How to find when the voltage gets zeroed

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The discussion focuses on determining the work done by net force between x=0m and x=6m using a provided graph of force versus displacement. The participant initially confused voltage with kinetic energy but clarified that they were working with kinetic energy (Ek). They successfully calculated the work done as 48J and sought guidance on identifying where the velocity reaches zero. It was concluded that the velocity becomes zero at x=18m, after which the force changes. The conversation emphasizes the importance of correctly interpreting the graph and equations related to work and force.
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Homework Statement


Basically, I was given this graph: http://i48.tinypic.com/qrdabs.png
In addition I was asked to find out what's the work of the net force between x=0m and x=6m.

Homework Equations



I'm not sure but I believe this one(?): WƩF=ΔEk

The Attempt at a Solution



Well...as I said I was trying to find it with the equation of WƩF=ΔEk but it doesn't seem to be relevant...
I was also trying to use the graph, yet I have no idea how to reach the answer with it.
Thanks in advance.
 
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Distance said:
Basically, I was given this graph: http://i48.tinypic.com/qrdabs.png
In addition I was asked to find out what's the work of the net force between x=0m and x=6m.
Hi Distance! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

The title of your thread is something about voltage. I see nothing pertaining to voltage in the thread. You have a graph of Force vs. distance and are attempting to determine Work Done.

There is an equation: Work = F ⋅ s
for F constant, and s is distance.

Otherwise, for changing F, use Work = ∫[/size][/color]F ⋅ ds

Homework Equations



I'm not sure but I believe this one(?): WƩF=ΔEk
Did you make this up? I presume Ek is kinetic energy?
 
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I wrote voltage instead of velocity...No idea why. And yes, Ek stands for kinetic energy. Can you/anybody else please help with my question(when does the velocity get zeroed?). Thanks
 


Distance said:

Homework Statement


Basically, I was given this graph: http://i48.tinypic.com/qrdabs.png
In addition I was asked to find out what's the work of the net force between x=0m and x=6m.

Homework Equations



I'm not sure but I believe this one(?): WƩF=ΔEk

The Attempt at a Solution



Well...as I said I was trying to find it with the equation of WƩF=ΔEk but it doesn't seem to be relevant...
I was also trying to use the graph, yet I have no idea how to reach the answer with it.
Thanks in advance.
Is that a graph of the net Force vs. displacement?

If it's not net Force, then you can't do the problem.

First: Answer "what's the work of the net force between x=0m and x=6m."

Then work on the title question ... although you can't answer "when", you can answer "where".
 
Distance said:
I wrote voltage instead of velocity...No idea why. And yes, Ek stands for kinetic energy. Can you/anybody else please help with my question(when does the velocity get zeroed?). Thanks
Is the mass of the body known?
 
SammyS said:
Is that a graph of the net Force vs. displacement?

If it's not net Force, then you can't do the problem.

First: Answer "what's the work of the net force between x=0m and x=6m."

Then work on the title question ... although you can't answer "when", you can answer "where".
Yes, it is a graph of the net Force vs. displacement.
Sorry for not mentioning it, but I already found the work of the net Force(48J).
How can I answer where(I think my teacher was mistaken)?

NascentOxygen said:
Is the mass of the body known?
No, there's another question to find the mass of the body and it's after this question so it's not relevant.
 
Hint: If we assume the body starts from rest at x=0, then from x=0 to x=10 it has been continuously acted on by an accelerating force, so its speed reaches a maximum at x=10.
 
So all I can say is that the speed will be zeroed at x>10m?
Is there an absolute answer?
 
There is a precise numerical answer. :wink:

After 10m, the force undergoes a change...
 
  • #10
Alright I reached to the right answer(x=18m).
Thank you very much!
 
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