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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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.hello478 said:
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!kuruman said:
Of course. I confess hasty reading.erobz said:
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:
You have to distinguish between work done on and work done by.hello478 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.hello478 said:
Who said it is? Looks to me that no force is needed in that stage.hello478 said:
Why do you think that law is to be used?hello478 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.haruspex said:
Strange. I did a search for newton in the thread and it didn’t find it; now it does.erobz said:
thank you!!!erobz said:
ok ill try to think more on it...haruspex said:
You know there is no net force because if there were the case would accelerate.hello478 said:
how do we know its not accelerating?haruspex said:
how can i know if they balance?haruspex said:
Read options of the test.hello478 said:
but like there is no other hint of it in the question, right?MatinSAR said:
Yes. These conditions (constant or changing speed) are stated in options so you can compare them.hello478 said:
so, in 16m, its moving at a constant speed... which means no change in speed, and no KEMatinSAR said:
I know you meant net work.haruspex said:
Is it’s velocity changing?hello478 said:
the question doesnt say it... but the options do...erobz said:
You aren’t reading B carefully.hello478 said:
yeah it does say constant speed, like i said above the options tell me that... but not the questionerobz said:
what does this mean?haruspex said:
Do you know about Work and Energy theorem? Use it.hello478 said:
that energy = work?MatinSAR said:
Those give you the conditions to analyze each section. They are part of the problem statement. It’s unanswerable without them.hello478 said:
ok thank you for making it sense to meerobz 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.hello478 said:
ok and can you please answer this?erobz said:
or is it because there is no change in KE?hello478 said: