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Tendons. Tendons are strong elastic fibers that attach muscles to bones. To a reasonable approximation, they obey Hooke’s law. In laboratory tests on a particular tendon, it was found that, when a 250-g object was hung from it, the tendon stretched 1.23 cm. (a) Find the force constant of this tendon in N>m. (b) Because of its thickness, the maximum tension this tendon can support without rupturing is 138 N. By how much can the tendon stretch without rupturing, and how much energy is stored in it at that point?

Respuesta :

Answer:

Explanation:

Given

mass of object(m)=250 gm

Stretching(x)=1.23 cm

Suppose force constant to be k

then kx=mg

[tex]k\times 1.23\times 10^{-2}=250\times 10^{-3}\times 9.8[/tex]

[tex]k=\frac{9.8\times 100}{4\times 1.23}[/tex]

[tex]k=199.18 N/m[/tex]

(b)Maximum Tension is 138 N

138=kx

[tex]x=\frac{138}{199.18}=0.692\approx 69.2 cm[/tex]

Energy stored[tex]=\frac{kx^2}{2}=\frac{199.18\times (0.692)^2}{2}=47.80 J[/tex]

Lanuel
  1. The force constant of this tendon is equal to 199.19 N/m.
  2. The extension of the tendon and the quantity of energy stored at this point are 0.69 meter and 47.42 Joules respectively.

Given the following data:

  • Mass of object = 250 grams to kg = 0.25 kg.
  • Extension = 1.23 cm to m = 0.0123 meter.
  • Maximum tension = 138 Newton.

Scientific data:

  • Acceleration due to gravity = 9.8 [tex]m/s^2[/tex].

a. To find the force constant of this tendon in N/m:

How to calculate the force constant.

Since the tendon obeys Hooke’s law, we have:

[tex]kx = mg\\\\k = \frac{mg}{x}[/tex]

Where:

  • k is the force constant.
  • m is the mass.
  • g is acceleration due to gravity.
  • x is the extension.

Substituting the given parameters into the formula, we have;

[tex]k=\frac{0.25 \times 9.8}{0.0123} \\\\k=\frac{2.45}{0.0123}[/tex]

k = 199.19 N/m.

b. To determine the extension of the tendon and the quantity of energy stored at this point:

For the extension:

[tex]T = kx\\\\138= 199.19x\\\\x=\frac{138}{199.19}[/tex]

x = 0.69 meter.

For the energy:

[tex]E = \frac{1}{2} kx^2\\\\E = \frac{1}{2} \times 199.19 \times 0.69^2\\\\E=99.595 \times 0.4761[/tex]

Energy = 47.42 Joules.

Read more on force constant here: https://brainly.com/question/25313999

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