Why is the formula for potential energy in a spring 1/2(kd^2) instead of kd^2?

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The formula for potential energy in a spring is E = 1/2(kd^2) because it represents the area under the force vs. displacement graph, which forms a triangle, not a rectangle. The force exerted by the spring, according to Hooke's law, is F = kx, and work done (W) is calculated as W = Fd. Since the graph is triangular, the area (work done) is calculated as 1/2(base)(height), leading to the factor of 1/2 in the potential energy equation. The discussion clarifies that while the formula E = kd^2 might seem plausible, it does not account for the nature of the force's change over distance. Understanding the integral of the force function is essential for deriving the correct potential energy formula.
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I just watched this http://www.youtube.com/watch?v=eVl5zs6Lqy0" video...

If W = KE ---> Fd = [mv^2] / 2

If in a spring, force is equal to a constant (k) times how far its been stretched (d), then subbing kd for F in the first equation...

Kd^2 = [mv^2]/2
Rearranging, E = kd^2

But in my textbook, it says that E = 1/2 [kd^2]
Where did the 1/2 come from? Even in the video, W = Fd, so from the graph, you times F by d, getting a SQUARE, not a triangle, but he says you times F by d, and divide by 2 since your looking at the area (triangle) underneath...

Any help?
 
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sodr2 said:
I just watched this http://www.youtube.com/watch?v=eVl5zs6Lqy0" video...

If W = KE ---> Fd = [mv^2] / 2

If in a spring, force is equal to a constant (k) times how far its been stretched (d), then subbing kd for F in the first equation...

Kd^2 = [mv^2]/2
Rearranging, E = kd^2

But in my textbook, it says that E = 1/2 [kd^2]
Where did the 1/2 come from? Even in the video, W = Fd, so from the graph, you times F by d, getting a SQUARE, not a triangle, but he says you times F by d, and divide by 2 since your looking at the area (triangle) underneath...

Any help?

U = \frac{1}2kx^2

is the potential energy of the spring. You're talking about the kinetic energy.
 
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Okay...

If on a F vs x graph, you want to find work...

F = k x
W = F d

Then shouldn't work = k xd = kx^2?
 
sodr2 said:
Okay...

If on a F vs x graph, you want to find work...

F = k x
W = F d

Then shouldn't work = k xd = kx^2?

Yes...but the net work is only equal to the change in kinetic energy. What your book is saying (spring's potential energy) has nothing to do with what you're saying. For a conservative force (ie. only a position-dependent force) like Hooke's law, the force will equal the negative gradient of the potential. So now we have:

F = -kx

Take integral and get:

U = \frac{1}2kx^2
 
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whats that big s line thing for lol
 
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