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Work/Energy graph of car problem?

by QuarkCharmer
Tags: graph, work or energy
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QuarkCharmer
#1
Oct11-11, 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 radio-controlled model car parallel to the x-axis as it moves along a straight track. The x-component of the force varies with the x-coordinate 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 work-energy 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.
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PhanthomJay
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Oct11-11, 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)?
cepheid
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Oct11-11, 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 work-energy 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?

QuarkCharmer
#4
Oct11-11, 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.
QuarkCharmer
#5
Oct11-11, 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
QuarkCharmer
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Oct11-11, 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?
cepheid
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Oct11-11, 11:07 PM
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Quote Quote by QuarkCharmer View Post
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?
Just do the math. If the area is negative, the work done is negative. So the total area under the curve will actually be less than what it was up to x = 3.0 m.
QuarkCharmer
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Oct11-11, 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.
PhanthomJay
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Oct12-11, 05:34 AM
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Quote Quote by QuarkCharmer View Post
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.
Although you did not show your calculations for the work done and the speed of the thing at x = 7 m, if you followed cepheid's advice, you should have arrived at the correct answer. However, your analogy
since I can run 10 miles north and 10 miles south and do no work
is not a good one. If a force is applied to an object in the direction of the displacement, the work done is positive.


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