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

  • Thread starter Thread starter sodr2
  • Start date Start date
  • Tags Tags
    Energy Forces
AI Thread Summary
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.
sodr2
Messages
26
Reaction score
0
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?
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
Last edited by a moderator:
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
 
Last edited:
https://www.physicsforums.com/latex_images/24/2483850-2.png
whats that big s line thing for lol
 
Last edited by a moderator:

Thread 'Rolling without slipping on a curved surface'

Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
View full post »
Thread 'Correct statement about a reservoir with an outlet pipe'

Thread 'Correct statement about a reservoir with an outlet pipe'

The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
View full post »
Thread 'A bead-mass oscillatory system problem'

Thread 'A bead-mass oscillatory system problem'

I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
View full post »

Similar threads

Spring deflection problem using energy equations
Replies
2
Views
2K
Question about the conservation of mechanical energy
Replies
7
Views
2K
Calculating Work Done by a Spring
Replies
2
Views
4K
Energy conservation equation to find equation for final velocity
Replies
4
Views
1K
Why is the height of the table not necessary to solve this problem?
Replies
6
Views
2K
Spring with mass hangs from the ceiling
Replies
22
Views
2K
What is the meaning of the intercept of a particular solution to damped oscillator?
Replies
7
Views
930
Finding Spring Constant & Energy w/ Doubt in Exercise
Replies
7
Views
2K
Energy stored in a double spring system
Replies
17
Views
1K
Expression for magnitude of the constant friction force
Replies
7
Views
1K

Hot Threads

Recent Insights

Back
Top