The forum discussion centers on calculating work done on a suitcase on a conveyor belt with friction. The correct answer to the posed problem is option C, indicating that the work done is zero due to the absence of net force when the suitcase moves at constant velocity. Key equations referenced include the work-energy theorem, which states that net work is equal to the change in kinetic energy (KE). The discussion emphasizes the importance of considering resistive forces, such as static friction, and clarifies that work done is not solely dependent on distance but also on the net forces acting on the object.
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The net work is 0 because there is no change in KE( or visa-versa)hello478 said:
work done is 0 because there is no net force so w=16*0 ... so this explanation is wrong?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.hello478 said:
sorry for bothering too much but the question says to ignore them... so it would be correct, right?erobz said:
I just didn’t want you to be confused about this later on.hello478 said:
ok thank you soo much! ill keep that in minderobz said:
@erobz Answered them clearly.hello478 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…hello478 said:
For a particle total work is related to change in KE. Not PE.hello478 said:
will get back to you in a whileerobz said:
but isnt PE also energy?MatinSAR said:
Read option a again. Think of work and energy theorem. Do you think work done on the object is nonezero?hello478 said:
will do and let you know... :)MatinSAR said:
Are you working out of a textbook?hello478 said:
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.hello478 said:
There is a fundamental problem with “A” because it is a frictionless belt…probably best just skip it for the time?MatinSAR said:
yeah we can address it later onerobz said:
no, using google and the previous explanations you and everyone gave...erobz said:
The problem says no "resistive" forces. A forward force from the conveyor belt is not resisting the motion. It is assisting it.erobz said:
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.jbriggs444 said:
Read post #6 again.hello478 said:
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 said:
I agree, but I think they meant there is friction between the belt and the box, and there is no “drag” type forces.haruspex said:
Oh yes, that is what was meant, no question.erobz said: