Question

The tension in a ligament in the human knee is approximately proportional to the extension of the ligament, if the extension is not too large.If a particular ligament has an effective spring constant of 150 N/mm as it is stretched, what is the elastic energy stored in the ligament when stretched by 0.780 cm?

Answer 1

Given that the spring constant of the ligament is

$\begin{gathered} K=150\text{ N/mm} \\ =\frac{150N}{10^{-3}\text{ m}} \\ =150\times10^3\text{ N/m} \end{gathered}$

The displacement of the ligament is

$\begin{gathered} x=0.78\text{ cm} \\ =0.78\times10^{-2}\text{ }m \end{gathered}$

We have to find elastic energy.

Elastic energy is given by the formula

$U=\frac{1}{2}kx^2$

Substituting the values, the elastic energy will be

$\begin{gathered} U=\frac{1}{2}\times(150\times10^3)\text{ }\times(0.78\times10^{-2})^2 \\ =\text{ 5.85 J} \end{gathered}$

Thus, the elastic energy of the ligament is 5.85 J.