# Question on forces and energy

1. Dec 9, 2009

### sodr2

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: Apr 24, 2017
2. Dec 9, 2009

### Cryxic

$$U = \frac{1}2kx^2$$

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

Last edited by a moderator: Apr 24, 2017
3. Dec 9, 2009

### sodr2

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?

4. Dec 9, 2009

### Cryxic

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: Dec 9, 2009
5. Dec 9, 2009

### sodr2

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

Last edited by a moderator: May 4, 2017