Elastic potential energyEdexcel International A Level Further Maths: Revision notes

The tension in an elastic string or spring with natural length $l$ and modulus of elasticity $\lambda$ is $T=\frac{\lambda x}{l}$ when the extension (or compression) is $x$. The tension is not constant, so the work done in stretching is the area under the tension–extension graph, found by integration: $\text{work}=\int_0^x\frac{\lambda s}{l}\,ds=\frac{\lambda x^2}{2l}.$ This work is stored as elastic potential energy (EPE), and is recovered when the string returns to its natural length. For a stretched string $x$ is the extension, found from the stretched length minus $l$. EPE is zero when the string is slack.

Since $T=\frac{\lambda x}{l}$, the energy can be written $\frac12Tx$, i.e. half the final tension multiplied by the extension. This is often quicker when the tension is already known. A spring behaves in the same way, and it can also be compressed: for compression $x$ the thrust is $\frac{\lambda x}{l}$ and the stored energy is the same $\frac{\lambda x^2}{2l}$. A string can only pull, so it has no energy when slack.

To find the work done when the extension changes from $x_1$ to $x_2$, subtract the energies: $\frac{\lambda x_2^2}{2l}-\frac{\lambda x_1^2}{2l}=\frac{\lambda}{2l}(x_2^2-x_1^2).$ Do not use $\frac{\lambda(x_2-x_1)^2}{2l}$: energy depends on the square of each extension, not on the change. Example: $\lambda=30$, $l=0.6$, extension $0.3$ m to $0.5$ m. Work done $=\frac{30}{1.2}(0.25-0.09)=4$ J.

For a particle moving under gravity and an elastic string or spring, the work-energy principle states that the work done by external forces (such as friction or resistance) equals the change in the total of kinetic, gravitational potential and elastic potential energies: $\left(\tfrac12mv^2+mgh+\frac{\lambda x^2}{2l}\right)_{\text{final}}-\left(\tfrac12mv^2+mgh+\frac{\lambda x^2}{2l}\right)_{\text{initial}}=\text{work done by other forces}.$ If there is no friction or resistance, the total mechanical energy is conserved. Write the energy at the start and end, count every term that is non-zero, and equate. Energy is a scalar, so no resolving is needed, but the vertical distance fallen is needed for gravitational energy.

A particle of mass $2$ kg hangs from a point $A$ by a light string of natural length $1.5$ m and modulus $60$ N. It is released from rest at $A$. At the lowest point the extension is $x$, so the particle has fallen $1.5+x$. Energy: $2g(1.5+x)=\frac{60x^2}{2(1.5)}=20x^2$. This gives $20x^2-19.6x-29.4=0$ and $x=1.80$ m. At a point $2$ m below $A$ the extension is $0.5$ m and the speed satisfies $\frac12(2)v^2=39.2-5$, so $v=5.85$ m s$^{-1}$.

On a smooth slope at angle $\alpha$, a particle moving a distance $d$ down the slope loses $mgd\sin\alpha$ of gravitational energy. If a string only becomes taut after the particle has moved a certain distance, the elastic energy is zero until then. The maximum speed occurs where the resultant force is zero, that is where the tension equals the component of weight along the slope. The greatest extension occurs where the speed is zero.