Why x^2 for PE?

1. Jul 29, 2013

JFS321

Hi all,

I notice the patterns such as v = 0.5at^2 and PE(spring)=0.5kx^2, etc...but, all examples I have seen show the slope (acceleration, or the spring constant, k) as being a 45 degree angle. Thus, the area of the triangle underneath the graph makes good sense (x^2 or t^2).

But, let's say we have an example where the spring constant is much larger or much smaller. Why is x^2 still valid as the base x height, if the two are not equal values?

2. Jul 29, 2013

DrewD

Maybe I'm just tired, but I'm really confused about your question. So that people don't start answering every question except what you meant, would you mind clarifying? (also, x, not v in your first one)
What do you mean a or k is a 45 degree angle?
I'm guessing that the answer you are looking for is going involve integration. What level of math are you comfortable with?

3. Jul 29, 2013

JFS321

Sorry for the confusion. I'm evidently tired too.

I have it answered. All it took was writing PE = 0.5 k(x) times x. I couldn't intuitively see the x^2 mentally.

4. Jul 29, 2013

Staff: Mentor

Those terms come from calculus derivations. Are you familiar yet with differential and integral calculus?

5. Jul 29, 2013

PhanthomJay

that is s = 1/2at^2
when you plot v vs.t, and the acceleration is constant, then you have a linear equation v = at, and the slope of the line is the acceleration. Thus, the area of the triangle underneath the graph is 1/2 at^2. Or since F=kx, the slope of the line is k. and the PE is 1/2kx^2. But the slope is not always 45 degress, it could be much higher , say 60 degrees , but the area under the curve (straight line) is still the same, the area of the triangle.
Tey don't have to be equal. The area of the triangle is still 1/2kx^2 for the spring PE case, whether k is 1 (straight line graph for f = kx, k=1, 45 degree slope, or k is greater than 1 (higher slope and thus greater angle) or less than 1, (less steep slope , angle less than 45 degrees).