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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

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


F=kx

Where:


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

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


F=(150\frac{N}{\operatorname{mm}})(0.78\operatorname{cm})

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


1\operatorname{cm}=10\operatorname{mm}

Now, we multiply by the conversion factor:


F=(150\frac{N}{\operatorname{mm}})(0.78\operatorname{cm}*\frac{10\operatorname{mm}}{1\operatorname{cm}})

Now, we solve the operations:


F=1170N

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


1000N=1kN

Multiplying by the conversion factor we get:


F=1170N*(1kN)/(1000N)=1.17kN

Therefore, the tension is 1.17 kN.

Now, the elastic energy is given by:


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

Substituting the values we get:


U=(1)/(2)(150\frac{N}{\operatorname{mm}})(0.78\operatorname{cm}*\frac{1\operatorname{cm}}{10\operatorname{mm}})^2

Solving the operations:


U=0.456Nmm

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


1000\operatorname{mm}=1m

Multiplying by the conversion factor:


U=0.456\text{Nmm}*\frac{1m}{1000\operatorname{mm}}

Solving the operations:


U=0.000456J

Therefore, the elastic energy is 0.000456 Joules.

User Tomasz Mularczyk
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