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 tension in this ligament when it is stretched by 0.780 cm? kNIf 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? J

Respuesta :

We are given that the tension of a ligament is proportional to its extension. Therefore, we have from Hook's law:

[tex]F=kx[/tex]

Where:

[tex]\begin{gathered} F=\text{ tension} \\ k=\text{ spring constant} \\ x=\text{ distance stretched} \end{gathered}[/tex]

Now, we are asked to determine the tension of a string given that is stretched a distance of 0.780 cm:

[tex]F=(150\frac{N}{\operatorname{mm}})(0.78\operatorname{cm})[/tex]

Now, we need to convert the centimeters into millimeters. To do that we will use the following conversion factor:

[tex]1\operatorname{cm}=10\operatorname{mm}[/tex]

Now, we multiply by the conversion factor:

[tex]F=(150\frac{N}{\operatorname{mm}})(0.78\operatorname{cm}\times\frac{10\operatorname{mm}}{1\operatorname{cm}})[/tex]

Now, we solve the operations:

[tex]F=1170N[/tex]

Now, we convert the "N" into "kN" using the following conversion factor:

[tex]1000N=1kN[/tex]

Multiplying by the conversion factor we get:

[tex]F=1170N\times\frac{1kN}{1000N}=1.17kN[/tex]

Therefore, the tension is 1.17 kN.

Now, the elastic energy is given by:

[tex]U=\frac{1}{2}kx^2[/tex]

Substituting the values we get:

[tex]U=\frac{1}{2}(150\frac{N}{\operatorname{mm}})(0.78\operatorname{cm}\times\frac{1\operatorname{cm}}{10\operatorname{mm}})^2[/tex]

Solving the operations:

[tex]U=0.456Nmm[/tex]

Now, to covert the energy into Joules we need to covert the "mm" into "m". To do that we use the following conversion factor:

[tex]1000\operatorname{mm}=1m[/tex]

Multiplying by the conversion factor:

[tex]U=0.456\text{Nmm}\times\frac{1m}{1000\operatorname{mm}}[/tex]

Solving the operations:

[tex]U=0.000456J[/tex]

Therefore, the elastic energy is 0.000456 Joules.

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