# Spring Potential Energy [closed]

A spring whose spring constant is $k$, having an initial "free" length is $l$, is being pressed by $2$ hoops on a metal(the parabola) on his both sides. (see image below)

I want to calculate the potential energy of the hoops due to the spring.

If we take the normal formula of a spring's potential energy, we get: $$U(x) = \frac{1}{2}k(2x-l)^2 = 2kx^2-2kxl+0.5kl^2$$

But when I try to calculate the potential energy expression by an integral on the force I get:

The force:

$$F = -k(2x-l)$$

I consider the potential energy to be in relation to $x = l/2$.

$$U(x) = -\int_{0.5l}^x -k(2x-l)dx = kx^2-kxl+0.25kl^2$$

Which is exactly half of the previous expression.

Where am I mistaken?

Thank you very much.

## closed as off-topic by ACuriousMind♦, user36790, John Rennie, Sebastian Riese, HDE 226868Dec 21 '15 at 21:51

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• Your result is lacking a factor of two, so Ockham's razor tells us: Your calculation is lacking a factor of two. Hint: Can your force term be true? Is the center of mass moving? (Or more direct: how many forces are acting here) – Bort Dec 21 '15 at 13:46
• @Bort If I take a look at the right hoop, the spring is pushing the right hoop outside, and the right hand side of the metal is pushing it towards the inside (Normal force 1, though not on the same angle as the spring). But there's also the left hand side of the metal pushing back the spring via the left hoop, on the same angle so it cancels the X-projection of the normal force 1 but gives a twice as big Y-projection of the normal force... – Taru Dec 21 '15 at 13:59
• Your last expression for $U(x)$ seems to compute the potential energy for just one half of the spring (look at the limits of your integral). – Floris Dec 21 '15 at 14:52

$$U(x) = \frac{1}{2}k(2x-l)^2$$

$$\frac{dU(x)}{dx}=\frac{1}{2}2k(2x-l)(2x-l)'$$

$$\frac{dU(x)}{dx}=k(2x-l)2=2k(2x-l)$$

So:

$$F=-2k(2x-l)$$.

You simply forgot to derive $(2x-l)$ because:

$$d(y^2)=2ydy$$

Edit:

To answer the OP's question in the comment section, I don't think potential energy was defined properly here. If we assume $x_0$ to be the equilibrium position, then each spring has potential energy:

$$U(x)=\frac{1}{2}k(x-x_0)^2$$

So double that for the entire system.

• But how could I reach that this is the expression of the force from the image? – Taru Dec 21 '15 at 14:33
• @Taru: edit made. – Gert Dec 21 '15 at 14:54