MC5, MC6 - Variable forces and elastic energy

Work as an integral

A $1\,\mathrm{kg}$ particle is released from rest, hanging from a vertical elastic string of natural length $1\,\mathrm{m}$ and modulus $20\,\mathrm{N}$. The string initially has extension $0.8\,\mathrm{m}$. Is the particle fastest when the string first becomes slack? Use $g=10\,\mathrm{m\,s^{-2}}$ and ignore air resistance. To answer, we need the work of a force that changes with extension.

Suppose a force along a straight line has value $F(x)=2x$ newtons, with $x$ measured in metres. Between $x=1$ and $x=3$, the force increases from $2\,\mathrm{N}$ to $6\,\mathrm{N}$. Using $6\times2$ would treat the largest force as if it acted throughout.

Split the displacement into tiny intervals. On one interval of width $\Delta x$, the work is approximately $F(x)\Delta x$. Summing these narrow rectangles and taking the limit gives the integral.

Area above the axis contributes positively when the coordinate increases; area below contributes negatively. If the endpoints are reversed, the integral changes sign. A force opposing the actual motion does negative work whichever coordinate direction you choose.

[DIAGRAM: asset_name: Work as area under a variable force; asset_slug: aqa7367-variable-work; caption: For F equals 2x, work from x equals 1 to x equals 3 is the shaded area, 8 joules; description: A straight force-displacement graph F equals 2x passes through the origin. The region below it and above the horizontal axis between x equals 1 and x equals 3 is shaded pale blue. Dashed vertical lines mark both endpoints. The displayed integral evaluates to 3 squared minus 1 squared equals 8 joules.]

Change only the starting coordinate to $2\,\mathrm{m}$, keeping release from rest. The upper endpoint is unchanged, but less work is done.

Multiplying the mean endpoint force by the distance worked here because the force was linear. For a curved force graph, endpoint averaging generally fails; the integral remains valid. Also, since the force changes with position, acceleration is generally not constant.

Use a curved force law in this fresh check.

Elastic potential energy

Stretching a string from its natural length takes work. The force starts at zero and increases as the string stretches, so using the final force over the whole extension would overestimate that work. In an ideal elastic model, the work put in during a slow stretch is stored as elastic potential energy.

At an intermediate extension $z$, a slowly applied force balances the restoring force $kz$. Integrate using $z$ as a temporary variable so that the final extension $x$ stays a fixed upper limit:

The same result is the area of a triangle on the force-extension graph: half of final force times extension. The spring itself does the opposite work during stretching, $-\frac12kx^2$; the positive stored energy is not that signed work.

To move between two stretched states, calculate the difference between their stored energies. The work needed is $\frac12k(x_1^2-x_0^2)$, not $\frac12k(x_1-x_0)^2$. Stretching an already taut string begins against a nonzero force.

Keep the same string and final length, but begin from an already stretched state. This changes the energy transferred, although the final stored energy is unchanged.