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

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SUMMARY

The formula for potential energy in a spring is given by U = 1/2 kx², where k is the spring constant and x is the displacement from the equilibrium position. This factor of 1/2 arises from the integration of the force exerted by the spring, which follows Hooke's Law (F = kx). The work done on the spring is calculated as the area under the force vs. displacement graph, which forms a triangle, hence the division by 2. Understanding this relationship clarifies the distinction between potential energy and kinetic energy in spring systems.

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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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https://www.physicsforums.com/latex_images/24/2483850-2.png
whats that big s line thing for lol
 
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