Work done by a variable force

When the force does not stay the same

Stretch a spring and it pulls back harder and harder. The force changes as you go, so $Fd$ no longer works. Instead, cut the path into tiny steps $\Delta x$. Over each step the force is nearly steady, and the work is $F\,\Delta x$. Add them all up:

On a graph of $F$ against $x$, each $F\,\Delta x$ is a thin strip, and the work is the area under the graph.

The work to stretch a spring

To stretch a spring slowly you pull with $F = kx$. The graph is a straight line through the origin, so the area is a triangle (or a trapezium):

From zero, this is $\tfrac12 kx^2$. Try it.

You should find that the same $5\ \text{cm}$ costs more and more work as it moves further out, because the force is bigger out there.

Watch out. While you stretch a spring, the spring pulls back, against the motion. Its own work is negative, $-\tfrac12 k(x_2^2 - x_1^2)$. Your work and the spring's work are equal and opposite.

A worked example

A trampoline spring has $k = 3200\ \text{N/m}$. A child's landing stretches it from $2.0\ \text{cm}$ to $6.0\ \text{cm}$.

1. In metres. $x_1 = 0.020\ \text{m}$, $x_2 = 0.060\ \text{m}$.
2. Work. $W = \tfrac12 \times 3200\ \text{N/m} \times ((0.060\ \text{m})^2 - (0.020\ \text{m})^2) = 1600\ \text{N/m} \times 0.0032\ \text{m}^2 = 5.12\ \text{J}$.
3. Compare. From $0\ \text{cm}$ to $2.0\ \text{cm}$ took only $0.64\ \text{J}$.

Think first. A graph is often the quickest way. Break the shape under it into rectangles and triangles, and add their areas.

Practise with new numbers

Try it in Python

The program finds the area under $F = 3x^2$ by adding thin strips, and compares it with the integral.

Remember these

• Work of a changing force: $W = \int F\,dx$, the area under the graph of $F$ against $x$.
• Spring, from $x_1$ to $x_2$: $W = \tfrac12 k(x_2^2 - x_1^2)$.
• The spring's own work while it is stretched is negative.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A graph of $F$ against $x$ is a flat line at $12\ \text{N}$ from $x = 0$ to $x = 3.0\ \text{m}$. How much work is done?

2. How much work stretches a spring with $k = 150\ \text{N/m}$ by $0.20\ \text{m}$ from its natural length?

3. Stretching a spring from $10\ \text{cm}$ to $20\ \text{cm}$ takes three times the work of stretching it from $0\ \text{cm}$ to $10\ \text{cm}$. Why?

4. A spring pushes a toy car as it unwinds back to its natural length. Is the spring's work positive or negative?