Question on forces and energy

26
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?
 
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52
1
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?
[tex] U = \frac{1}2kx^2[/tex]

is the potential energy of the spring. You're talking about the kinetic energy.
 
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26
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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?
 
52
1
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:

[tex] F = -kx[/tex]

Take integral and get:

[tex]U = \frac{1}2kx^2[/tex]
 
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26
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https://www.physicsforums.com/latex_images/24/2483850-2.png [Broken]
whats that big s line thing for lol
 
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