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Work/Energy graph of car problem? 
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#1
Oct1111, 10:11 PM

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1. The problem statement, all variables and given/known data
A force is applied to a 3.5 kg radiocontrolled model car parallel to the xaxis as it moves along a straight track. The xcomponent of the force varies with the xcoordinate of the car as shown in figure .Suppose the model car is initially at rest at x=0 and vector F is the net force acting on it. Use the workenergy theorem to find the speed of the car at x=3.0m 2. Relevant equations [tex]W_{tot} = \frac{1}{2}mv_{f}^{2}\frac{1}{2}mv_{i}^{2}[/tex] [tex]F = ma[/tex] 3. The attempt at a solution At x=3 is the force 0 or 2? I have no clue how to approach this one. 


#2
Oct1111, 10:38 PM

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Well, the force is 2 N an instant before 3 seconds, and no force the instant after 3 seconds (the force stops being applied at t = 3 s). You have to use the definition of work to find the work done between 0 and 3 seconds (HINT: look at the area under the curve...what does it represent)?



#3
Oct1111, 10:39 PM

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Hi QuarkCharmer,
At x = 3.0 m, the force is both 2 N and 0 N, because it instantaneously changes from one value to another. This is an unphysical situation. In reality, the force would take a finite amount of time to change value, and the line at x = 3.0, instead of being perfectly vertical, would have some slope to it. However, it doesn't matter that they've given you a physically unrealistic force vs. position function, because you can still solve the problem. In order to use the workenergy theorem as you've been instructed, you need to first find the work done up to x = 3.0 m. How would you do that? Hint: what is the relationship between work, force, and distance? 


#4
Oct1111, 10:54 PM

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Work/Energy graph of car problem?
Edit: The area under the curve from 0 to 3 is 4. It represents Work.



#5
Oct1111, 10:59 PM

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Now can i say that:
4 = .5(3.5)(v^2)  .5(3.5)(0) 4 = .5(3.5)(v^2) v = 1.51186 


#6
Oct1111, 11:03 PM

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Okay, that works out. I see now.
How would I find the speed of the thing at x = 7? How does the negative area (under the axis) effect the work? 


#7
Oct1111, 11:07 PM

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#8
Oct1111, 11:12 PM

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That's what my intuition said too. I was just not sure if it was like one of those "areas bound by some functions" style problem, where you need the abs() of each part and all that to get a total. Now it sounds silly. Of course it's negative, since I can run 10 miles north and 10 miles south and do no work.
Thanks guys. 


#9
Oct1211, 05:34 AM

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