} \times D$$
In my opinion, this problem assumes the latter approach. We are treating gravity as just another force. A force that can contribute to the net work done.
 
  • #46
MatinSAR said:
Read option an again. Think of work and energy theorem. Do you think work done on the object is nonezero?
There is a fundamental problem with “A” because it is a frictionless belt…probably best just skip it for the time?
 
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  • #47
erobz said:
There is a fundamental problem with “A” because it is a frictionless belt…probably best just skip it for the time?
yeah we can address it later on
 
  • #48
erobz said:
Are you working out of a textbook?
no, using google and the previous explanations you and everyone gave...
but as im working on many questions simultaneously so it will take me a few hours to figure out the answer and reply
 
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  • #49
erobz said:
There is a fundamental problem with “A” because it is a frictionless belt…probably best just skip it for the time?
The problem says no "resistive" forces. A forward force from the conveyor belt is not resisting the motion. It is assisting it.
 
  • #50
jbriggs444 said:
The problem says no "resistive" forces. A forward force from the conveyor belt is not resisting the motion. It is assisting it.
Well it was pointed out early ( by @kuruman , and maybe @haruspex) that at the very least the wording is unclear. On the incline it can be argued that as it climbs the ramp friction must be there to resist the boxes component of weight down the ramp.

I don’t think this is the intention, but it seems like a point of contention. We should probably abandon it for the sake of the OP and say they didn’t mean frictionless belt.
 
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