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*10^3\text{ N/m} \end{gathered}`

The displacement of the ligament is

`\begin{gathered} x=0.78\text{ cm} \\ =0.78*10^(-2)\text{ }m \end{gathered}`

We have to find elastic energy.

Elastic energy is given by the formula

`U=(1)/(2)kx^2`

Substituting the values, the elastic energy will be

`\begin{gathered} U=(1)/(2)*(150*10^3)\text{ }*(0.78*10^(-2))^2 \\ =\text{ 5.85 J} \end{gathered}`

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