91Ó°ÊÓ

The given data can be listed below as,

• The mass of an object is, m = 250 g .
• The stretched in the tendon is, $∆\mathrm{x}=1.23\mathrm{cm}$.
• The amount of maximum tension that the tendon can stretch without rupturing is, T = 138 N .

In this question, the concept of Hooke’s law can be used to obtain the stiffness/force constant value of the tendon by using the value extension value of the tendon. The relation between the stiffness and the extension value is an inverse one.

According to Hooke’s law, the expression to calculate the force constant of the tendon is expressed as,

$$\begin{gathered} \mathrm{W}=\mathrm{k}∆\mathrm{x} \\ \mathrm{mg}=\mathrm{k}∆\mathrm{x} \\ \mathrm{k}=\left(\frac{\mathrm{mg}}{∆\mathrm{x}}\right) \end{gathered}$$

Here, k is the force constant of the tendon, W is the weight of the object, g is the gravitational acceleration whose value is $9.81\mathrm{m}/{\mathrm{s}}^2$.

Substitute all the known values in the above expression.

$$\begin{gathered} \mathrm{k}=\left[\frac{\left(250\mathrm{g}\right)\left({\displaystyle \frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}}\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}{\left(1.23\mathrm{cm}\right)\left({\displaystyle \frac{{10}^{-2}\mathrm{m}}{1\mathrm{cm}}}\right)}\right] \\ ≈1.99×{10}^2\mathrm{kg}/{\mathrm{s}}^2 \\ ≈\left(1.99×{10}^2\mathrm{kg}/{\mathrm{s}}^2\right)×\left(\frac{1\mathrm{N}/\mathrm{m}}{1\mathrm{kg}/{\mathrm{s}}^2}\right) \\ ≈1.99×{10}^2\mathrm{N}/\mathrm{m} \end{gathered}$$

Thus, the force constant of the tendon is$1.99×{10}^2\mathrm{N}/\mathrm{m}$ .

The expression of the amount of stretch in the tendon without rupturingis expressed as,

$$\begin{gathered} \mathrm{T}=\mathrm{k}∆\mathrm{x}\text{'} \\ ∆\mathrm{x}\text{'}=\left(\frac{\mathrm{T}}{\mathrm{k}}\right) \end{gathered}$$

Here, $∆\mathrm{x}\text{'}$is the amount of stretch in the tendon without rupturing.

Substitute all the known values in the above equation.

$$\begin{gathered} ∆\mathrm{x}\text{'}=\left(\frac{138\mathrm{N}}{1.99×{10}^2\mathrm{N}/\mathrm{m}}\right) \\ ≈0.693\mathrm{m} \end{gathered}$$

The expression of the amount of energy stored in tendon at the point without rupturing is expressed as,

$\mathrm{U}=\frac{1}{2}\mathrm{k}{\left(∆\mathrm{x}\text{'}\right)}^2$

Here, U is the amount of energy stored in tendon at the point without rupturing.

Substitute all the known values in the above equation.

$$\begin{gathered} \mathrm{U}=\frac{1}{2}\left(1.99×{10}^2\mathrm{N}/\mathrm{m}\right){\left(0.693\right)}^2 \\ ≈47.8\mathrm{N}.\mathrm{m} \\ ≈\left(47.8\mathrm{N}.\mathrm{m}\right)×\left(\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\right) \\ ≈47.8\mathrm{J} \end{gathered}$$

Thus, the amount of stretch in the tendon without rupturing is 0.693 m and the amount of energy stored in tendon at the point without rupturing is 47.8 J.