Spring Potential Energy

When you compress or stretch a spring, you're doing work against the spring force, and that work is stored as elastic potential energy. Releasing the spring converts this PE into kinetic energy of whatever it's pushing.

The Formula

For a Hookean spring with stiffness $k$ and displacement $x$ from equilibrium:

$U_{\text{spring}} = \dfrac{1}{2}\,k\,x^{2}$

Notice $x$ is squared — so doubling the compression quadruples the stored energy. Compress to $4×$ the distance, and you store $16×$ the energy.

Step-by-Step Solution

Given: $k = 200\;\text{N/m}$, compression $x = 0.3\;\text{m}$.

Find: The stored elastic potential energy $U$.

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Step 1 — Plug into the formula.

$U = \dfrac{1}{2}\,k\,x^{2} = \dfrac{1}{2}(200)(0.3)^{2}$

Step 2 — Compute $x^{2}$.

$(0.3)^{2} = 0.09\;\text{m}^{2}$

Step 3 — Multiply through.

$U = \dfrac{1}{2}(200)(0.09) = (100)(0.09) = 9.00\;\text{J}$

Step 4 — Equivalent computation as work done.

The force varies linearly from 0 (at equilibrium) to $F_{\max} = kx = 200(0.3) = 60\;\text{N}$ (at full compression). The average force is $\bar F = (0 + 60)/2 = 30\;\text{N}$. Work = average force × distance:

$W = \bar F \cdot x = 30 \times 0.3 = 9.00\;\text{J}\;\;\checkmark$

Both methods give the same answer, as they should — the work done compressing the spring is exactly the energy stored.

Step 5 — What if the spring is released?

That stored 9 J converts to kinetic energy in whatever the spring pushes. For example, if it pushes a 0.5 kg block:

$\dfrac{1}{2}(0.5)v^{2} = 9$

$v^{2} = 36 \;\Longrightarrow\; v = 6\;\text{m/s}$

The block would shoot off at 6 m/s — almost 22 km/h.

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Answer: The elastic potential energy stored in the spring is

$\boxed{\;U = \dfrac{1}{2}kx^{2} = 9.00\;\text{J}\;}$

This is the energy required to compress the spring from equilibrium to 0.3 m, and equivalently the energy that would be released if you let the spring rebound.

Try It

• Adjust the spring constant $k$ — energy scales linearly.
• Adjust the compression $x$ — energy scales quadratically, so small changes near zero matter much less than changes at large compressions.
• The graph on the right shows the parabolic relationship between $x$ and $U$.
• Watch the spring physically compress as you move the slider.