Elastic energy and the work-energy principleEdexcel A-Level Further Maths: Revision notes

A light elastic string of natural length $l$ and modulus of elasticity $\lambda$ (measured in newtons) has tension $T=\frac{\lambda x}{l}$ when its extension is $x$ (Hooke's law). The extension is measured from the natural length, never from zero. A string can only pull: if its length is less than $l$ it is slack and $T=0$. A spring obeys the same law but can also be compressed, when it pushes with a thrust $\frac{\lambda x}{l}$, where $x$ is now the compression. Example: $\lambda=40$ N, $l=0.8$ m, $x=0.2$ m gives $T=\frac{40\times0.2}{0.8}=10$ N.

The work done in stretching a string from natural length to extension $x$ is the area under the tension–extension graph: $\text{EPE}=\int_0^x\frac{\lambda y}{l}\,dy=\frac{\lambda x^2}{2l}=\frac12Tx.$ The energy stored when $\lambda=40$ N, $l=0.8$ m and $x=0.2$ m is $\frac{40\times0.04}{1.6}=1$ J. To stretch from extension $x_1$ to $x_2$ the work done is $\frac{\lambda}{2l}\left(x_2^2-x_1^2\right)$. The same formula holds for a spring, with $x$ the compression or the extension.

The work done by forces other than gravity and the elastic force (friction, a driving force, a pull) equals the change in total mechanical energy: $W_{\text{other}}=\Delta(\text{KE}+\text{GPE}+\text{EPE}).$ If only gravity and the elastic force do work, mechanical energy is conserved: $\text{KE}+\text{GPE}+\text{EPE}$ is constant. Work done against friction is $F\times$ distance, with $F=\mu R$.

1. Pick two positions: the start and the position you want, such as when the string goes slack, at equilibrium, or at the lowest point.
2. Write KE, GPE (from a chosen level) and EPE at each. EPE is zero whenever the string is slack.
3. Add any work done against friction on the side that loses the energy.
4. Solve. A greatest-extension problem gives a quadratic in $x$; reject the negative root.

In a vertical problem the particle falls through the natural length plus the extension.

Example: a particle of mass $0.5$ kg on a string with $l=1$ m and $\lambda=49$ N is released from rest at the fixing point $O$. At greatest extension $x$: $0.5(9.8)(1+x)=\frac{49x^2}{2}$, so $5x^2-x-1=0$ and $x=\frac{1+\sqrt{21}}{10}=0.558$ m.

A spring can store energy when compressed, so a vertical spring problem may involve both compression and extension. A ball released from a compressed spring gains height equal to the compression plus any extension it reaches above the natural length. When the spring passes through its natural length it stores no energy, so check whether the ball is still attached when this happens. Check that $v^2$ is positive and the extension is less than any stated limit.