How Is Work Calculated on a Conveyor Belt with Friction?

[erobz]

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)

 

[hello478]

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?

 

[erobz]

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.

 

[hello478]

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?

 

[erobz]

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.

 

[hello478]

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

 

[MatinSAR]

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.

 

[erobz]

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…

 

[MatinSAR]

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.

 

[hello478]

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)

 

[hello478]

For a particle total work is related to change in KE. Not PE.

but isnt PE also energy?
and so energy = work?

 

[MatinSAR]

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?

 

[hello478]

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... :)

 

[erobz]

will do an let you know... :)

Are you working out of a textbook?

 

[jbriggs444]

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.

 

[erobz]

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?

 

[hello478]

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

 

[hello478]

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

 

[jbriggs444]

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.

 

[erobz]

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.

 

[haruspex]

for a work is done because it gains GPE

Read post #6 again.
In a, work mgh is done by the belt and work -mgh is done by gravity. But the question asks about work done on the case: mgh+(-mgh)=0.

 

[haruspex]

they didn’t mean frictionless belt

and, indeed, did not say frictionless belt. They wrote no "resistive forces opposing motion", which is ambiguous. The motion could mean the actual velocity of the case up the ramp (the friction does not oppose that) or it could be taken as the motion that would occur were there no friction: acceleration down the ramp.

 

[erobz]

and, indeed, did not say frictionless belt. They wrote no "resistive forces opposing motion", which is ambiguous. The motion could mean the actual velocity of the case up the ramp (the friction does not oppose that) or it could be taken as the motion that would occur were there no friction: acceleration down the ramp.

I agree, but I think they meant there is friction between the belt and the box, and there is no “drag” type forces.

 

[haruspex]

I agree, but I think they meant there is friction between the belt and the box, and there is no “drag” type forces.

Oh yes, that is what was meant, no question.