Work Done by a Variable Force (Spring)

Compressing or stretching a spring requires more force the further you push — the spring "fights back" harder. So the work done isn't a simple Force × Distance product; it requires integrating the variable force over the displacement.

The Physics

Hooke's Law says the spring force is proportional to its displacement from equilibrium:

$F(x) = k\,x$

where $k$ is the spring constant (in N/m) and $x$ is the compression (or extension). The work done against the spring as you compress it from $0$ to $x_{0}$ is:

$W = \int_{0}^{x_{0}} F(x)\,dx = \int_{0}^{x_{0}} k\,x\,dx$

Step-by-Step Solution

Given: $k = 100\;\text{N/m}$, compression $x_{0} = 0.5\;\text{m}$.

Find: The work $W$ done in compressing the spring.

---

Step 1 — Write the spring force.

$F(x) = 100\,x \quad\text{(in newtons)}$

At $x = 0$ the force is $0$ N. At $x = 0.5$ m the force is $50$ N.

Step 2 — Set up the work integral.

$W = \int_{0}^{0.5} 100\,x\,dx$

Step 3 — Pull the constant out and find the antiderivative.

$W = 100\int_{0}^{0.5} x\,dx = 100 \cdot \dfrac{x^{2}}{2}\bigg|_{0}^{0.5}$

Step 4 — Evaluate at the bounds.

$= 100 \cdot \left(\dfrac{(0.5)^{2}}{2} - 0\right)$

$= 100 \cdot \dfrac{0.25}{2}$

$= 100 \cdot 0.125$

$= 12.5\;\text{J}$

Step 5 — Recognize the elastic-PE formula.

That entire calculation collapses into the famous formula:

$W = \dfrac{1}{2}k\,x_{0}^{2} = \dfrac{1}{2}(100)(0.5)^{2} = \dfrac{1}{2}(100)(0.25) = 12.5\;\text{J}$

This is also the elastic potential energy stored in the spring.

Step 6 — Sanity check via average force.

Since the force is linear in $x$, the average force is $\bar F = (0 + 50)/2 = 25\;\text{N}$. Multiplying by displacement $0.5\;\text{m}$:

$W = \bar F \cdot \Delta x = 25 \times 0.5 = 12.5\;\text{J} \;\;\checkmark$

For a constant-force problem you'd just write $W = F \cdot d$, but for a linear force you can use $\bar F$. For more complex forces you must integrate.

---

Answer: The work done in compressing the spring by 0.5 m is

$\boxed{\;W = 12.5\;\text{J}\;}$

This is also the elastic potential energy stored — and it would be released as kinetic energy if the spring were allowed to push something free.

Try It

• Adjust the spring constant widget — see the work scale linearly.
• Adjust the compression distance — work scales quadratically ($x_{0}^{2}$), so doubling the compression quadruples the work.
• The right panel shows the force-vs-displacement graph; the shaded triangle under the line equals the work done.
• Notice the formula $W = \tfrac{1}{2}kx^{2}$ is exactly the area of a triangle with base $x$ and height $kx$.